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05/03/2305/03/23 11
Termodinamica de Termodinamica de HidrocarburosHidrocarburos
More EOS ApplicationsMore EOS ApplicationsTuning EOS for Reservoir Tuning EOS for Reservoir
SimulationSimulation
05/03/2305/03/23 22
Petroleum Engineering Petroleum Engineering Applications of Phase EquilibriaApplications of Phase Equilibria
High Pressure Phase Equilibria High Pressure Phase Equilibria Applications (Reservoir).Applications (Reservoir).
Equations of State Models (EOS). Cubic Equations of State Models (EOS). Cubic EOS. Root Selection.EOS. Root Selection.
Evaluation of Fugacity Coefficients from Evaluation of Fugacity Coefficients from Equations of State.Equations of State.
Evaluation of Phase Boundaries (Dew and Evaluation of Phase Boundaries (Dew and Bubble Points) and Flash Equilibrium with Bubble Points) and Flash Equilibrium with EOS. Tuning of Equations of State (EOS). EOS. Tuning of Equations of State (EOS). Miscible Gas Injection. Swelling Tests.Miscible Gas Injection. Swelling Tests.
05/03/2305/03/23 33
Petroleum Engineering Petroleum Engineering Applications of Phase EquilibriaApplications of Phase Equilibria
Constant Volume Depletion Studies Constant Volume Depletion Studies (Gas Condensates and Volatile (Gas Condensates and Volatile Oils).Oils).
Determination of Oil and Gas in Determination of Oil and Gas in Place by Recombination. Place by Recombination.
Additional Reading: Selected SPE Additional Reading: Selected SPE paperspapers
05/03/2305/03/23 44
Instructional ObjectivesInstructional ObjectivesAfter seeing this module the student After seeing this module the student
should be able to:should be able to: Evaluate volume roots from a cubic Evaluate volume roots from a cubic
equation of state when two-phases equation of state when two-phases coexist.coexist.
Derive and evaluate fugacity coefficients Derive and evaluate fugacity coefficients from cubic EOS.from cubic EOS.
Evaluate phase boundaries (dew and Evaluate phase boundaries (dew and bubble points) and flash separations bubble points) and flash separations using EOS.using EOS.
05/03/2305/03/23 55
Equations of State (EOS)Equations of State (EOS)
Cubic equations are EXPLICIT in Cubic equations are EXPLICIT in pressure and can be written as the pressure and can be written as the sum of a term indicating repulsion sum of a term indicating repulsion forces and a term indicating attraction forces and a term indicating attraction forces forces
attrrep PPP
05/03/2305/03/23 66
Dew and Bubble functions Dew and Bubble functions behavior on a PT diagrambehavior on a PT diagram
.
Pb
Pd
Pressure
Temperature
CP
Bubble-Curve
Dew-C
urve
11
Nc
i i
i
Kz
11
Nc
i i
i
Kz
11
Nc
i i
i
Kz
11
Nc
iiiKz
11
Nc
iiiKz
11
Nc
iiiKz
2-phases
Tr
A
B
05/03/2305/03/23 77
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS The general expression to evaluate the The general expression to evaluate the
fugacity coefficient for component “fugacity coefficient for component “ii” in ” in any phase (vapor, liquid-I, liquid-II, etc.) any phase (vapor, liquid-I, liquid-II, etc.) isis
fixedT
P
ivi dP
PRTVRT
0
ˆln
05/03/2305/03/23 88
The partial molar volume of The partial molar volume of component “i” is defined as:component “i” is defined as:
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS
inPT
v
vv
vj
i
t
i nV
V
,,
05/03/2305/03/23 99
Following the (-1) rule of thermodynamics and rules of Following the (-1) rule of thermodynamics and rules of differentiationdifferentiation
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS
in,Tv
n,T
v
in,P,Tv
v
vj
ivt
t
vj
i
t
nP
P
V
n
V
vt
t
vj
i nT
v
inT
vv
i PV
nPV
,,
05/03/2305/03/23 1010
Note that we have changed the limits of the Note that we have changed the limits of the first integral accordingly.first integral accordingly.
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS
dPP
RTdPP
VnPRT
P
nT
v
inT
v
Pvi
vt
t
vj
i
0,,0
ˆln
dPP
RTdVnPRT
Pv
t
inT
v
Vvi
vj
i
vt
0,
ˆln
05/03/2305/03/23 1111
The second integral in the right hand The second integral in the right hand side can be expanded using the following side can be expanded using the following identity:identity:
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS
)/()/(
t
t
VZRTVZRTd
PdP
tVdZdPd lnlnln
t
t
VdV
ZdZ
PdP
05/03/2305/03/23 1212
The final expression to evaluate the The final expression to evaluate the fugacity coefficient using an EOS is.fugacity coefficient using an EOS is.
vv
tv
inT
v
Vvi ZRTdV
VRT
nPRT
tvj
i
vt
lnˆln,
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS
05/03/2305/03/23 1313
And the K-value isAnd the K-value is
vi
li
i
ii x
yKˆˆ
t
inT
V
Vi dV
nPKRT
ji
vt
lt
,
ln
Evaluation of Fugacity Evaluation of Fugacity Coefficients and K-values from Coefficients and K-values from
an EOSan EOS
05/03/2305/03/23 1414
ExampleExampleEvaluation of the fugacity Evaluation of the fugacity
coefficient from van der Waals coefficient from van der Waals EOSEOS
We will illustrate the evaluation of We will illustrate the evaluation of a K-value for species “i” in a a K-value for species “i” in a multicomponent mixture using the multicomponent mixture using the Van der Waals EOS. Van der Waals EOS. You will do it with another EOS.You will do it with another EOS.
