B49CE
Multiphase Thermodynamics
Topics in VLE, LLE and Miscibility
Topic 1
Pure Species Phase Equilibrium
1
Learning Objectives
• After studying this “Pure Species Phase Equilibrium” topic
you should be able to:
• Write down phase equilibrium criterion for a pure species.
• Identify and differentiate between the potentials for heat
transfer, work transfer and interphase mass transfer.
• Define (for a pure species) the chemical potential, the
fugacity and the fugacity coefficient and write down
relationships between them.
• Show how the fugacity may be determined from measurable
data.
2
Learning Objectives
• Explain the background to Corresponding States Principle
and compressibility factor. Outline and distinguish between
the two-parameter and the three-parameter Law of
Corresponding States.
• Review and solve problems using generalised
compressibility factor charts.
• Demonstrate how charts may be used to determine the
fugacity coefficient and fugacity knowing only reduced
temperature, reduced pressure and acentric factor.
• Show how the fugacity coefficient and the fugacity may be
calculated for both a saturated and a subcooled liquid.
3
Recommended Textbooks
• Smith, J.M., van Ness, H.C., Abbott, M.M., Introduction to
Chemical Engineering Thermodynamics. McGraw-Hill. 7th
ed.,2005.
• S.I. Sandler. Chemical and Engineering Thermodynamics.
Wiley, 3rd ed., 1999.
• R.E. Sonntag, C. Borgnakke. Fundamentals of
Thermodynamics. Wiley, 5th ed., 1998.
• M.J. Moran, H.N. Shapiro. Fundamentals of Engineering
Thermodynamics. Wiley 2nd ed., 1993.
• J.R. Howell, R.O. Buckius. Fundamentals of Engineering
Thermodynamics. McGraw-Hill, SI ed., 1987.
1.1 Introduction
In previous courses (Process Engineering B) a general chemical
reaction was written as follows:
aA  bB  cC  dD
• At chemical equilibrium the concentration of each species
A,B,C,D taking part in the reaction was related to the
equilibrium constant K through the expression:
C c Dd
K
Aa Bb
• This provides the criterion for “chemical equilibrium” but it
only applies (in this form) to an ideal system.
>>> Later it will be shown how this may be adapted and
applied to non-ideal systems.
5
What is also important is the degree to which a given species
will transfer from one phase to another. This type of problem
is known as “phase equilibrium”.
• Important operations, such as distillation, gas absorption
and liquid-liquid extraction, are concerned with the transfer
of species between different phases.
• The rate at which a species is transferred from one phase to
another depends on how far the system is from equilibrium
• This degree of departure from equilibrium is usually
referred to as the driving force or simply the “potential”.
• Far from equilibrium the potential is high with high rates of
transfer – the opposite applies close to equilibrium.
>>> One of the first things that needs to be done is to identify
the “potential” that controls phase equilibrium/ interphase
mass transfer.
6
1.2 Criterion for Phase Equilibrium
Thermal equilibrium implies no temperature gradients; thus,
temperature differences across an uninsulated boundary is the
“potential” for heat transfer.
• Mechanical equilibrium implies no pressure gradient; thus,
pressure differences across a moveable boundary (caused
by an unbalanced force) is the “potential” for work transfer.
• The “potential” that drives interphase mass transfer is not
immediately obvious.
• However, it is only when this potential approaches zero that
the system may be assumed to be in phase equilibrium and
only then will net interphase mass transfer approach zero.
>>> This subject deals mainly with mixtures. However, best
place to start is phase equilibrium involving a single species;
an example might be a steam-water system.
7
1.2.1 Pure Species Phase Equilibrium
Consider a closed system with two phases but only one
component (this could be a steam-water system):
• Take it that both phases within this closed system are at a
constant T & P and that any heat or work, Q or W , that
is transferred takes place reversibly.
• Initially the system is not in phase equilibrium. The entropy
change of the surroundings dSSURR is given by
dS SURR 
dQSURR
TSURR
>> But>>
dQ SURR  dQSYS
• If heat transfer is reversible, then TSURR  TSYS and
dS SURR  
dQSYS
TSTS
8
The second law (Process Engineering B) is given by
dSSYS  dSSURR  0
• Substitute the entropy change of the surroundings intothe
second law to get
dS SYS
dQ
SYS
TSYS
0
dQSYS  TdS
Substitute for dQSYS in here
• The non-flow energy equation (Process Engineering B) is
given by
dQSYS  dU  PdV
To
get
dU  PdV  TdS  0
Note, these are all SYSTEM
properties.
9
If T & P remain constant this may be written
dU T ,P  d PV T , P  d TS T , P  0
 d U  PV  TS T ,P  0
Substitute for G inhere
• The Gibb's energy (Process Engineering B) is defined as
G  H  TS
>>>
G  U  PV  TS
To
dGT ,P  0
get
• The inequality dGT ,P  0 applies to irreversible processes
when the system is NOT in equilibrium
>>> The equality dGT ,P  0 applies to reversible processes
when the system IS in equilibrium
10
The inequality dG  0 is the direction for spontaneous
change. But when the system is at equilibrium
The Gibbs free energy is an
EXTENSIVE property (kJ/K) (i.e.
depends on quantity of material)
dG  0
• Take two phases: the  phase and the  phase (this could
be water and steam respectively):


