PETE 310
Lectures # 32 to 34
Cubic Equations of State
…Last Lectures
Instructional Objectives
 Know the data needed in the EOS to evaluate
fluid properties
 Know how to use the EOS for single and
multicomponent systems
 Evaluate the volume (density, or z-factor)
roots from a cubic equation of state for
 Gas phase (when two phases exist)
 Liquid Phase (when two phases exist)
 Single phase when only one phase exists
Equations of State (EOS)
Single Component Systems
Equations of State (EOS) are mathematical
relations between pressure (P) temperature
(T), and molar volume (V).
Multicomponent Systems
For multicomponent mixtures in addition to
(P, T & V) , the overall molar composition
and a set of mixing rules are needed.
Uses of Equations of State
(EOS)
Evaluation of gas injection processes
(miscible and immiscible)
Evaluation of properties of a reservoir
oil (liquid) coexisting with a gas cap (gas)
Simulation of volatile and gas condensate
production through constant volume
depletion evaluations
Recombination tests using separator oil
and gas streams
Equations of State (EOS)
One of the most used EOS’ is the PengRobinson EOS (1975). This is a threeparameter corresponding states model.
RT
a
P

V  b V (V  b)  b(V  b)
P  Prep  Pattr
PV Phase Behavior
Pressurevolume
behavior
indicating
isotherms for
a pure
component
system
Tc
Pres sure
CP
T2
v
P
1
T1
L
2 - P has es
V
L
V
Mo lar V olum e
Equations of State (EOS)
The critical point conditions are used
to determine the EOS parameters
 P 

 0
 V  Tc
 P
 2   0
 V  Tc
2
Equations of State (EOS)
Solving these two equations
simultaneously for the Peng-Robinson
EOS provides
2
2
c
RT
a  a
Pc
and
RTc
b  b
Pc
Equations of State (EOS)
Where
and
 a  0.45724
 b  0.07780