05/03/2305/03/23 1515
Van Der Waals EOS Van Der Waals EOS
05/03/2305/03/23 1616
Relations to keep in mind:Relations to keep in mind:
multicomponent: multicomponent:
lv nnn
lll nnn 21
li
Nc
il nn
1
Total number of moles:Total number of moles:
Number of liquid molesNumber of liquid molesbinary:binary:
05/03/2305/03/23 1717
Relations to keep in mind:Relations to keep in mind:
Number of vapor molesNumber of vapor moles
binary:binary:vv
v nnn 21
vi
Nc
iv nn
1
multicomponent: multicomponent:
05/03/2305/03/23 1818
Relations to keep in mind:Relations to keep in mind:
Partial derivatives:Partial derivatives: nnnz
vi
li
i
1
in
l
l
lj
inn
1
in
v
v
vj
inn
0
in
l
lj
vj
inn
0
in
v
vj
vj
inn
Overall mole fractionOverall mole fraction
05/03/2305/03/23 1919
Relations to keep in mind:Relations to keep in mind:
Recall that the volume used in the Recall that the volume used in the EOS is a molar volume:EOS is a molar volume:
nVV t
05/03/2305/03/23 2020
Mixture parametersMixture parametersThe mixture parameters used in the The mixture parameters used in the
EOS if linear mixing rules are used are:EOS if linear mixing rules are used are:Attraction term (liquid)Attraction term (liquid)
Nc
iil
lNc
iii
l ann
axa i
11
Nc
iil
lNc
iii
l bnn
bxb i
11
Repulsion term (liquid)Repulsion term (liquid)
05/03/2305/03/23 2121
Mixture parametersMixture parametersAttraction term (vapor)Attraction term (vapor)
Nc
i
Nc
iiv
v
iiv a
nn
aya i
1 1
Nc
iiv
vNc
iii
v bnn
byb i
11
Repulsion term (vapor)Repulsion term (vapor)
05/03/2305/03/23 2222
The van der Waals EOSThe van der Waals EOS
Write EOS in terms of nWrite EOS in terms of n
2
2
2tt V
anbnV
RTnVa
bVRTP
05/03/2305/03/23 2323
ExampleExampleTo evaluate the K-values we need To evaluate the K-values we need
to evaluate,to evaluate,
init
inittinVTi
j
jj
nana
Vn
nbnb
bnVRTn
bnVRT
nP
2
2,,
2
05/03/2305/03/23 2424
ExampleExample
For linear MR these partial For linear MR these partial derivatives arederivatives are
naa
na i
inij
nbb
nb i
inij
05/03/2305/03/23 2525
ExampleExample
Replacing these expressions …Replacing these expressions …
naana
Vn
nbbnb
bnVRTn
bnVRT
nP
i
t
i
ttinVTij
2 2
2,,
05/03/2305/03/23 2626
ExampleExample
Now put the expression back using Now put the expression back using the molar volume the molar volume
nVV t
ndVdVt
ndV
naa
Vnn
Vnna
nbVRTnb
nbVRTKRT i
tt
iV
Vi
v
l
22
2
2222
2ln
and replaceand replace
05/03/2305/03/23 2727
ExampleExample
After cancellations and integration After cancellations and integration between the final expression for between the final expression for the K-value of species “i” isthe K-value of species “i” is
li
l
vi
v
llvvill
vv
i
Vaa
Vaa
bVbVRTb
bVbVRTKRT
11lnln
05/03/2305/03/23 2828
Relations to keep in mind:Relations to keep in mind:
Mole fraction of component “i” in Mole fraction of component “i” in liquid:liquid:
l
li
i nnx
v
vi
i nny
Mole fraction of component “i” in Mole fraction of component “i” in vapor:vapor:
05/03/2305/03/23 2929
ExampleExample
The same algebraic procedure is The same algebraic procedure is used for all EOSused for all EOS
Suggested exercise: Practice with Suggested exercise: Practice with PR and with quadratic MRPR and with quadratic MR
05/03/2305/03/23 3030
Fugacity Coefficients from Fugacity Coefficients from other Cubic EOSother Cubic EOS
See notesSee notes
One exercise will be to derive the One exercise will be to derive the expression forexpression for
i
05/03/2305/03/23 3131
Peng-RobinsonPeng-Robinson
RTPbB
RTPbB
RTPaA
BzBzay
aBB
BA
BzzBB
ii
vv
vv
vv
vv
jijj
vv
i
v
v
vvvv
ivi
2
)12()12(ln2
22
ln1ˆln
05/03/2305/03/23 3232
Tuning of an EOS to VLE Tuning of an EOS to VLE datadata
(Example with Binary data)(Example with Binary data)
05/03/2305/03/23 3333
Experimental VLE Data - COExperimental VLE Data - CO22/C/C33
Binary VLE Data on the CO2/C3 System at T = 4.4oC
Pressure (atm) x(CO2) y(CO2)6.81 0.0247 0.2056
10.21 0.0884 0.467613.61 0.1602 0.603617.01 0.2402 0.686420.41 0.3316 0.743123.81 0.4361 0.787627.22 0.5532 0.830930.62 0.6714 0.868834.02 0.7956 0.910237.43 0.9401 0.9702
05/03/2305/03/23 3434
Tuning GoalTuning Goal
Determine (adjust) parameter (s)Determine (adjust) parameter (s) in in the EOS such that predicted values the EOS such that predicted values from the EOS are as close as from the EOS are as close as possible to the experimental valuespossible to the experimental values
05/03/2305/03/23 3535
Tuning EOS ExampleTuning EOS ExampleThe values chosen for prediction The values chosen for prediction
can be saturation pressures and can be saturation pressures and either the liquid or the gas either the liquid or the gas compositions. compositions.
