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Volumetric Properties of Pure Fluids

THERMODYNAMICS 1

Department of Chemical Engineering, Semarang State University

Dhoni Hartanto S.T., M.T., M.Sc.

Introduction

Pressure, Volume, and Temperature (PVT) are important propertiessuch purposes as the metering of fluids

and the sizing of vessel and pipelines.

Quantitive description of real fluids

Equation of State (EoS)

Generalized correlation are used to predict PVT behaviourof fluid which has no experimental data

PVT Behaviour of Pure Substances

PT Diagram

C : critical point (critical pressure Pc and Tc)

the highest temperature and pressure at which a purechemical species can exist invapor/liquid equilibrium

In critical condition, fluid is classified as liquid or gas(the two phase becomeindistinguishable)

Gas in left side of dashed line can be condensed as vapor

Gas in right side of dashed line(T >TC) is supercritical condition

C

Triple point

Solid

Liquid

VaporP

TTc

Pc

A

B

Gas Region

Fluid3

2

1Sublimation curve


C

Liquid+Vapor

V

P

TcT>TcT<Tc

VLVV

PV DiagramLiq

uid

Vc

Pc

Fluid

Solid+

Liq

uid

Solid

Solid+VaporB

D

B to C line single-phase saturated liquids at their boling temperature

C to D line single-phase saturatedvapor at their condensation temperature

Under curve BCD :Subcooled-liquid (T below T boiling)Superheated-vapor (T above T boil.)


0),,( TVPf

dPP

VdT

T

VdV

TP

PT

V

V

1

TP

V

V

1

dPdTV

dV

Thus:

Equation of State (EoS) : Functional equation to express the relation between P, V, and T

Partial derivatifve :

Partial derivatifve related to 2 properties:

Volume expansivity : Isothermal compressibility:


1212

1

2

V

Vln PPTT

Simple EoS

The value of dan has been commonly tabulated

PVT relation

If and are constant (for liquid approximation)

Is almost always positive, is necessarily positive


Example :Acetone at 293.15 K and 1 bar has =1.487 x 10-3 K-1 , = 62 x 10-6 bar-1, and V = 1.287 x 10-3 m3 kg-1

a) Find the value ofb) The pressure generated when acetone is heated at constant volume from

293.15 K and 1 bar to 303.15 Kc) The volume change when acetone is changed from 293.15 K and 1 bar to

273.15 K and 10 bar

Solution :a)

V is constant dV = 0,

VTP )/(

dPdTV

dV

dPdT 0

1

6

3

241062

10487.1

Kbar

T

P

V


Solution :b) The value of and can be assumed constant at interval temperature 10 K

The equation in (a) :

and

c)

barTP 240)10()24(

barPPP 241240112

1212

1

2

V

Vln PPTT

0303.09)1062(20)10487.1(V

Vln 63

1

2

1333

12

1333

2

1

2

)10()038.0()10()287.1249.1(

10249.1)10287.1(9702.0;9702.0

kgmVVV

kgmVV

V

Virial Equation of State

• Equation of State (EoS)

0),,( TVPf

• Gas Ideal (Simplest EoS)

RTPV

- Volume individual = 0- No interaction- Valid in low pressure

• Real Gas

Compressibility factor (Z)

RT

PVZZRTPV ;

For ideal gas, Z = 1


Virial EoS

.........V

D

V

C

V

BZ

321

.........PDPCPBZ 321

:,2V

C

V

B2-body interaction and 3-body interaction between pairs of molecules

Virial expansions

:'V

BandB Second virial coefficients

:'2V

CandC

For a given gas the virial coefficients are functions of temperature only

Third virial coefficients

3

3

2

2

)(

23';

)(';'

RT

BBCDD

RT

BCC

RT

BB


Truncated Virial EoS

V

BZ 1 PBZ 1

P

Z 1

PBZ 1

The B value has been tabulated for various gases

Application :a) Gas phase only (satisfactory results for vapor at

subcritical T)b) Significantly molecul interactionsc) For low pressure gas (up to a pressure about 5 bar)

Application of the Virial Equation

Originate value at Z = 1 for P = 0

The tangent to an isotherm at P = 0 is good approximation

Tangent line eq. : PBZ 1

BPTP

Z

Z

;

V

B

RT

PVZ

RT

BP

RT

PVZ

1

1


The pressure above the range eq.in previous slide but below critical pressure three term virial equation give excellent result

21

V

C

V

B

RT

PVZ

Value of C and B depend on the gas and on temperature

In figure 3.11, the trends are similar


Ideal gas Equation of State

.........V

D

V

C

V

BZ

321

If:

or 0P V

1Z or RTPV

Virial EoS

Assumption : no molecule interaction

Good approximation for gas : in very low pressure or high

temperature (big Volume)


P,TUU

Internal energy for ideal gas

P depend on molecule interaction

for real gas

In ideal gas, no molecule interaction occured (V= infinite)

TUU

Enthalpy for ideal gas

PVUH

RTUH )T(HH


Heat capacity for ideal gas

)PV(ddUdH

RdTdTCdTC Vp

V

VT

UC

TUU )T(CC VV

P

PT

HC

THH

)T(CC PP

The relation between CP and CV for ideal gas

RCC Vp


Extended virial equation (Benedict/Webb/Rubin Equation)

2223632

2

000 exp1/

VVTV

c

V

a

V

abRT

V

TCARTB

V

RTP

Where : are all constant for a given fluid

This complex equation are used in the petrolium and natural-gas industries(light hydrocarbon and a few gas)

,,,,,,, 000 cbaCBA


Example

The virial coefficients of isopropanol vapor at 200 oC :

B = -388 cm3 mol-1 ; C = -26000 cm6 mol-2

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

a) The ideal-gas equation

b) Eq.3.37 (Smith Van Ness Handbook 6th ed)

c) Eq.3.39 (Smith Van Ness Handbook 6th ed)

Solution :

The absolute temperature is T = 473.15 K, gas constant (R) = 83.14 cm3 bar mol-1 K-1

a) Ideal gas, Z = 1

393410

)15.473)(14.83(

P

RTV cm3 mol-1


b) Solving eq. 3.37 for V : 3546)388(3934 BP

RTV cm3 mol-1

9014.03934

3546

/

PRT

V

RT

PVZ

c) Solving eq. 3.39, rearrange equation to facilitate iteration, yield :

21 1

ii

iV

C

V

B

P

RTV

2

2600038813934

VVV

Using goal seek in M. Excel to obtain V , then V = 3488 cm3 mol-1

8866.03934

3488

/

PRT

V

RT

PVZ

The ideal gas value is 13% too high and no (b) is 1.7% too highcompare with this result

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