l10 tuning eos simulations

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:


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


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.


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:


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,


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


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


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


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


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


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


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


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


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


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


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


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


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



(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


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



(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



(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


3) 2.515 .678 1.837


3) 25.404 4.499 33.117



(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


3) 3.225 1.339 1.887


3) 21.752 3.011 35.048



(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


3) 139.930 138.367 1.564


3) .569 .082 43.661



(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.

l10 tuning eos simulations

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