If the liquid compositions are If the liquid compositions are selected as independent the type selected as independent the type of calculation will be a bubble point of calculation will be a bubble point and the vapor compositions will be and the vapor compositions will be evaluated. evaluated.
05/03/2305/03/23 3636
Tuning EOS ExampleTuning EOS Example If the gas compositions are selected as If the gas compositions are selected as
independent variable the type of independent variable the type of calculation will be a dew point and the calculation will be a dew point and the liquid compositions will be evaluated.liquid compositions will be evaluated.
This choice is arbitrary, but it is normally This choice is arbitrary, but it is normally constrained by selecting as constrained by selecting as “independent” variables those for which “independent” variables those for which one has the highest precision.one has the highest precision.
05/03/2305/03/23 3737
Experimental PX diagram Experimental PX diagram for Propane/COfor Propane/CO22 at 4.4 ºC at 4.4 ºC
0
10
20
30
40
Pres
sure
(At m
)
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
x (CO2 ) , y(CO2)
T = 4.4 o C
Liquid
Vapor
2-phases
05/03/2305/03/23 3838
Tuning EOS ExampleTuning EOS ExampleMathematically the minimization Mathematically the minimization
problem is expressed such that problem is expressed such that
MinPPFM
j
cali
datai
2
1
function is minimized.function is minimized.
05/03/2305/03/23 3939
Tuning EOS ExampleTuning EOS ExampleThe Redlich-Kwong EOS isThe Redlich-Kwong EOS is
)(2/1 bVVTa
bVRTP
05/03/2305/03/23 4040
Tuning EOS ExampleTuning EOS ExampleStep 1.Step 1.
Evaluate pure component EOS Evaluate pure component EOS constants (a & b)constants (a & b)
ci
cii P
TRa5.22
42748.0
ci
cii P
RTb 08664.0
05/03/2305/03/23 4141
Tuning EOS ExampleTuning EOS Example
211 COaa
2
1.52
11 mol KAtm l7537.63a
2
1.52
22 molK Atm l48.180
3aaC
Verify thesecalculations
05/03/2305/03/23 4242
Tuning EOS ExampleTuning EOS ExampleThe interaction constant of attraction isThe interaction constant of attraction is
12221112 1 kaaa
AtmK
K moll Atm
85.7219.30408205.008664.01b
moll02968.01 b
moll0062682 b
And the “repulsion constants” are:And the “repulsion constants” are:
andand
05/03/2305/03/23 4343
Tuning EOS ExampleTuning EOS ExampleNow evaluate fugacity coefficientsNow evaluate fugacity coefficients
PyPx vii
lii ˆˆ
05/03/2305/03/23 4444
Tuning EOS ExampleTuning EOS ExampleStep 2.Step 2.
Evaluate the fugacity coefficients using Evaluate the fugacity coefficients using the RK EOS. These expressions are,the RK EOS. These expressions are,
bVb
VbV
bRTab
RTPV
VbV
bRT
az
bVb
bVV i
ii
ln
lnln2
lnˆln
25.11
5.1
2
11
11
05/03/2305/03/23 4545
Tuning EOS ExampleTuning EOS ExampleTo demonstrate the procedure, we To demonstrate the procedure, we
will choose one data point:will choose one data point: TPyx ,,, 11
K 55.27715.2734.4Atm 61.13
6036.01602.0
1
1
TPyx
For this example:For this example:
05/03/2305/03/23 4646
Tuning EOS ExampleTuning EOS ExampleTo begin the iterations we will To begin the iterations we will
assume the interaction parameter:assume the interaction parameter:
06.012 k
05/03/2305/03/23 4747
Tuning EOS ExampleTuning EOS ExampleSelect dependent and independent Select dependent and independent
variables:variables:
Given Variables
Independent
Unknown Variables
To evaluateProblem Type
T, zi = xi
T=277.55 K
x1=0.1602
P,yi Bubble Point
05/03/2305/03/23 4848
Tuning EOS ExampleTuning EOS ExampleTo evaluate the fugacity To evaluate the fugacity
coefficients we need:coefficients we need:Assume an initial guess for the gas Assume an initial guess for the gas
composition and pressure. We can use composition and pressure. We can use the experimental data as the initial the experimental data as the initial guess.guess.
Evaluate the interaction constant a12, Evaluate the interaction constant a12, and the mixture parameters for both and the mixture parameters for both gas and liquid phases. gas and liquid phases.
