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THIS.1SA PREPRINT--- SUE!JECTTO CORRECTION

PAPER SPE 4271

SIMULATION OF GAS-CONDENSATE

RESERVOIRS

by

A. Spivak* and T. N. Dixon, Members, AIME, The U. of Texas,Austin. Texas

@ Copyright 1973American Institute of Mining, Metsllurgical, and Petroleum Engineers, Inc.

This paper was preparedfor the 3rd NumericalSimulationof ReservoirPerformanceSymposiumof the Societyof PetroleumEngineersof AIME, to be held in Houston,Tex.,Jan. 10-12,1973.Permissionto copy is restrictedto an abstractof not more than 300 words. Illustrationsmay notbe copied. The abstractshouldcontainconspicuousacknowledgmentof where and by whom the paperis presented. Publicationelsewhereafterpublicationin the JOURNALOF PETROLEUMTECHNOLOGYorthe SOCIETYOF PETROLEUMENG&ERS JOURNALis usuallygrantedupon requestto the Editorof theappropriatejournalprovidedagreementto give proper creditis made.

Discussionof this paper is invited. Three copiesof any discussionshouldbe sent to theSocietyof PetroleumEngineersoffice. Such discussionmay be presentedat the above meetingand,with the paper, may be consideredfor publicationin one of the two EWE magazines.

ABSTRACT temperature of the oil. It is reason-able to assume, therefore, that

This paper describes a new method for reservoir fluids on the other side ofsimulating gas-condensate reservoirs. the critical point (e.g., gas-The simulator accounts for both condensate fluids) could be treatedretrograde condensation and vapori.za- in an analogous manner. In othertion of condensed liquid as well as words, one would expect that gas-arbitrary field shapes, well patterns condensate reservoirs could be treatedand heterogeneities. The formulation by a formation volume factor approachof the simulator is based upon a with fluid properties determined fromformation volume factor or Beta-type laboratory depletion data. Theanalysis which is analogous to that simulator described in this paperused in black-oil simulation models. employs such an approach.In the gas-condensate analogy, masstransfer between the gas and liquid The simulator handles three phases andhydrocarbon phases is handled by an consists of numerical solution of the

‘$ term which has units of STB continuity or mass balance equaticnsllquid/MCF dry gas and is similar to for gas, liquid hydrocarbon and waterthe Rs term in black-oil simulation. in one, two and three dimensions.

Fluid properties for back-oil simula- INTRODUCTIONtion models are usually obtained fromlaboratory PVT cell depletion data. The proposed method assumes aThis type of model is of questionable condensate-gas to be a pseudo two-reliabili.ty only in special cases such component system. The two pseudo-as volatile-oil reservoirs where thereservoir pressure and temperature are

components are, at standard conditionsof pressure and temperature, a dry gas

close to the critical pressure and and a hydrocarbon liquid. Each%Presently with Chevron Oil Field pseudo-component is itself a multi-

Research Co., La Habra, California component hydrocarbon fluid. TheReferences and illustrations at end of water phase, if present, constitutespaper. a third “component”.

STMIJI,ATTONOF GAS-CONDENSATE RESF,RVOTRS SPE 477’--. .-—--- ----- ----- -.-. ——-.-.--— “.—-—------.- —.— .-..

The basic assumption for the pseudo In many condensate-gas reservoirs, thetwo-component hydrocarbon system is retrograde liquid is immobile and thusthat dry gas holds hydrocarbon liquid its viscosity is of no importance.as a single-valued function of pressure , Also, for the retrograde liquidsThis assumption is essential to the associated with condensate-gas fluids,characterization of a condensate-gas the variations in composition willas a two-component system and is have negligible effect on liquidbased on the following facts which compressibilities . Thus the assumptio]are universally observed for of a constant retrograde liquidcondensate-gases : composition is justified from a

volumetric and flow point of view.a. liquid condenses from a condensate-

gas by retrograde condensation when It should be noted that the assumptionthe pressure is reduced isothermally of a pseudo two-component system wherefrom the dew point. the dry gas component holds liquid as

a single valued function of pressureb. retrograde liquid is “picked up” does not imply a constant vapor phase

or vaporized by dry gas. composition. The vapor phase composi-tion at any pressure can be anything

A convenient way of expressing the from the dry gas composition toholding tendency of the dry gas for the composition represented by thethe liquid hydrocarbon is in STB equilibrium liquid content of theliquid per MCF of dry gas. To conform gas at that pressure. Also, thewith observed experimental data on assumption that the equilibrium liquidthe depletion of a condensate-gas content is a single-valued fmction offluid, a liquid content vs. pressure pressure is entirely consistent withcurve appears qualitatively as shown the behavior exhibited by real, two-in Figure 1. The liquid content of component systems..a gas 1s designated as rs.