G  ng
G  ng
Total both phases:
G  n g   n  g 
In which g and g are the molar Gibbs energies (kJ/mol) of
each phase separately (intensive properties, i.e. not dependant
on quantity of material) and n and n are the number of
moles in each phase.
11
Thus, at phase equilibrium
d ng  ng  0
gdn  ndg

 gdn  n dg  0


• But it has been shown (Process Engineering B)
dg  vdP sdT
• If T & P remain constant, then in each phase
and dg  0 thus it follows
To
get
dg  0
g dn  g  dn  0
dn
Must arrive
here
 gdn  gdn  0
 dn 
1 mole from
here
12
Thus, for phase equilibrium between these two phases
g  gdn  0
 g  g
• Only when this driving force is zero g  g   0 , willnet
mass transfer stop and only then can the two phases coexist in equilibrium.
• The molar Gibbs free energy IS the “potential” for pure
species phase equilibrium. This potential must be the same
in each phase for interphase mass transfer to stop.
• Water and steam at 101.325 kPa have exactly the same
molar Gibbs energy at 100oC. (and only at 100oC)
>>> This means that pure water and steam can only co-exist in
equilibrium at 1 atm when the temperature is 100oC (NBP).
13
1.2.2 Pure Species Chemical Potential
This “potential” is the so called “chemical potential” and, for a
pure species, it is denoted by the symbol  .
• In Process Engineering B it was shown that, for a pure
species, the chemical potential is equal to the molar Gibbs
energy. Thus, for a pure species
g
• Notice that the chemical potential, like all driving forces
must be intensive (just like T & P ).
• Thus, for phase equilibrium between these two phases
 
Thus, two phases in equilibrium leads to
this phase constraint and hence F=1.
>>> The in either phase depends only on T & P . Satisfying
the phase constraint means if T is independent , then P must
be a dependent variable and vice versa.
14
1.2.3 Pure species Gibbs Energy Pressure
dependence
See notes for its derivation, but it has already been shown
(Process Engineering B) that, in terms of total properties
G, V and S , the total Gibbs energy depends on T & P :
dG  VdP  SdT
• It follows that the partial change in total Gibbs energy w.r.t.
pressure must be given by
 G 

 V
 P T
• In terms of purely intensive properties these become
dg  vdP sdT
&
 g 
  v
 P T
15
1.2.4 Pure species Chemical Potential
Pressure Dependence
But the Gibbs free energy of a pure species is its chemical
potential g  ,thus
  
  v
 P T
• Hence, the isothermal variation of  with P
is then
d  vdP
• Integrate the above expression, at constant temperature,
between some very low pressure state P* and any higher
pressure state P .