05/03/2305/03/23 4949
Tuning EOS ExampleTuning EOS ExampleWith these constants, we now must With these constants, we now must
solve the cubic EOS for Vl,solve the cubic EOS for Vl,
)(2/11lll
l
ll
data
bVVTa
bVRTP
05739.055.27705.156
05739.055.27708205.061.13 5.0
lll VVV
moll08785.0lV
05/03/2305/03/23 5050
Tuning EOS ExampleTuning EOS ExampleNow we can evaluate the fugacity Now we can evaluate the fugacity
coefficient for CO2 (component 1) coefficient for CO2 (component 1) of the liquid.of the liquid.
ll
l
l
ll
l
l
l
l
ll
l
iii
llll
ll
bVb
VbV
bRTba
RTPV
VbV
bRT
ax
bVb
bVV
ln
lnln2
lnˆln
25.11
5.1
2
11
11
05/03/2305/03/23 5151
Tuning EOS ExampleTuning EOS ExampleReplacing the results obtained, we Replacing the results obtained, we
obtain:obtain:
43228.0ˆ 8387.0ˆln
7116.2ˆ 9975.0ˆln
22
11
ll
ll
Now, repeat the same procedure Now, repeat the same procedure for the vapor phase.for the vapor phase.
05/03/2305/03/23 5252
Tuning EOS ExampleTuning EOS ExampleNow use the RK EOS with bv, av, Now use the RK EOS with bv, av,
and Pi data and solve for the vapor and Pi data and solve for the vapor volume. This provides;volume. This provides;
moll4251.1vV
05/03/2305/03/23 5353
Tuning EOS ExampleTuning EOS ExampleCalculate the fugacity coefficients Calculate the fugacity coefficients
for both components in the vapor for both components in the vapor phase. This provides.phase. This provides.
78483.0ˆ93093.0ˆ
2
1
v
v
05/03/2305/03/23 5454
Tuning EOS ExampleTuning EOS ExampleNext we evaluate the K-values and Next we evaluate the K-values and
a new set of vapor compositionsa new set of vapor compositions
v
l
xyK
1
1
1
11 ˆ
ˆ
9128.29309.07116.2
ˆˆ
1
11
v
l
K
5508.078483.043228.0
ˆˆ
2
22
v
l
K
05/03/2305/03/23 5555
Tuning EOS ExampleTuning EOS ExampleThe new vapor compositions are:The new vapor compositions are:
4626.05508.08398.04666.09128.21602.0
222
111
KxyKxy
the sum is not = 1, Whoops…!the sum is not = 1, Whoops…!
05/03/2305/03/23 5656
Tuning EOS ExampleTuning EOS ExampleThe material balance constraint The material balance constraint
must be satisfied. Therefore we must be satisfied. Therefore we must normalize the new vapor must normalize the new vapor compositions such that its sum is compositions such that its sum is equal to one.equal to one.
05/03/2305/03/23 5757
Tuning EOS ExampleTuning EOS Example
9292.04626.04666.021 yy
5022.09292.04666.0
2
1
1,1
i
i
k
y
yy
4978.09292.04626.0
2
1
2,2
i
i
k
y
yy
RenormalizationRenormalization
05/03/2305/03/23 5858
Tuning EOS ExampleTuning EOS Example The New pressure is evaluated asThe New pressure is evaluated as
2
11
iiikk KxPP
Atm65.129292.061.13 kP Now return to Step 1. And using these Now return to Step 1. And using these
new values for P and vapor compositions new values for P and vapor compositions repeat the procedure until the bubble repeat the procedure until the bubble point criteria is satisfied.point criteria is satisfied.
05/03/2305/03/23 5959
Tuning EOS ExampleTuning EOS ExampleThis is a minimization problem. The This is a minimization problem. The
final answer to this exercise gives final answer to this exercise gives an optimum an optimum kkijij of 0.07605. of 0.07605.
Verify it !!!Verify it !!!
05/03/2305/03/23 6060
Flow Chart Flow Chart SummarySummary
G i v e n d a t a :
T e m p e r a t u r e
V a p o r C o m p o s i t i o n s
L i q u i d C o m p o s i t i o n s
P r e s s u r e c o m p o s i t i o n d a t a
M = # o f d a t a s e t s
),...,...,,( 21 Nci yyyyy
),...,...,,( 21 Nci xxxxx
MjyxP j ,...1 ,,
P u r e f l u i d p r o p e t i e s
H y d r o c a r b o n C h a r a c t e r i z a t i o n S c h e m e s
...),,,,( MwTTP bcc
M i x i n g R u l e s - E O S
S e l e c t d e p e n d e n t a n d i n d e p e n d e n t v a r i a b l e s
( i . e . , D e p e n d e n t : - I n d e p e n d e n t )yP , xT ,
F o r a l l d a t a s e t s : k = 1 , … M d o
S e t i n i t i a l g u e s s e s a n d i n t e r a c t i o n p a r a m e t e r s ( j = 1 )
kyy exp1 kPP exp1 1
ijij kk
1
05/03/2305/03/23 6161
Flow Flow Chart Chart
SummarSummaryy
E v a l u a t e f o r ( i = 1 , … N c ) k
kli
vilv VV ˆ,ˆ,,
E v a l u a t e f o r ( i = 1 , … N c ) k
kvi
li
iK
ˆˆ
R e - e v a l u a t e f o r ( i = 1 , … N c ) k
k
Nc
jjj
iii
Kx
Kxy
1
I s
f o r ( i = 1 , … N c ) k ?k
Nc
jjj Kx
01
1N
U p d a t e
k
Nc
iiijj KxPP
11
Y
P r i n t
ijk kxyTP ,,,,
I s
o p t i m a l ?
