Generalized Correlations for
Gases (Lee-Kesler)
By: Santosh Koirala
Manoj Joshi
Lauren Billings
Cory Klemashevich
The Generalized Correlation for Gases
•
The generalized Pitzer’s correlation is a three-parameter corresponding
states method for estimating thermodynamic properties of pure, nonpolar
fluids . For the compressibility factor Z, it takes the form
Z = Z0 + ω Z1
where,
Z0 = Compressibility factor for fluids of nearly spherical molecules,
ω = Pitzer's acentric factor, and
Z1 = Corrects for nonspherical intermolecular forces.
•
Appendix E provides values of Z0 and Z1 (Lee-Kessler correlations), from
which Z can be calculated and, hence, the molar volume can be computed.
Virial
equations
The Virial equation (up to the second Virial coefficient) provides an
approximation of Z and the equation is:
BP
Z  1
RT
0
1
ˆ
B  B  B
BPc
ˆ
B
RTc
0.422
B  0.083  1.6
Tr
Bˆ Pr
 Z  1
Tr
0
0.172
B  0.139  4.2
Tr
1
Cont……
The virial equation (up to the Third coefficient) also provides an approximation of Z.
Such virial equation is:
ˆ  B 0  B1
B
B C
Z  1  2
V V
ˆB  BPc
RTc
2
c
2 2
c
CP
ˆ
C
RT
2
ˆBP
 Pr 
r
ˆ

 Z  1
 C 
Tr Z
 Tr Z 
B
0
0.422
 0.083 
Tr1.6
0.172
B  0.139 
Tr4.2
Cˆ  C 0  C 1
1
0.2432 0.00313
C  0.01407 

Tr
Tr10.5
0
0.05539 .00242
C  0.02676 
 10.5
2.7
Tr
Tr
1
Example Problem
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,
c) Generalized correlation for B̂using eq. 3.61,
d) Equation 3.68 the third virial coefficient equation.
a
A. By the Ideal Gas equation:
RT (83.14)(510)
3
1
V

 1,696.1cm mol
P
25
B. From the values of Tc and Pc given
in Table B.1 of App. B
Z  Z  Z  0.873
0
1
ZRT (.873)(83.14)(510)
3
1
V

 1,480.7cm mol
P
25
C. Using the Second Virial Equation
P
T
Pc
Tc
Pr
Tr
ω
B0
B1
B̂
Z
R
V
25bar
510K
37.96bar
425.1K
0.658587987
1.199717713
0.2
-0.232344991
0.058943546
-0.220556282
0.878925087
83.14cm3bar/molK
1490.706167cm3/mol
Here, Excel was used to calculate
the volume using second virial
coefficient equation.
BP
Z  1
RT
BP c
ˆ
B 
RT c
Bˆ P r
 Z  1
Tr
Bˆ  B 0   B 1
B
0
B
1
0 . 422
 0 . 083 
T r1 . 6
0 . 172
 0 . 139 
T r4 .2
D. Using Second and Third Virial coefficient
P
T
Pc
Tc
Pr
Tr
ω
B0
B1
C0
C1
Ĉ
R
V
25bar
510K
37.96bar
425.1K
0.658587987
1.199717713
0.2
-0.232344991
0.058943546
-0.220556282
0.03312865
0.006760336
0.034480717
83.14cm3bar/molK
1484.438039cm3/mol
B
C
 2
V
V
ˆ  BPc
B
RTc
Z  1
Z
Z (Guess) (Calculated)
1
0.889316
0.889316
0.876994
0.876994
0.875453
0.875453
0.875258
Ĉ
0.875258
0.875233
0.875233
0.87523
0.87523
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
0.875229
2
ˆ  CPc
C
R 2Tc2
ˆP
B
Pr
r
ˆ

 Z  1
C

Tr Z
 Tr Z




2
ˆ  B 0  B1
B
B 0  0.083 
0.422
Tr1.6
B1  0.139 
0.172
Tr4.2
ˆ  C 0  C 1
C
C 0  0.01407 
0.2432 0.00313

Tr
Tr10.5
C 1  0.02676 
0.05539 .00242

Tr2.7
Tr10.5
This is the third coefficient virial
equation for the same problem.
Again Excel was used to obtain
the solution. The solution is
again very close to the value
obtained by the Lee-Kesler
method.
Where the Virial Equation Applies
•
•
This graph shows the difference obtained for the
Z0 value for the Lee/Kesler correlation vs. the
virial coefficient equation.
The second coefficient virial
equation works at low
pressures where Z is a
linear function. It is used
when an approximation of a
non ideal gas is needed,
but at non extreme
temperatures and
pressures.
The third virial coefficient
equation provides another
correction to the virial
equation.