P
*
P*
 d   vdP
16
Integrating this leads to
P
*   vdP
P*
• There are no absolute values for chemical potential (just
like internal energy, enthalpy and Gibbs free energy).
• Thus, if is finite at some specified pressure P , then *in
some lower pressure state
P* will depend on the
integral on the right of the equation.
• It is clear that as
P*  0 , then v   and the integralon
the right also tends to infinity.
>>> Thus if  is finite at somespecified pressure P , then
the chemical potential at some very low pressure * ,
i.e. at low pressure *tends to minusinfinity.
17
April 2018 v4
© Heriot-Watt University, B49CE,
Multiphase Thermodynamics
18
The variation in chemical potential versus pressure is also
highly non-linear.
• Both these features make the chemical potential “badly
behaved” (mathematically speaking) at low pressure.
• The value of properties as pressure approaches zero is very
important as gases become ideal at this low-pressure limit
and the ideal gas state is used as a reference state.
• It would be better to define the chemical potential in terms
of some new property that increasingly tended to some ideal
gas property at low pressure.
• A good idea would be to define an “auxiliary property”, one
that depends directly on but one that behaves differently
at low pressure.
dg i g  v ig dP
To seek this new auxiliary property consider
how Gibbs energy varies with pressure for
an ideal gas.
19
The superscript “ig” denotes that the gas is an ideal gas.
Substitute the ideal gas EOS into the previous expression.
d ig  RT dP
P
 dig  RTd ln P
In which is the temperature- dependent
constant of integration.
>Integrate>
ig  RT ln P
>> For an ideal gas, these expressions show how the isothermal
change in chemical potential depends only on the pressure.
1.3 Pure Species Fugacity
The same expression may be applied to a real fluid so long as
the real pressure P is replaced by some new property which
allows the equation to hold for real fluids.
>>> This new property is given the symbol f , has the same
units as pressure (kPa) and is called “fugacity”.
20
Thus, for a real fluid changes in the chemical potential are
given by
In which is the temperature- dependent
constant of integration.
d  RTd ln f
>Integrate>
 RT ln f 
• The fugacity may be considered an auxiliary property of
the chemical potential, changes in this auxiliary property
are linked directly to changes in chemical potential.
• The next step is to see how the fugacity behaves at low
pressure….. remember the chemical potential was badly
behaved at low pressure.
• Subtract these last two equations to get


d    ig  RT d ln


f
P
 d ig  RT d ln 
Where  is the
dimensionless fugacity
coefficient
21
Thus, the “fugacity coefficient”  is defined as:
f
P
• However, take the integrated expression  RT ln f ,
and subtract ig RT ln P  , then we see that the
constant of integration  cancels which leads to

  RT ln 
ig
• Then at:
P 0
  0
ig
• And whenever:
P0
 1
As P  0
then
 1
>>> Thus, the low pressure limit (ideal gas value) of the
fugacity coefficient is ONE.
22
The expression that connects the chemical potential to its
auxiliary property (fugacity) is given by
  RT ln


f
ig  RT ln 
<< >>
P
• The term on the left is the difference between the chemical
potential of a real fluid and that of an ideal gas – both at the
same T & P . If the real fluid is an ideal gas
ig
   0
ig
• And, because the constant of integration cancels it follows
that, for an ideal gas:
 ig  1
f ig  P
Notice as P  0 , f  P .
Whereas as P  0 ,   .
23
The “fugacity coefficient”  was defined asfollows:

f
P
• But since f ig  P , it follows that the fugacity coefficient of
a pure species could equally well be defined by

f
f ig
The fugacity coefficient does exact the
same job for the fugacity.
• The fugacity coefficient is a correction factor that corrects
the fugacity between the ideal gas state and real fluid state
(at the same T & P ).
• Notice the fugacity coefficient does a very similar job to the
compressibility factor Z
Z  Pv  v
RT vig
The compressibility factor corrects the
molar volume v (m3/mol) between ideal
gas state and real fluid state (at same T&P)
24
1.3.1 Fugacity as a Criterion for Equilibrium
For a pure species the criterion for phase equilibrium between
solid, liquid and vapour phases at the Triple Point is as follows:
  