M
k
calc PP1
2exp NC h a n g e i n t e r a c t i o n p a r a m e t e r s Y P r i n t
ijk kxyTP ,,,, E n d
1
05/03/2305/03/23 6262
Evaluation of Reservoir Evaluation of Reservoir Engineering PVT Engineering PVT
Properties from an EOSProperties from an EOS Oil and gas formation volume factors (Oil and gas formation volume factors (BBoo’s ’s
and and BBgg’s’s), ), PPbb, and solution gas oil ratios , and solution gas oil ratios ((RRss’s’s) can be obtained from flash ) can be obtained from flash vaporization, differential liberation, and vaporization, differential liberation, and separator separation tests. These are separator separation tests. These are standard PVT experimental procedures standard PVT experimental procedures for the oil industry, although they may for the oil industry, although they may not always be economically affordable. not always be economically affordable.
05/03/2305/03/23 6363
Evaluation of Reservoir Evaluation of Reservoir Engineering PVT Engineering PVT
Properties from an EOSProperties from an EOSThe underlying assumption for The underlying assumption for
having all these tests is that the having all these tests is that the reservoir process can be simulated reservoir process can be simulated from a differential vaporization, from a differential vaporization, while the process from the bottom while the process from the bottom of the well to the stock tank can be of the well to the stock tank can be simulated with a separator test. simulated with a separator test.
05/03/2305/03/23 6464
Evaluation of Reservoir Evaluation of Reservoir Engineering PVT Engineering PVT
Properties from an EOSProperties from an EOSThere are a variety of correlations There are a variety of correlations
for these standard PVT properties for these standard PVT properties BBoo, , RRss, , BBgg..
Equations of State provide an Equations of State provide an alternative method of estimating alternative method of estimating these properties provided that the these properties provided that the reservoir fluid composition is reservoir fluid composition is known. known.
05/03/2305/03/23 6565
Phase Equilibrium Phase Equilibrium ProblemProblem
GivenVariables
(independent)
UnknownVariables(dependent)
ProblemType Governing Equations Example Application
P , zi = xi T, yi 011
Nc
iii Kz Separator-Distillation
tests
T, zi = xi P,yi
BubblePoint
iii Kzy Gas injectionProduction
P, zi = yi T,xi 011
Nc
i i
i
Kz
Gas Condensates
T, zi = yi P,xi
Dew Point
i
ii K
yx Production
P, T, zi xi, yi, fv Flash
0)1(1
)1(
1
cN
i iv
ii
KfKz
)1(1
iv
iii Kf
Kzy
i
ii K
yx
Production
Separation Processes
PVT Tests
Evaluation ofBo, Rs, Rv, Bg
05/03/2305/03/23 6666
Simulated PVT Simulated PVT Experiments Experiments
DepletionStage
StartingMoles fv fl
Moles of Gas Produced(nv)
Moles of Liquidin PVT ce ll (nl)
0 1 0 1 0 11 1 fv1 f l1 fv1*1 f l1*12 f l1*1 fv2 f l2 fv2*f l1*1 f l2*fl1*13 f l2*fl1*1 fv3 f l3 fv3*f l2*f l1*1 f l3*fl2*f l1*1… … … … … …
Nd 11
1Õ
dN
ilif fvNd f lNd 1
1
1Õ
li
N
ivN ff
d
d1
1Õ
dN
ilif
05/03/2305/03/23 6767
Simulated PVT Simulated PVT Experiments: Experiments: BBoo
BBoo at at P > PP > Pbb
Õ
dN
iilSCSCo
Roo
fPTV
PTVB
1,),(~
),(~
Õ
Õ
dN
ii,lSCSCo
j
ii,ljRo
o
f)P,T(V~
f)P,T(V~
B
1
1
BBoo at at P< PP< Pbb
05/03/2305/03/23 6868
Simulated PVT Simulated PVT Experiments: Experiments: BBoo
it can bee seen that the depletion it can bee seen that the depletion stages up to pressure stages up to pressure jj cancel out cancel out leaving the following expressionleaving the following expression
Õ
dN
jii,lSCSCo
jRoo
f)P,T(V~
)P,T(V~B
1
05/03/2305/03/23 6969
Simulated PVT Simulated PVT Experiments: Experiments: RRss
Initial RsiInitial Rsi
Õ
Õ
d
d
N
iilSCSCo
N
j
j
kkljvSCSCg
si
fPTV
ffPTVR
1,
1
1
1,,
),(~
),(~
Õ
Õ
d
d d
N
ii,lSCSCo
N
Ej
N
Ekk,lj,vSCSCg
sE
f)P,T(V~
ff)P,T(V~
R
1
1
1
at a pressure stage ‘at a pressure stage ‘EE’ below ’ below PbPb
05/03/2305/03/23 7070
Expansion of ExpressionsExpansion of Expressions
Example of math symbols usedExample of math symbols used
54321
5
1
ffffffj
j Õ
5
154321
kk ffffff
Õ
3
1321321211
1k
k
jjk gggfggfgfgf
05/03/2305/03/23 7171
Simulated PVT Properties Simulated PVT Properties for a Black Oil and a for a Black Oil and a
Volatile OilVolatile Oil
3160
3180
3200
3220
3240
3260
3280
3300
Sol
utio
n G
as O
il R
atio
at P
b (R
sb)
2.76
2.78
2.80
2.82
2.84
2.86
Form
atio
n V
olum
e Fa
ctor
at P
b (B
ob)
50 100 150 200 250 300
Separator 2 Pressure (psia)
High Pressure Separator @ 900 psia and
TSep2 = 75oF
100 ºF
Optimization of separator 2 conditions for a volatile oil.