As is done with compositionalThe “J” shape of the liquid content simulators , the proposed approachVs , pressure curve reflects the assumes that there is equilibrium atdropping out of liquid as pressure all times wherever two phases aredeclines below the dew point and present. This assumption is generallythe revaporization of a portion of accepted as being valid in hydrocarbonthe dropped out liquid as pressure reservoirs . The assumption simplycontinues to decline below a certain says that the rate of mass transferminimum value. of components between phases is much

greater than the rate at whichAn additional assumption is that the individual components travel withinretrograde liquid composition remains the phases themselves. There isconstant, which implies that the expe i ental evidence confirmingliquid pseudo-component does not hold this~,~93

dry gas ..

This assumption is notconsistent with the behavior exhibited FORMULATION OF THEby both two-component and multi- MATHEMATICAL MODELcomponent systems. However it mustbe looked at in the light of its The Flow Equationseffect on volumetric calculations.If only the volumes of liquid phase The mathematical model consists of ain the reservoir and their movement mass balance on each cf the threeunder flow potential gradients are component~ of the system:of interest, the composition of liquid hydrocarbon, and dryw~~~~’the liquid is important only insofar Darcy’s Law gives the volumetric flowas it is related to the viscosity and at a point for each phase in terms ofcompressibility of the liquid phase. the potential gradient:Thus, the assumption of constant liquidcomposition, as far ~s fluid flow in far the water phase:the reservoir is concerned, impliesthat the viscosity and compressibility krwof the liquid phase are functions of ;=-k— V*W . . . . , . (la)pressure only. w

k

DV h977 A. SPTVAK and T. N. T)TXC)NLuT&r.b . . . -. A, . . . . ---- - . . . . . . . . . . .

for the liquid hydrocarbon phase: for water:

k a4 -QV*L‘t = ‘k /J& (

~bk.* **** (lb) w ~~) + ;(bwk~>) +

a

(

km a~

)

b’~ bwk —— -

()~ azw qvw ‘~~ bwsw

k-1-k~ V@ (lC) “*’*”””*”**””””” ●

. (3a‘g = pg g“””””

for the vapor phase:

I for liquid hydrocarbon:

The continuity equation expressesconservation of mass. Introducingformation volume factors, the con-tinuity equations for water, liquidhydrocarbon, and dry gas are expressedas follows:

for water:

for liquid hydrocarbon:

V*(b4;L)+ V*(bgrs;g)+ q =v4/

= d% ( )bLS2+b r S . . . . . . . (3bgsg

for dry gas:● ***. . (2b)

for dry gas:

k a+V“(bg:g)+q =-~ a

Vg; (bg Sg) . .

(2C) z ( ) ()bgkX~ - q = # b S

/Jg Vg atgg

The second term on the left side ofequation 2b represents the liquidhydrocarbon being transported inthe vapor phase since the dry gascan hold hydrocarbon liquid.Equations 2 assume that porosity isconstant and therefore rockcompressibility is zero. Thisassumption is not a necessary oneand is made only for convenience indescribing the formulation of thesimulator.

. ...0 . . . . . . ...* ● ** (3C

where the flow potential for anyphase, m, is defined as:

S Pm-ymh=m

Additional equations are:

Substituting+for the phase volumetric P -P4=P()s (4)

flow rates (u), g .**** ● *the conservation of cl% g

mass equations in Cartesiancoordinates become:

()‘4-PW=PC4WSW ● “”*””” (5)

SW+S4+S = 1g

. ..0. .*O (6)

SIMULATION OF GAS-CONDENSATE RESERVOIRS SPE 4271——---—--- —-.. _- —---

The Role of r , the? Parameter

point and the gas content of the oil

Liquid Conten becomes a dependent variable to bedetermined by mass balance or

The liquid content> rs~ is either a continuity.function of pressure or an unknown(i.e., primary dependent variable). Although this feature is present inThis feature is fundamental to the almost any oil reservoir that under-discussions that follow. If liquid goes primary depletion and is sub-hydrocarbon is present at a point, sequently repressured, the literaturethen the liquid hydrocarbon and vapor on three-phase oil reservoir simulationphases are in equilibrium and rs is contains no discussions of how thedetermined from the equilibrium rs vs. problem should be handled. The methodpressure curve (see Figure 1). If described here for a condensate-gas-liquid hydrocarbon is not present in water system should be directlythe form of a liquid phase, it can applicable to a black-oil system.still be transported in the vaporphase and the liquid content at a When all three phases are present, thepoint must be determined by continuity primary dependent variables or unknownsor mass balance. are:

This can be better visualized in terms @w, $ + s s and Sof cycling gas through a reservoir 4’ ~’ w’ t’ gwhich has undergone pressure depletionand has a retrograde liquid hydrocarbonphase present. The first dry gasinjected will vaporize liquid accord- When there is no liquid hydrocarboning to the ~s (equilibrium liquid present, the unknowns are:content) curve. Eventually, all ofthe liquid will be vaporized. Thedry gas that vaporizes the last drops @w, $ s s and rof liquid will have a liquid content g’ w’ g’ s

less than ;s. As this gas flowsthrough the reservoir, it will comeinto contact with either “wetter” or“dryer” gas and its liquid content The cases of liquid hydrocarbonwill change accordingly. Thus, for present and no liquid hydrocarbonthis gas, rs is not simply a function present will subsequently be referredof pressure but depends upon the path to as Case 1 and Case 2, respectively.it takes and the gas with which it It should be noted that for Case 2,comes into contact. since S

L = O, equation 2b becomes:

An analogous situation can exist inblack-oil simulation for an oil v ● (bg r~ ~g) + qv~= ‘dreservoir where the gas content of ~ (bg % q

the oil (R ) can be either a pressuredependent ?unction or a dependentvariable. This can be visualized if Equations 4, 5, and 6 cm be used toone considers a steeply dipping oil eliminate two potentials and onereservoir initially above its bubble saturation from equations 3a, b, andpoint. If the field is produced by c. As a result, we obtain three secondpressure depletion, gas is released order nonlinear partial differentialfrom the nil. and portions of the equations in three unknowr,s. Forreleased gas will migrate updip, Case 1 the unknowns are one potentialIf subsequently pressure is increased and two saturations. For Case 2 theby, say, water injection, the gas unknowns are one potential, oneremaining in the oil zone will go back saturation, and rs.into solution in the oil. However,because of the updip migration of some In addition to the spatial derivatives,of the gas, when all of the gas is the terms on the left side of equationsredissolved, the gas content of the 3 contain quantities that are functionsoil will be less than the gas content of either the dependent or independentgiven by the original Rs vs pressure variables. The absolute permeabilitycurve. Thus the oil has a new bubble k is a function of the spatial

D17 IL971 A C!DTtl AK and T M nlvnhl f.J# T&r* ‘>. “. .I. ,’.’. u.’= .L . . . ● JJ J.’>”..

independent variables only (i.e. , k = where kmk (X, Y,Z)). Relative permeabilities,k

AW=bWk —are functions of saturation. For Mw

tfi;ee-phase relative permeabilitycalculations, relative permeabilitiesto water and gas are functions of water Backward time differences are employedand gas saturations, respectively. for the right hand side time deriva-Relative permeability to liquid hydro- tives .carbon is a function of two of thethree saturations.

()& ~~ ~

Formation volume factor, viscosityand density for the water and liquidhydrocarbon phases are functions of @ (bw ‘w)i,j,k,n+l - @ (bw ‘W)i,j,k,ntheir respective phase pressures.Formation volume factors, viscosities, Atand densities for the gas phase aredifferent for Cases 1 anti 2, Inthe case of equilibrium gas (Case 1), Using these difference approximationsthese quantities are functions of in equations 3 and multiplying eachpressure only. For case 2, they term in the equations by Ax Ay Az/must be functions of both rs and 5.6146 yields:pressure.

The reason for the differences in the ATw A@W-qW=&&(bw %) (7a;calculation of bg, p , and p

. .between

Cases 1 and 2 is tha%, in ge~eral,these quantities are functions of AT&A@4+ATgr~A~g- q=both pressure and liquid content. 4For equilibrium gas, the liquidcontent is known as a single-valued &’t(bLs4+ bg%sg) . . . . . (7b;function of pressure but for non-equilibrium gas the liquid contentand can vary from that of dry gas tothat of the equilibrium gas at any A~gAeg-qg= fiw% %) . . (7C:given pressure. Table 1 summarizesthe dependence of the variousquantities in the flow equations 3.

where:

Finite Difference Approximationsto the Partial Differential AT A@= Ax TAX@+Equations 3