S
L
Substitute for  in here
V
• But it has been shown:
   RT ln
ig
>>>
 ig RT ln
To
ig  RT ln  S  ig  RT ln  L  ig  RT ln  V 
get
• But at equilibrium all phases are at the same T & P.
 S   L  V
f
S
 f
L
 fV
The criterion for phase equilibrium is
the same in terms of fugacity as it
was in terms of chemical potential.
25
1.3.2 Fugacity in terms of Residual Volume
When trying to find numerical values for the fugacity it is
better to first find the fugacity coefficient, since it is
dimensionless.
• Once  is known f may be calculated from the definition:

f
P

or
f
f ig
• It has been shown that Gibbs energy depends only on T & P
dg  vdP sdT
• For an isothermal process
d vdP
dig  vig dP
For an ideal gas

 

Subtract one from the other
d    ig  v v ig dP
26
Integrate the previous equation from P  0 , where    ig  0
to any other higher pressure P ,where  ig  0 , leads to
   ig   v  v ig dP
P
0
• The term on the right v  v ig is known as the “residual
volume” (m3/mol) and in many texts this is given the
Aside
symbol v R.
R
• Any residual property m is just the difference between
the real fluid property m and the ideal gas property mig,
both measured at the same T & P.
R
• Thus, any residual property m ( property m is generic
and could be v, u, h, s, g ) is defined as follows;
m R  m  mig
27
The previous expression in terms of residual volume is now
P
 ig  v R dP
0
Equate these two
Expressions…..
• But it was also shown that
   RT ln
ig
To
get
P
1
R
v
dP
ln  

RT 0
• This expression is VERY important because it links the
fugacity coefficient (thus also the fugacity) to measurable
properties.
28
1.3.3 Worked Example 1.1
Estimate the fugacity of gaseous propane at 2.4 MPa and
344 K using the following data for propane at 160o F
(complete this example by hand in the workbook).
Molar Mass of propane is 44.097 kg/kmol
P
v
vig
(MN/m2)
(m3/kg)
(m3/kg)
0
0.1724
0.3447
0.6895
--0.3691
0.1810
0.08652
-
1.0342
1.2066
1.3790
2.4000
0.05512
0.04598
0.03903
0.01783
P
v R P
(m3/kg)
(kN/m2)
(kJ/kg)
-
-
-
vR
1
RT
v
R

P
29
Estimate of fugacity coefficient at 24 bar and 344 K:  ….............
Estimate of fugacity at 24 bar and 344 K:
f  ….............
30
ig
R

Pave
PvR3RP
vv
10
P
1.3.3 Worked Example 1.1
32
ig
R

Pave
PvR3RP
vv
10
P
1.3.3 Worked Example 1.1
P
1
1
R
Ln  
v
dP


 18303.9 J / kg  0.2822

8314
J
/
kmol
.
K
RT 0
 344 K
44.097kg / kmol
f
Ln  0.2822
p
f
 0.7541
p
f  0.7541  2.4MPa  1810kPa
33
1.3.4 Fugacity in terms of Z-Factor
The fugacity coefficient relation below may be successively rearranged as follows:
P
1
R
v
dP
ln  

Using residual
RT 0
property definition
ln   1
RT
 v  v dP
P
ig
0
 v
v ig 
 dP
ln    