05/03/2305/03/23 7272
Simulated PVT Properties Simulated PVT Properties for a Black Oil and a for a Black Oil and a
Volatile OilVolatile Oil
3360
3380
3400
3420
3440
3460
Solu
tion
Gas
Oil
Rat
io a
t Pb
(Rsb
)
2.90
2.92
2.94
2.96
Form
atio
n Vo
lum
e Fa
ctor
at P
b (B
ob)
400 500 600 700 800 900 1000Separator 1 Pressure (psia)
Low Pressure Separator @ 300 psia and
TSep1 = 160oF 75ºF
One Separator Stage Comparison between one
and two separator
stages for a volatile oil.
05/03/2305/03/23 7373
Simulated PVT Properties Simulated PVT Properties for a Black Oil and a for a Black Oil and a
Volatile OilVolatile OilEOS simulated
oil formation volume factors obtained from a
differential depletion test at T = 180 oF for a
black and a volatile oil.
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
Form
atio
n V
olum
e Fa
ctor
(Bo)
0 1000 2000 3000 4000 5000Pressure (psia)
Black OilVolatile OilBlack Oil Correlation
TR =180 oF
PbPb
05/03/2305/03/23 7474
Simulated PVT Properties Simulated PVT Properties for a Black Oil and a for a Black Oil and a
Volatile OilVolatile OilEOS simulated solution gas oil ratios obtained
from a differential
depletion test at T = 180 oF for a
black and a volatile oil.
0
1000
2000
3000
4000
Solu
tion
Gas
Oil
Rat
io (R
s)
0 1000 2000 3000 4000 5000Pressure (psia)
Black OilVolatile OilBlack Oil Correlation
TR =180 oF
PbPb
05/03/2305/03/23 7575
Simulated PVT Properties Simulated PVT Properties for a Black Oil and a for a Black Oil and a
Volatile OilVolatile OilGas specific gravities obtained from a differential depletion test at T = 180 oF of a black and a volatile oil.
0.6
0.8
1.0
1.2
1.4
Gas
Spe
cific
Gra
vity
0 1000 2000 3000 4000 Pressure (psia)
Black OilVolatile Oil
T=180oF
05/03/2305/03/23 7676
Simulated PVT Properties Simulated PVT Properties for a Black Oil and a for a Black Oil and a
Volatile OilVolatile OilGas formation volume factors
from a differential
depletion test at T = 180 oF
for a black and a volatile oil.
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0.18
Bg
(cu
ft /
SCF)
0 1000 2000 3000 4000Pressure (psia)
Black OilVolatile Oil
T = 180oF
scfft cu0282.0.0
PZTBg
05/03/2305/03/23 7777
The Less Popular The Less Popular Volatized Oil or Solution Volatized Oil or Solution
Oil-Gas Ratio, Oil-Gas Ratio, RvRv The The RRvv of a reservoir gas at of a reservoir gas at PP11,T,T11 (reservoir (reservoir
pressure and temperature) is obtained pressure and temperature) is obtained by expanding a fixed volume of this gas by expanding a fixed volume of this gas from to from to PPSCSC, T, TSCSC. .
The volumes of surface gas and stock The volumes of surface gas and stock tank oil obtained are recorded, and tank oil obtained are recorded, and RRvv is is calculated as the number of calculated as the number of STBSTB of of surface oil obtained per cubic feet of surface oil obtained per cubic feet of surface gas at standard conditions. surface gas at standard conditions.
05/03/2305/03/23 7878
The Less Popular The Less Popular Volatized Oil or Solution Volatized Oil or Solution
Oil-Gas Ratio, Oil-Gas Ratio, RvRvThis oil that appeared at surface was This oil that appeared at surface was
volatilized at reservoir conditions. volatilized at reservoir conditions. For pressures above the saturation For pressures above the saturation
pressure, pressure, RRvv is just the inverse of is just the inverse of RRss.. But it is not at pressures But it is not at pressures PP << PPsaturationsaturation
MMSCFSTBRv
conditions standardat gas surface offeet cubic ofmillion conditions standardat oil of barrels
05/03/2305/03/23 7979
separator oil
stock tank oil
P > Pb P, T P = PSC
Reservoir Separator SurfaceConditions Conditions Conditions
TR
T= TSC
Schematics of the Schematics of the additional oil produced additional oil produced
from a volatile oilfrom a volatile oil
05/03/2305/03/23 8080
Schematics of the oil Schematics of the oil produced from a gas produced from a gas
condensatecondensate
surface oil
P > Pd P = P SC
ReservoirSurface
Conditions Conditions
TR
T = TSC
05/03/2305/03/23 8181
ExampleExample Evaluation of Reservoir Evaluation of Reservoir
Engineering Fluid Properties using Engineering Fluid Properties using an Equation of State (EOS)an Equation of State (EOS)
The following example illustrates a The following example illustrates a set of fluid property calculations set of fluid property calculations using the Soave-Redlich-Kwong using the Soave-Redlich-Kwong EOS. The fluid is a hypothetical EOS. The fluid is a hypothetical reservoir oil (BigBucks) with 4 reservoir oil (BigBucks) with 4 components. components.