In the derivations that follow, theAYTAYWAZTAZ4

subscripts i, j, k, and n refer tothe x, y, z, and t independentvariables. The reservoir is dividedinto a number of grid blocks by gridlines parallel to the x, y, and z AXTAX$ = ‘i+~,j,k,naxes . The standard central differenceapproximation is employed for thespatial derivatives: (

~i+l,j,k,n+l -1’)

i,j,k,n+1 -

& (Aw~) ‘& [%i++,j,k‘i-&,j,k,n

(‘i,j,k,n+l

)- ii-l, j,k,n+l

2

(‘wi+x,j,k

)- $wi,j,k -

‘i+~,j,k =(

bkkrAYAz.—

2 w 5.6146&: ) i+~,j,k

‘wi-~,j,k(

‘Wi,j ,k )1- %i-~,j,k 2

2

—

I SIMULATION OF GAS-C

q = production rate from block (i,j,k)in STB/D or MCF/D

PV = 0. . i Ax Av Az = Dore volume of“l,],K .

r . -1,, block (i,_j,k)P

The terms Tw, T~, and Ttransmissi.bilitles for $h%%ebetween grid blocks, of the water,liquid hydrocarbon, and gas phases,respectively. In addition, the termTgrs represents the transmissibilityfor the flow between grid blocks ofliquid hydrocarbon being transportedin the gas phase. Thus Twi+~,j ,k

(Qwi+l,j ,k-awi.,”,k) is the rate of

$flow in STB/D o water in thex-direction from block i+l,-j,k toblock i,j,k.

The solution of the finite differenceapproximations to the flow equationsis discussed in Appendix A.

FLUID PROPERTIES

Compositions of the LiquidHydrocarbon and Dry GasPseudo-Components

The method assumes that a condensate-gas is a pseudo two-component systemfor which each component may be amulti-component mixture. The liquidphase composition must remain constant;however, the vapor phase cap holdvarying amounts of liquid and thus itscomposition can change. In reality,the compositions of the liquid andvapor phases both in the reservoirand at the surface are constantlychanging. As a result there is adegree of freedom in the choice ofliquid and dry gas compositions forthe two pseudo-components.

Most wells in a condensate-g<s fieldare tested soon after com~letion todetermine deli.verabilities. At leastone well is extensively tested throughsurface separation facilities to obtairfluid samples for laboratory analysis.The compositions of the separatorliquid and separator gas obtainedduring this sampling provide convenientliquid hydrocarbon and dry gas pseudo-compositions . The initial liquidcontent (rs) in this case is simplythe field measured liquid-gas ratio.These data are always availablewhenever a condensate-gas well isproduced to obtain fluid samples forrecombination and analysis.

DENSATE RESERVOIRS SPE 4271

Alternately, the calculated ormeasured compositions of the dry gasand liquid products from the fieldsurface treatment facilities duringactual production may be used as thepseudo-compositions. The productcompositions will change with time;however, they can always be determinedfor the time of initial productionwhen the composition of the producedgas (i.e., well effluent) is known.For example, in the case where thegas is to be processed through aplant which will strip out all of theC~+ components, appropriate composi-tions for the liquid hydrocarbon anddry gas are:

a. for the liquid hydrocarbon--ahydrocarbon liquid which containsonly C3 and heavier components inthe same relative proportions asin the initial reservoir fluid.

b. for the dry gas--a gas whichcontains Cl and C , as well asC02, N2, He, ~nd i2S if theyare present, m the same relativeproportions as in the initialreservoir fluid.

In the case where the produced gasis to be field separated only,appropriate compositions for theliquid hydrocarbon and dry gaspseudo-components are the separatorgas and liquid compositions.

If the liquid and dry gas compositionsare always based on some split of thedew point composition and if theinitial liquid content (rs) is suchthat it gives a vapor phase composi-tion equal to the dew point composi-tion, then at lea-t initially theformation volume factor model iscompositionally correct. As liquidcondenses out in the reservoir andas cycled dry gas is introduced,the compositions of the liquid andvapor phases in the pseudo two-component system will deviate fromthe actual phase compositions.

Properties

The fluid properties of interest aredensity, viscosity, and formationvolume factor. As was discussedpreviously, under isothermal conditionsthese properties are functions ofeither pressure or, in the case ofnon-equilibrium gas, pressure, and rs.

.