RT RT 
0
P
 Pv Pv ig  dP

ln    

RT
RT  P
0
P
P
dP
 ln    Z 1
P
0
Pv
Z
RT
Pvig
RT
 Z ig 1
34
1.3.5 Compressibility Factor Charts
If no experimental data is available, Z-factors may be obtained
from charts or tables. This approach has been discussed in
Process Industries C.
• A three-parameter generalised correlation for the Z-factor
takes the general form
Z  Z(Tr , Pr , )
•
Where two of the parameters are reduced temperature Tr
and reduced pressure Pr, given by
T
Tr 
TC
P
Pr 
PC
Remember to use absolute
units for BOTH temperature
and pressure
>>> The third correlating parameter is the “acentric factor”
and is usually given the symbol .
35
Three-parameter correlations are based on the threeparameter Law of Corresponding States:
“All fluids with the same acentric factor  and at the
same Tr and Pr deviate from ideality to the same
extent and, therefore, have the same compressibility
factor Z ”.
• The acentric factor is easily determined by a single vapour
pressure measurement Psat at a specified reduced
temperature of Tr  0.7 :
sat
(P
  log 10 r ) Tr 0.7  1
• Where,
Prsat 
P sat
PC
The Z-factor is usually
estimated from two charts:
1. The “simple” fluid chart,
giving Z(0) .
2. The “non-simple” fluid
chart giving Z(1).
36
These two component parts of the Z-factor are combined as
follows:
Z  Z 0  Z 1
• The superscripts are used to identify the component parts
of the Z-factor; they are NOT powers of “Z”.
• The “Z0” component assumes uniform charge distribution.
So-called simple fluids, such as Ar, Kr and Xe have these
characteristics.
• The “Z1” component is a correction for non-uniformcharge
distribution. All other non-simple fluids require this
adjustment.
• Thus, the acentric factor for Ar, Kr, Xe must be ZERO.
>>> The Z0 and Z1 charts are on the next slide – to use them all
that is needed are the critical properties of the fluid.
37
The Z0 Simple FluidChart:
•
The Z-factor is usually
estimated from two charts:
1. The “simple” fluid chart,
giving Z(0) .
2. The “non-simple” fluid chart
giving Z(1).
The
Z1 Non-Simple FluidChart:
Z  Z 0  Z 1
38
1.3.6 Worked Example 1.2
Estimate the fugacity of gaseous propane at 2.4 MPa. and
344 K. Estimate the compressibility factor Z (at each Pr )
from the charts in Appendix A.
Complete this example by hand in this workbook using the
following critical and other properties for propane:
Molar Mass 44.097 kg/kmol
Critical Temperature 369.8 K
Critical Pressure 42.48 bar
Acentric Factor 0.152
P
Z 0 Z 1
Pr
Tr
Z
(MN/m2)
0
0.1724
0.3447
0.6895
1.0342
1.2066
1.3790
2.4000
0
1
0
1
39
P
(MN/m2)
0
Z 1
P
P
(kN/m2)-1
(kN/m2)
0
 Z 1  P


P


-
0.1724
0.3447
0.6895
1.0342
1.2066
1.3790
2.4000
 ZP1P


Estimate of fugacity coefficient at 24 bar and 344 K:
 …………..
Estimate of fugacity at 24 bar and 344 K:
f  …………..
40
Tr 
T
344

 0.93
Tc 369.8
Pr 
P 0.1724

 0.0406
Pc
4.248
P
ln    Z  1
0
dP
 0.2995
P
  exp( 0.2995)  0.7412
f    P  0.7412  2.4MPa  1.779MPa  17.79bar  1779kPa
41
Example Zo:
April 2018 v4
© Heriot-Watt University, B49CE,
Multiphase Thermodynamics
42
1.3.7 Fugacity Directly from Charts
Rather than finding fugacity from Z-factor charts, it may be
found directly from its own charts. Start with the relationship
below:
P
ln    Z 1 
0
dP
P
• In terms of reduced properties this is given by
Pr
Substitute for Z in here
dPr

ln    Z 1
Pr
0
• Also,
Tr
Z  Z 0  Z 1
To
get
dP
ln    Z 0 1 r
Pr
0
Pr
Pr
 Z1
Tr
0
dPr
Pr
Tr
43
Examine this expression:
dPr
0

1
ln    Z
Pr
0
Pr
Pr
dPr
 Z
Pr
0
1
Tr
Tr
• The first term on the right can be evaluated from Z 0 data.
The second term can be evaluated from Z1 data.
• Each integral represents component parts of ln  . Label
these integrals ln0 and ln 1 respectively.
ln   ln  0   ln  1
    0  1 