05/03/2305/03/23 8282
Initial Reservoir Initial Reservoir CompositionComposition
ComponentName & Short Cut
Identification
MoleFraction Tc (K) Pc (atm) Mw
MethaneEthanePentaneNonane
C1
C2
C5
C9
0.30.20.20.3
190.6305.4469.6594.6
48.548.233.322.8
0.0080.0980.2510.440
16.04230.06872.146
128.259
05/03/2305/03/23 8383
Example...Example...Given initial reservoir pressure and Given initial reservoir pressure and
temperature are: temperature are:
T = 200.0 F & P = 1800.0 psiT = 200.0 F & P = 1800.0 psi
Assume we start with 1 lb-mole of Assume we start with 1 lb-mole of fluid fluid upscaling easily done upscaling easily done
05/03/2305/03/23 8484
Example...Example...Properties evaluated from the EOS Properties evaluated from the EOS
areareVolume (ft3) (one lb-mole) Z-Factor (liquid) Density (lb/ft 3) Molecular Weight
1.9820.504132.15163.73
05/03/2305/03/23 8585Temperature
Pre
ssur
eP = 1507.5 psi
T=200oF
CP
Temperature
Pre
ssur
eP = 1507.5 psi
T=200oF
CP
The bubble point pressure
corresponding to T=200.0oF is P=1507.5 psi
05/03/2305/03/23 8686
Fluid properties at the Fluid properties at the bubble-point are:bubble-point are:
Volume (ft3/ lbmol)Z-FactorDensity (lb/ft3)Viscosity (cP) Molecular Weight
2.018.429831.585.127363.73
05/03/2305/03/23 8787
Flash LiberationFlash Liberation
Simulated Constant Mass Simulated Constant Mass Expansion.Expansion.
This experiment is used to obtain This experiment is used to obtain isothermal compressibilities.isothermal compressibilities.
05/03/2305/03/23 8888
First ExpansionFirst ExpansionT=200.0ºF & P=1500.0 T=200.0ºF & P=1500.0
psiapsiaComponent
NameTotal
Moles %zi
Vapor %
yi
Liquid %
xI
C1 30.000 72.017 29.848C2 20.000 22.507 19.991C5 20.000 4.499 20.056C9 30.000 .977 30.105
05/03/2305/03/23 8989
First ExpansionFirst ExpansionT=200.0ºF & P=1500.0 T=200.0ºF & P=1500.0
psiapsiaProperty Total
Mixture Vapor Liquid
Mole Fraction 1.000 .004 .996
Volume (ft3) 2.027 .014 2.013
Volume (ft 3 /lb mol)Volume% 100.0 .7 99.3
Density (lb/ft3) 31.438 5.713 31.622
Z-Factor .4296 .8465 .4281Viscosity (cP) .0163 .1278
Molecular Weight 63.73 22.82 63.88Surface Tension
(dyne/cm)2.667
05/03/2305/03/23 9090
Second expansionSecond expansionT=200.0T=200.0ooF & P=1200.0 F & P=1200.0
psiapsiaComponent
Name
TotalMoles %
zi
Vapor %yi
Liquid %xI
C1 30.000 70.211 23.663C2 20.000 24.399 19.307C5 20.000 4.597 22.427C9 30.000 .793 34.603
05/03/2305/03/23 9191
Second expansionSecond expansionT=200.0T=200.0ooF & P=1200.0 F & P=1200.0
psiapsiaProperty Total Mixture Vapor Liquid
Mole Fraction 1.000 .136 .864Volume (ft
3) 2.524 .694 1.830
Volume (ft3/lbmol)
Volume% 100.0 27.5 72.5Density (lb/ft
3) 25.249 4.499 33.118
Z-Factor .4280 .8643 .3592Viscosity (cP) .0153 .1469
Molecular Weight 63.73 22.93 70.16Surface Tension
(dyne/cm)4.008
05/03/2305/03/23 9292
Example...Example... Verify the evaluation of gas and liquid Verify the evaluation of gas and liquid
densities and molecular weights using densities and molecular weights using the following equations:the following equations:
vv V
vMw )(
Nc
iii MwyvaporMw
1
)(
Similarly for the liquid phase use liquid Similarly for the liquid phase use liquid compositions (compositions (xxii) and liquid volume.) and liquid volume.
05/03/2305/03/23 9393
Another expansionAnother expansionT=200.0T=200.0ooF & P=800.0 F & P=800.0
psiapsiaComponent
NameTotal
Moles %zi
Vapor %yi
Liquid %xI
C1 30.000 66.387 15.284C2 20.000 27.699 16.886C5 20.000 5.246 25.967C9 30.000 .669 41.863
05/03/2305/03/23 9494
Another expansionAnother expansionT=200.0T=200.0ooF & P=800.0 F & P=800.0
psiapsiaProperty Total Mixture Vapor Liquid
Mole Fraction 1.000 .288 .712Volume (ft
3) 3.897 2.276 1.621
Volume (ft3 /lb mol)
Volume % 100.0 58.4 41.6Density (lb/ft
3) 16.352 2.988 35.115
Z-Factor .4405 .8934 .2574Viscosity (cP) .0142 .1780
Molecular Weight 63.73 23.62 79.96Surface Tension
(dyne/cm)6.408
05/03/2305/03/23 9595
Summary Constant Mass Summary Constant Mass Expansion at 200.0 Expansion at 200.0 ooFF
Pressurepsi
RelativeVolume
V/Vb
Compressibility(1/psi) Y-Factor
1800.0 .9824 .517E-04Pb = 1507.5 1.0000 .589E-04
1500.0 1.0047 1.071200.0 1.2509 1.02800.0 1.9315 .95
05/03/2305/03/23 9696
Simulated Differential Simulated Differential Depletion Test Depletion Test
This test allows the determination This test allows the determination of the oil and gas formation volume of the oil and gas formation volume factors (factors (BBoo & & BBgg) and solution gas-oil ) and solution gas-oil ratios (ratios (RRss))
Here we assume that ALL the gas Here we assume that ALL the gas generated from a flash expansion generated from a flash expansion is being produced. is being produced.