,V 11971 A. SPIVAK and T. N. DIXON 7

Properties of theHydrocarbon Liquid

Since the hydrocarbon liquid composi-tion is constant by assumption, thevariation in hydrocarbon liquidproperties as a function of pressureonly are required. A widely usedcorrelation for liquid hydrocarbondensities as a function of pressure,temperature, and com osition is given

$by Alani and Kennedy . A method fordetermining the viscosity of liquidhydrocarbons as a.function of pressure,temperature, and composition has beenpublished by Little and Kennedy5. Theliquid hydrocarbon formation volumefactor is related to the density asfollows:

where p~ is the liquid hydrocarbondensity at stock tank conditions inlb/cu ft.

Properties of theDry Gas

Like the hydrocarbon li.qui.d,the drygas has a constant composition and thu:only the variation in dry gasproperties with pressure is required.The density of a gas at any temperatur~and pressure is given by

w (lb./cu. ft.)p = ZRT

where p = pressure (psia)MW = molecular weight

z= gas compressibility factorR= gas constant = 10.73

(psi-cu ft/mole-OR)T= temperature (degrees

Rankinel

Gas compressibility factors can bedetermined as functions of reducedpressure and temperature using thecorrelation of Standing and Katz6.A commnly used correlation for gasviscosity as a function of pressuretemperature and composition is thatof Carr, Kobayashi, and Burrows7.The dry gas formation volume factoris related to the density as follo~s:

~bg = ox x 5.6146 (MCF/RB)

where P* is the dry gas density inlb/MCF % standard conditions ofpressure and temperature (e.g., 14.65psia and 60”F).

Properties ofEquilibrium Gas

Equilibrium gas compressibility factorsand compositions are measured atvarious pressures during laboratoryconsta.~t-volume depletion experiments.The measured compressibility factorscan be used directly to construct acurve giving equilibrium gas densityvs.”pressure. Equilibrium gasviscosity vs. pressure can be determineusing the measured compositions7andthe correlation of Carr, et al. ,

The equilibrium gas formation volumefactor is, at any pressure, the amountof dry gas contained in one reservoirbarrel of equilibrium gas at thatpressure and is a function of the gasdensity and liquid content. Considerone reservoir barrel of equilibriumgas at some pressure p and-reservoirtemperature. The density p and liquidcontent Fs are known as functions ofpressure. Let b be the number of MCF

fof dry gas conta ned in that reservoirbarrel (i.e., the equilibrium gasformation volume factor). Then theequilibrium gas density ~ in lb/cuft is

f5g p;i=— 5.6146 + ‘g ‘s ‘~ ‘lb’cu ‘t)

Therefore the formation volume factor~g is given by

Just as in black-oil formation volumefactor models, it can be seen that thedensities and formation volume factors

SIMULATION OF GAS-CO

are not independent. For black-oilmodels, however, formation volumefactors are available directly fromlaboratory fluid analysis. Densities,if required, are calculated from theformation volume factors. In thecondensate-gas analogy, densities aremore readily available and formationvolume factors are calculated fromdensities.

Determination of theEquilibrium Liquid Contentvs. Pressure Curve

Mass transfer between the liquid andvapor phases (i.e. , retrograde conden-sation and revaporization of theretrograde liquid) is accounted forby the equilibrium rs curve. Thelaboratory constant-volume depletiontest provides a basis for determiningthe equilibrium liquid content vs.pressure curve just”as the dissolvedgas (R ) VS.

8

pressure curve for a blackoil mo el is determined from alaboratory differential depletion teston an oil sample. The condensate-gasmodel discussed here is a volumetricmodel in which the volumes orsaturations of the different phasesin the reservoir are of primaryinterest. Therefore, the correctequilibrium liquid content vs. pressurtcurve is that which gives the correctretrograde liquid volumes in thereservoir during pressure depletion.As is discussed in the Introduction,it is assumed that the equilibriumliquid content curve is a single-valuecfunction of pressure so that the curveas determined for pressure depletionis applicable to revaporization,

Having chosen the dry gas and liquidhydrocarbon pseudo-components, theequilibrium r curve should give thesame retrogra~e liquid volume vs.pressure curve as obtained in thelaboratory when used in a zero-dimensional material balance calcula-tion which simulates the laboratoryconstant-volume depletion test. Thisis the basis for the determinationof the equilibrium rs curve. Asimilar procedure is often used incompositional simulators whereK-values are adjusted to match thelaboratory constant-volume depletiondata.