>>> The integrals have been evaluated as functions of reduced
temperature and reduced pressure and are available in the
form of 0 and 1charts (tables are also available).
44
The  0 Simple FluidChart:
>> These are “low
pressure” charts.
The  Non-Simple Fluid Chart:
1
   0  1 

>>> Both low & high
pressure charts are
available in Appendix B.
45
1.3.8 Worked Example 1.3
Estimate the fugacity of gaseous propane at 2.4 MPa. and
344 K using the charts supplied in Appendix B. Complete this
example by hand in this workbook using the following critical
and other properties for propane:
Molar Mass 44.097 kg/kmol
Critical Temperature 369.8 K
Critical Pressure 42.48 bar
Acentric Factor 0.152
Tr  …………..
Pr  …………..
  …………..
0
1  …………..
  …………..
f  …………..
46
Tr 
T
344

 0.93
Tc 369.8
Pr 
P
24

 0.565
Pc 42.48
Using the charts supplied in Appendix B:
    0.770.89
 
0
1 
0.152
 0.7565
f    P  0.7565  2400kPa  1816kPa
April 2018 v4
© Heriot-Watt University, B49CE,
Multiphase Thermodynamics
47
April 2018 v4
© Heriot-Watt University, B49CE,
Multiphase Thermodynamics
48
1.3.9 Fugacity of Pure Saturated Liquid
The P  v and P  Tdiagrams may be linked to each other:
CONSTANT “T”
P
Psat
P
C
4
3
4
2
LIQUID
1
A
B
v
C
2&3
1
SOLID
TP
VAPOUR
T
• Notice how the saturated vapour curve on the right hand,
P  T , diagram………
• Corresponds to two-phase envelope on the left hand, P  v ,
diagram.
>>> Points 1, 2, 3 and 4 all lie on an isotherm. Whether fluid is
liquid or vapour depends on the P in relation to Psat.
49
If P  P sat , then fluid is a single-phase vapour – see state “1”.
If P  P sat , then fluid is a single-phase liquid – see state “4”.
If P  P sat , then fluid state is anywhere within thetwo-phase
envelope (anywhere between state “2” and state “3”)
• State “2” corresponds to a saturated vapour state (dryness
fraction one) where no liquid is present.
• State “3” corresponds to a saturated liquid state (dryness
fraction zero) where no vapour is present.
• From state “2” and state “3” the dryness fraction varies
from one to zero respectively.
Within the two phase envelope “2-3 ” the state is continually
changing while both T & P remain constant, so that
dg  vdP sdT
>>>
dg  0
50
Integrating the previous expression between saturated liquid
and saturated vapour states leads to
g V2  g 3L  0
>>>
gV2  g 3L
• For a pure species the Gibbs free energy is the chemical
potential, thus the above may be written
V2  L30
>>>
V2  L3
• And, in view of the pure species phase equilibrium criterion
f 2V  f 3L  0
>>>
f 2V  f 3L
• And, since the pressure is constant from “2-3”
 V2   3L  0
>>>
V2  3L
>>> Thus, there is no change in Gibb’s energy, chemical
potential, fugacity or fugacity coefficient along a tie-line.
51
When a phase change occurs the fugacity is calculated in a
three-step process as follows:
1. The fugacity of the saturated vapour f satV is found, at the
target temperature, by integrating volumetric data from
zero pressure to the vapour pressure Psa.t
f