05/03/2305/03/23 9797
Depletion Stage # 1Depletion Stage # 1T=200.0T=200.0ooF & P=1500.0 F & P=1500.0
psiapsiaComponent
Name
TotalMoles %
zi
Vapor %yi
Liquid %xI
C1 30.000 72.017 29.848C2 20.000 22.507 19.991C5 20.000 4.499 20.056C9 30.000 .977 30.105
Properties as this stage are still the Properties as this stage are still the same (no production took place yet).same (no production took place yet).
05/03/2305/03/23 9898
Depletion Stage # 1Depletion Stage # 1T=200.0T=200.0ooF & P=1500.0 psiaF & P=1500.0 psia
Property Total Mixture Vapor LiquidMole Fraction 1.000 .004 .996Volume (ft
3) 2.027 .014 2.013
Volume (ft3 /lb mol)Volume% 100.0 .7 99.3
Density (lb/ft3) 31.438 5.713 31.622
Z-Factor .4296 .8465 .4281Viscosity (cP) .0163 .1278
Molecular Weight 63.73 22.82 63.88Surface Tension
(dyne/cm)2.667
05/03/2305/03/23 9999
Depletion Stage # 2Depletion Stage # 2T=200.0T=200.0ooF & P=1200.0 F & P=1200.0
psiapsia
Note that the composition using for flashing is identical to the liquid composition from the previous stage (i.e., ALL gas has been produced).
ComponentName
TotalMoles %
zi
Vapor %yi
Liquid %xi
C1 29.848 70.202 23.661C2 19.991 24.407 19.314C5 20.056 4.597 22.426C9 30.105 .794 34.599
05/03/2305/03/23 100100
Depletion Stage # 2Depletion Stage # 2T=200.0T=200.0ooF & F & P=1200.0psiaP=1200.0psia
Property Total Mixture Vapor LiquidMole Fraction 1.000 .133 .867Volume (ft
3) 2.515 .678 1.837
Volume% 100.0 27.0 73.0Density (lb/ft
3) 25.404 4.499 33.117
Z-Factor .4263 .8643 .3592Viscosity (cP) .0153 .1468
Molecular Weight 63.88 22.93 70.16Surface Tension
(dyne/cm)4.008
05/03/2305/03/23 101101
Depletion Stage # 3T=200.0oF & P=800.0psia
ComponentName
TotalMoles %
zi
Vapor %yi
Liquid %xi
C1 23.661 65.493 15.112C2 19.314 28.574 17.421C5 22.426 5.263 25.934C9 34.599 .670 41.533
05/03/2305/03/23 102102
Depletion Stage # 3T=200.0oF & P=800.0psia
Property Total Mixture Vapor LiquidMole Fraction 1.000 .170 .830Volume (ft
3) 3.225 1.339 1.887
Volume% 100.0 41.5 58.5Density (lb/ft
3) 21.752 3.011 35.048
Z-Factor .3646 .8916 .2568Viscosity (cP) .0142 .1769
Molecular Weight 70.16 23.75 79.64Surface Tension
(dyne/cm)6.346
05/03/2305/03/23 103103
Example...Example...In a differential depletion test a In a differential depletion test a
last stage is added at standard last stage is added at standard conditions to determine the conditions to determine the residual oil volume that will be residual oil volume that will be used to determine the oil formation used to determine the oil formation volume factor. volume factor.
05/03/2305/03/23 104104
Last Depletion StageT=60.0ºF & P=14.7psia
ComponentName
TotalMoles %
zi
Vapor %yi
Liquid %xi
C1 15.112 40.581 .247C2 17.421 44.305 1.731C5 25.934 14.901 32.373C9 41.533 .212 65.649
05/03/2305/03/23 105105
Last Depletion StageT=60.0ºF & P=14.7psiaProperty Total Mixture Vapor Liquid
Mole Fraction 1.000 .369 .631Volume (ft
3) 139.930 138.367 1.564
Volume% 100.0 98.9 1.1Density (lb/ft
3) .569 .082 43.661
Z-Factor .3695 .9915 .0065Viscosity (cP) .0095 .5637
Molecular Weight 79.64 30.85 108.12Surface Tension
(dyne/cm)21.432
05/03/2305/03/23 106106
Summary of Results from Summary of Results from Differential Depletion at Differential Depletion at
200.0 200.0 ooFF
Pressure (psi) Bod
(Bbl/STB)Rsd
(Scf/bbl) Bgo
(lb/ft3)Z g
1800.0 1.767 32.15Pb = 1507.5 1.799 1030.1 31.58
1500.0 1.795 1023.3 .010595 31.62 .846 .7881200.0 1.632 773.3 .013524 33.12 .864 .792800.0 1.453 496.8 .020934 35.05 .892 .820
14.7/60 1.000 43.66
05/03/2305/03/23 107107
EXERCISEEXERCISEVerify the calculation of formation Verify the calculation of formation
volume factors and solution gas-oil volume factors and solution gas-oil ratios using the formulas derived in ratios using the formulas derived in class. class.