ENSATE RESERVOIRS SPE 4271—

There is no way ~f directly determiningthe equilibrium rs curve from thelaboratory retrograde liquid volumecurve. This is because during aconstant-volume depletion test, liquidis both dropping out by retrogradecondensation and being produced in thevapor phase. A material balance onthe liquid at any pressure may bestated as

retrograde liquid+

liquid produced inthe vapor phase = initial liquid

+ in placeliquid remainingin the vapor phase

Both the produced liquid and the liquicremaining in the ~apor phase arefunctions of the rs curve. Thereforethe ?s curve cannot be determined bysimple material balance on the liqu?d.

Determination of the equilibrium rscurve involves a trial and errorprocedure. The zero-dimensionalmaterial balance on the dry gas andliquid hydrocarbon components duringa constant-volume depletion experimentmay be written as

for the hydrocarbon liquid:

~-qg r~ = V at (b.L‘I + bg r. Sg) . (8~;

and for the dry gas:—

-qg ‘~ k(bgsg) ● ● ● . . ● . (8b;

where V is the volume of the systemin barrels and bl, b , and r~ arefunctions of pressur ~ only since thegas is at equilibrium conditions. Fora laboratory pressure-volume cell whicl-is a zero-dimensional system with aporosity equal to 100%, the liquidsaturati~n Sl is the perter.tage liquidvolume. Thus the above set ofequations can be solved for S~ vs pfor different ~s curves until an Fcurve which matches the laboratory s

retrograde liquid volume curve isfound.

eor 1,971 A. SPIVAK and T. N. DIXON 9c)rL -?&I-!.

Properties of Non-Equilibrium Gas

The non-equilibrium gas propertiesare functions of both rs and PC Forthe pseudo two-component system,rs is a composition parameter? andtherefore the dependence of fluidproperties on P and r~ is Comparableto the general dependence of fluldproperties on pressure and composi-tion at isothermal conditions. Areasonable methcd of determiningthe non-equilibrium gas propertiesis to assume that they vary linearlyas a function of rs between theproperties of the dry gas and thoseof the equilibrium gas. Since rsfor dry gas is equal to zero, theassumption of linearity between thedry gas and equilibrium gas propertiesmay be expressed as

r~ _X(r~, p)=~X+(l - >) x d.y g,.

s s

. . . . . ● ☛☛✎☛ ● ☛☛☛☛ ● “ (9)

where X is any property such asdensity or viscosity and ~ is theequilibrium value of that property.

It should be noted that the methodof solution of the finite differenceapproximations to the flow equationsas outlined in Appendix A does notdepend upon the form,of the functionalrelationship between the propertiesof non-equilibrium gas and those ofdry and equilibrium gas. Thus thefunction could be a more complexone if this were justified.

A THREE-DIMENSIONALEXAMPLE

Three-dimensional calculations weremade for the quarter five-spot shownin Figure 2. The quarter five-spotis 5,000 ft by 5,000 ft in arealextent and 200 ft thick. There is anunderlying aquifer and the initialgas-water contact is 144 ft from thetop of the formation. The reservoirwas assumed to be homogeneous butanisotropic. Values of horizontaland vertical permeability were 10 mdand 1 md, respectively, and porositywas 10 percent. A 5x5x5 grid systemwas utilized so that each grid blockhad areal dimensions of 1,000 ft by1,000 ft and was 40 ft thick.

Fluid Properties

Fluid properties were determined foran actual condensate-gas fluid havinga dew point of 3632 psig at thereservoir temperature of 175°F. Thedetails of the fluid property calcula-tions are given in Appendix B. Watercompressibility and viscosity wereassumed to be constant at 0.000003psi-l and,l cp, respectively.

Relative Permeability andCapillary Pressure Data

Three-phase relative permeability datawere calculated using two-phasewater-oil and gas-oil data. Thetwo-phase water-oil and gas-oilrelative permeability curves thatwere used are shown in Figures 3 and4, respectively. Relative permeabilityto water was calculated from Figure 3as a single-valued function of watersaturation. Relative permeability togas was calculated from Figure 4 asa single-valued function of gassaturation. Relative permeabilityto hydrocarbon liquid was calculatedas

k~h X k*r~

1

where krh is a function of watersaturation and is equal to krodetermined from the two-~hasewater-oil data. k$L is determinedfrom the two-phase gas-oil relativepermeability data as a function ofl-swc-s~. This method of calculating three-phase relative permeabilityfrom two-phase d ta is disc ssed by