ln sat 

 P State 2
satV
Psat
dP
0 Z 1 P
Solve the integral on the right
hand side, then find f satV .
2. The fugacity of the saturated liquid f
must be the same.
f
satV
 f
satL
satL
in equilibrium
The fugacity of the saturated
liquid is now known.
3. The fugacity of the compressed (or subcooled) liquid f L
can then be estimate using the method given in the next
section.
52
1.3.10 Fugacity of a Compressed Liquid
The expression below was derived earlier:
d  vdP
• The fugacity was defined earlier by
d  RTd ln f
• Equating these two expressions leads to
v
d ln f 
dP
RT
This links the fugacity to
measurable properties
• Integrating (along an isotherm) from
Psat to any higher
pressure P
f
1
satd ln f  RT
f
P
 vdP
Psat
53
The left side of the equation becomes
fL
1
ln sat 
f
RT
P
 vdP
P sat
• Notice how the lower limit of integration is Psat and not
zero pressure; thus, the right hand integral does not tendto
infinity.
• Also, liquid molar volume v is a weak function of pressure
and, as a first-order approximation, it may be regarded as
constant
• The molar volume is often taken to be the saturated liquid
molar volume vsat
L ……an average value may also be used.
ln
fL
f
sat

vLsat
RT
P
 dP
Psat
54
Completing the integration on the right leads to
f L vLsat P Psat 
ln sat 
f
RT
• Sometimes written as
sat
sat


v

L
sat
L P  P
f  f exp

RT


• Since sat  f sat / Psat this is often written as
sat
sat


v

P

P

L
sat sat
L
f   P exp

RT


>>> The above derivation assumes liquids at low/moderate
pressure, where the pressure, P  PC ; above this the
assumption of near constant v starts to break down.
55
1.3.11 Worked Example 1.4
Estimate the fugacity of liquid propane at 4.8 MPa and 344 K
using the same gaseous P  v  T data for propane at 160oF
(344 K) as provided in Example 1.1. The vapour pressure of
propane at 344 K is 2.56 MPa.
Molar Mass of propane is 44.097 kg/kmol
v
v ig
vR
P
v R P
(MN/m2)
0
(m3/kg)
---
(m3/kg)
-
(m3/kg)
-
(kN/m2)
-
(kJ/kg)
-
0.1724
0.3691
0.3447
0.181
0.6895
0.08652
1.0342
0.05512
1.2066
0.04598
1.3790
0.03903
2.4000
0.01783
P
2.5600
1
RT
v
R

P
56
Fugacity coefficient of saturated vapour at 25.6 bar and 344 K
 sat  …………
Fugacity of saturated vapour at 25.6 bar and 344 K
f sat  …………
The specific volume and pressure data for the liquid (pastthe
point of saturation) is given below:
v
P
 10 3
(MN/m2)
(m3/kg)
2.551
2.446
2.758
2.446
3.447
2.404
4.800
2.338
Fugacity of liquid at 48 bar and 344 K f L  ………
Fugacity coefficient of liquid at 48 bar and 344 K  L  ………
57

Pave
Pv
3 PP
vv
10
From worked example 1,
2.56
2.56
    dP 
 18.3039kJ / kg  19.5242kJ / kg
2.4
2.4
P
1
1
R
Ln  
v
dP


 19524.2 J / kg  0.3010

8314 J / kmol.K
RT 0
 344 K
44.097kg / kmol
Ln
f
 0.3010
p
f
 0.7401
p
f  0.7401  2.56MPa  1895kPa
April 2018 v4
© Heriot-Watt University, B49CE,
Multiphase Thermodynamics
58
Ln
f
f
sat
1

RT
P
1
 5385.1J / kg 0.0830
sat vdP  8314 J / kmol.K
P
 344 K
44.097kg / kmol
f  1.0865  1895kPa  2059kPa
April 2018 v4
© Heriot-Watt University, B49CE,
Multiphase Thermodynamics
59
1.3.12 Fugacity Coefficient Using Cubic EOS
Please refer to Appendix C to see how the Peng-Robinson EOS
may be used to find the fugacity coefficient of a pure species.
• Also outlined in Appendix C is how to estimate the pure
species vapour pressure using an EOS.
• This approach is based on the fact that two co-existing
phases must have the same fugacity coefficient.
• The material in Appendix C may be needed later for Design
and Research Projects. Software packages, such as HYSYS,
often calculate thermodynamic properties this way.
Other Appendices:
• Appendix A includes compressibility factor charts, both
simple and non-simple fluid charts.
• Appendix B includes coefficient charts, both simple and
non-simple fluid charts (at high & low pressure).
60