  1  m 1  Tr

2
with
m  0.37464  1.54226  0.2699
2
EOS for a Pure Component
Pres sur e
CP
T2
4
v
P
1
3
1
L
A1
2
10
0
5
A2
 P 
 ~ >0
 V T
V
2 - P has
hases
L
7
1
2
V
Mo la r V o lum e
T1
6
Equations of State (EOS)
PR equation can be expressed as a
cubic polynomial in V, density, or Z.
Z  ( B  1) Z 
3
2
( A  3B  2 B ) Z 
2
( AB  B  B )  0
2
3
with
a P
A
2
RT 
bP
B
RT
Equations of State (EOS)
When working with mixtures (a)
and (b) are evaluated using a set of
mixing rules
 The most common mixing rules are:
 Quadratic for a
 Linear for b
Quadratic MR for a
 a m   xi x j  ai a ji j 
Nc Nc
i 1 j 1
0.5
1  k 
where the kij’s are called interaction
parameters and by definition
kij  k ji
kii  0
ij
Linear MR for b
Nc
bm   xibi
i 1
Example
For a three-component mixture (Nc
= 3) the attraction (a) and the
repulsion constant (b) are given by
 a m  2 x1x2  a1a21 2  (1  k12 )  2 x2 x3  a2a3 2 3  (1  k23 )
0.5
 2 x1 x3  a1a31 3  (1  k13 )  x 2  a11   x22  a2 2 
 x32  a3 3 
0.5
0.5
1
bm  x1b1  x2b2  x3b3
Equations of State (EOS)
The constants a and b can be
evaluated using
 Overall compositions zi with i = 1, 2…Nc
 Liquid compositions xi with i = 1, 2…Nc
 Vapor compositions yi with i = 1, 2…Nc
Equations of State (EOS)
The cubic expression for a mixture is
then evaluated using
 a m P
Am 
2
 RT 
bm P
Bm 
RT
Analytical Solution of Cubic
Equations
The cubic EOS can be arranged
into a polynomial and be solved
analytically as follows.
Z  ( B  1) Z 
3
2
( A  3B  2 B ) Z 
2
( AB  B  B )  0
2
3
Analytical Solution of Cubic
Equations
Let’s write the polynomial in the
following way
x  a1x  a2 x  a  0
3
2
3
Note: “x” could be either the molar
volume, or the density, or the zfactor
Analytical Solution of Cubic
Equations
When the equation is expressed in
terms of the z factor, the
coefficients a1 to a3 are:
a1  ( B  1)
a2  ( A  3B  2 B)
2
a3  ( AB  B  B )
2
3
Procedure to Evaluate the Roots
of a Cubic Equation Analytically
Let
3a2  a
Q
9
3
9a1a2  27 a3  2a1
R
54
2
1
S  R Q R
2
T  R Q R
2
3
3
3
3
Procedure to Evaluate the Roots
of a Cubic Equation Analytically
The solutions are,
1
x1  S  T  a1
3
1
1
1
x2    S  T   a1  i 3  S  T 
2
3
2
1
1
1
x3    S  T   a1  i 3  S  T 
2
3
2
Procedure to Evaluate the Roots
of a Cubic Equation Analytically
If a1, a2 and a3 are real and if D =
Q3 + R2 is the discriminant, then
 One root is real and two complex
conjugate if D > 0;
 All roots are real and at least two are
equal if D = 0;
 All roots are real and unequal if D <
0.
Procedure to Evaluate the Roots
of a Cubic Equation Analytically
If D  0 
where
1  1
x1  2 Q cos     a1
3  3
1
 1
x2  2 Q cos    120   a1
3
 3
1
 1
x3  2 Q cos    240   a1
3
 3
cos 
R
Q3
Procedure to Evaluate the Roots
of a Cubic Equation Analytically
x1  x2  x3  a1
x1 x2  x2 x3  x3 x1  a2
x1 x2 x3  a3
where x1, x2 and x3 are the three
roots.
Procedure to Evaluate the Roots
of a Cubic Equation Analytically
The range of solutions that are
used for the engineer are those for
positive volumes and pressures, we
are not concerned about imaginary
numbers.
Solutions of a Cubic Polynomial
From the shape
of the
polynomial we
are only
interested in
the first
quadrant.
Solutions of a Cubic Polynomial
http://www.uni-koeln.de/math-natfak/phchem/deiters/quartic/quartic
.html contains Fortran codes to
solve the roots of polynomials up to
fifth degree.
Web site to download Fortran source
codes to solve polynomials up to fifth
degree
EOS for a Pure Component
Pres sur e
CP
T2
4
v
P
1
3
1
L
A1
2
10
0
5
A2
 P 
 ~ >0
 V T
V
2 - P has
hases
L
7
1
2
V
Mo la r V o lum e
T1
6
Parameters needed to solve
EOS
Tc, Pc, (acentric factor for some
equations I.e Peng Robinson)
Compositions (when dealing with
mixtures)
 Specify P and T  determine Vm
 Specify P and Vm  determine T
 Specify T and Vm  determine P
Tartaglia: the solver of cubic
equations
http://es.rice.edu/ES/humsoc/Galileo/Catalog/Files/tartalia.html
Cubic Equation Solver
http://www.1728.com/cubic.htm
Equations of State (EOS)
Phase equilibrium for a single
component at a given temperature
can be graphically determined by
selecting the saturation pressure
such that the areas above and
below the loop are equal, these are
known as the van der Waals loops.
Two-phase VLE
The phase equilibria equations are
expressed in terms of the
equilibrium ratios, the “K-values”.
l
ˆ
yi i
Ki   v
xi ˆi
Dew Point Calculations
Equilibrium is always stated as:
l
v
ˆ
ˆ
xii P  yii P
(i = 1, 2, 3 ,…Nc)
with the following material balance
constrains
Nc
x
i 1
i
 1,
Nc
y
i 1
i
 1,
Nc
z
i 1
i
1
Dew Point Calculations
At the dew-point
l
v
ˆ
ˆ
xii  zii
xi K i  zi
(i = 1, 2, 3 ,…Nc)
Dew Point Calculations
Rearranging, we obtain the DewPoint objective function
Nc
zi
1  0

i 1 K i
Bubble Point Equilibrium
Calculations
For a Bubble-point
Nc
z K
i 1
i
i
1  0
Flash Equilibrium Calculations
Flash calculations are the work-horse
of any compositional reservoir
simulation package.
The objective is to find the fv in a
VL mixture at a specified T and P
such that
zi ( Ki  1)
0

i 1 1  f v ( K i  1)
Nc
Evaluation of Fugacity Coefficients
and K-values from an EOS
The general expression to evaluate
the fugacity coefficient for
component “i” is


RT


v
ˆ
RT
ln

dP


i   Vi 

P  T  fixed
0 

P
Evaluation of Fugacity Coefficients
and K-values from an EOS
The final expression to evaluate
the fugacity coefficient using an
EOS is.
Vtv 



P 
RT  v
v
ˆ


RT ln i   
 v dVt  RT ln Z v
v
 n 


V
v

i
t
T , n j  i