# 8Peery and Herron and Coats ,

Figures 3 and 4 show the followingend point saturations:

a. irreducible water saturation = 10%b. residual hydrocarbon by water

drive = 30%c. residual liquid by gas drive = 30%

From Figure 4 and the method ofcalculating liquid hydrocarbonrelative permeability it can be seenthat the relative permeability toliquid hydrocarbon is equal to zerountil the liquid hydrocarbon satura-tion reaches 20 percent.

o SIMULATION OF GAS-C

For this example it was assumed thatgas-liquid hydrocarbon and liquidhydrocarbon-water capillary pressurewere linear functions of gas and watersaturations, respectively, The gas-liquid hydrocarbon capillary pressure(Pcgz) varied from 2 psi at maximumgas saturation to O psi at zero gassaturation and the liquid-hydrocarbon-water capillary pressure (Pc ~) varied

$from 3 psi at irreducible wa ersaturation to O psi at 100 percentwater saturation.

Initial Pressure andQuantities in Place

The initial gas pressure was specifiedto be 3750 psig at the gas watercontact. This is 58 psig above thedew point. The initial quantitiesof water, hydrocarbon liquid, andgas in place were 30,620,000 STB,12,076,000 STB, and 69,400,000 MCF,respectively, Since there was nohydrocarbon liquid phase presentinitially, all of the initial liquidhydrocarbon was contained in the gas.

Production and Injection

A constant dry gas production of7,000 MCF/D was taken from grid block(i=l, 5=1, k=l). Hydrocarbon liquidproduction varied with the rs liquidcontent of this grid block. Dry gaswas injected into grid block (5,5,1)at 4,000 MCF/D. On a reservoir barrelbasis , the initial ratio of inject~onto production was 53%.

Results

The simulation was carried out to1,500 days. Retrograde liquid beganto drop out in the producing gridblock after about 10 days when thepressure in this grid block d~oppedbelow the dew point. Tables 2 through5 show the gas pressures, watersaturations , liquid hydrocarbonsaturations, and ??sliquid content,respectively, for each grid blockafter 1,500 days or about 4 years.

Values for each quantity are shown bylayers from top to bottom with gridblock (1,1) for each layer being atthe upper left. It should be notedthat all quantities are symmetricabout the injection and productiongrid blocks.

DENSATE RESERVOIRS SPE 4271

Table 2 shows that after 4 years thepressure in the producing grid block(1,1,1) has decreased from an initialvalue of 3,735 psi to 3,219 psi. Thepressure in the injection grid blockhas decreased to 3,428 psi resultingin a gradient of 209 psi from theinjectiun to the production gridblocks .

The water saturations in Table 3 showa slight increase in the top threelayers from an initial irreduciblewater saturation of 10 percent. Thisis due to expansion of the connatewater with pressure decline. It canbe seen from this table that the watersaturation in layer 4 has increasedfrom an initial value of 40 percentto about 45 percent directly belowthe producing grid block. This isa coning effect and is due to thepressure sink in this region.

Table 4 shows that a retrogradeliquid saturation of about 9.5percent has built up in the producinggrid block. Retrograde liquid ispresent in all grid blocks wherethere was initially gas except inthe vicinity of the injection gridblock where the dry gas is sweepingout the initial reservoir fluid andretrograde liquid is being vaporized.

The rs liquid content values inTable 5 show that at 1,500 days,the injection grid block has beenalmost entirely swept out by drygas . The grid blocks adjacent toand below the in-jection grid blockhave also been swept to a lesserdegree . The fact that the in-jetteddry gas has swept some grid blocksbelow the injection grid block showsthe presence of a vertical downwardgas potentia~ gradient in this region.

Figure 5 shows the hydrocarbon liquidproducing rate vs time for thisexample. The liquid rate droppedfrom an initial value of 1218 STB/Dto 876 STB/D at 1,500 days. Thedecrease in hydrocarbon liquidproduction with time reflects thedecrease in rs liquid content ofthe producing grid block.

Time Steps and Number ofIterations to Convergence

The time step sequence used for thisexample was as follows:

.—— ---- A CDT(7AV .mrl T N T)TXON 11

At = 1 day from O to 1 day MCF/D = thousands of standard cubic

At = 2 days from 1 to 5 days feet per day

At=5 days from 5 to 20 daysAt = 20 days from 20 to 1,500 days MW = molecular weight

For this example, ADIP was used tosolve the simulator equations. Theaverage number of iterations toconvergence was about 10 and overallmaterial balances remained correctto within 0.5 percent. The totalcomputing time for this example was107 seconds on a CDC 6600 computer.

