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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
PVT Behaviour of Pure Substances
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.)
PVT Behaviour of Pure Substances
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:
PVT Behaviour of Pure Substances
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
PVT Behaviour of Pure Substances
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
PVT Behaviour of Pure Substances
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
PVT Behaviour of Pure Substances
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
PVT Behaviour of Pure Substances
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
Application of the Virial Equation
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
Application of the Virial Equation
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)
Application of the Virial Equation
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
Application of the Virial Equation
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
Application of the Virial Equation
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
Application of the Virial Equation
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
Application of the Virial Equation
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