CONCLUSIONS

1. Gas-condensate reserv~irs can bemathematically modeled using aformation volume factor orBeta-type simulator which isanalogous in formulation to ablack-oil simulator,>

2. Since the formulation of theproposed simulator is analogous tothat of a conventional black-oilsimulator, it is of’comparablecomputational efficiency.

NOMENCLATURE

b ❑ water formation volumewfactor (STB/RB)

bl = hydrocarbon liquid formationvolume factor (STB/RB)

bg

= gas formation volume factor(MCF/RB)

BBL = barrel (5.6146 cu. ft.)

h = elevation with respect tosome arbitrary reference levelmeasured positive downward(ft)

HCPV = hydrocarbon pore volume

P = pressure (psia)

Pc = capillary pressure (psia)

Pv = grid block pore volume(barrels)

qv = production term (STB orMCF/D/BBL of reservoir)

q = production term for an entire<rid block (STB/D or MCF/D)

RB = reservoir barrels

RB/D = reservoir barrels per day

r = liquid content of gas phases(STB/MCF)

R~ = gas content. of liquidhydrocarbon phase (MCF/STB)

s = saturation (fractional)

s = irreducible water saturationWc

STB = stock tank barrel

STB/D = stock tank barrels per day

t ❑ time (days)

T = transmissibility (MCF/D/psi orSTB/D/psi)

T = temperature (degrees Rankine)

+u = darcy velocity vector

(BBL/ft2/D)

z = gas compressibility factor

k = absolute permeability Greek(md x .00633)

Y = specific weight = pg/144 gckr = relative permeability (psi/ft)

!3 = acceleration due to gravity P = density (lb/cu ft)(ft/sec/see)

D = viscosity (cp)

gc = gravitational constant(32.12 ft/sec/see) @ = flow potential = p-yh (psi)

MCF = thousands of standard cubic I 0 = porosity (fraction)feet

2 SIMULATION OF GAS-

Difference Notation

AtU=U(t+At) -U(t)

AXU=U(X+A X)-U(X).

AT AU= AX TAX U+ AYTAYU+ Az T&u

AX TAXU= ‘i~, j,k (Ui+i, j,k - ‘i, j,k )-

Ti-~,j,k (%,j,k ‘Ui-i, j,k )

Subscripts and Superscripts

Subscripts w, L, and g refer to water,liquid hydrocarbon, and gas,respectively. Subscripts i,’j,and krefer to the x,y~ and z cartesianco-ordiriate directions, respectively.Subscript n refers to time level[e.g., At U = U (t+At) - U (t) =Un+l-unl. A bar over fluid property

variable (e.g. , F5) refers to anequilibrium quantity.

ACKNOWLEDGMENT

The authors gratefully acknowledgefinancial support for this projectfurnished jointly by Atlantic RichfielCompany, Continental Oil Company andSb,ellOil Company through a researchgrant.

REFEREiiCES

1.

2*

3.

4*

5*

Smith, L. R., and Lyman Yarborough:“Equilibrium Revaporization ofRetrograde Condensate bv Drv GasInjec~ion, “ Trans. AIME” (1968),243, 87.

Oxford, Charles W., and R. L.Huntington: “Vaporization cfHydrocarbons from an UnconsolidatedSand,” Trans. AIME (1953), 198,—. —318.

Raimondi, P. , and M. A. Torcaso:“Mass Transfer Between Phases ina Porous Medium,” Sot. Pet. ~.~. (1965), 5_, 51. — —

Alani, G. H., and H. T. Kennedy:“Volumes of Liquid Hydrocarbonsat ~’ighTemperatures and PressuresTrans.; AIME” (1360), 219, 288.— —.

Little, J. ‘E..and H. T. Kennedv:“A Correlation of the Viscosity-of Hydrocarbon Systems withPressure, Temperature and

NDENSATE RESERVOIRS SPE 4271

Composition ,“ Trans. AIME (1968),243, 157.

6. Standing, M. B., and D. L. Katz:“Density of Natural Gases,”Trans. AIME (1942), 146, 140.—.

i’,C?lrr, N, L,, R. Kobayashi, andD. b. Burrows: “Viscosity ofHydrocarbon Gases Under Pressure,”Trans. AIME (1954), 201, 264.— .

8. Peery, James H., and E. H. Herron?Jr.: “Three-Phase ReservoirSimulation,” J. Pet. Tech. (Feb.1969), ~, 211. —

9. Coats, K. H.: “An Analysis forSimulating Reservoir PerformanceUnder Pressure Maintenance by Gasand/or Water Injection,” Sot. Pet.Eng. ~. (Dec. 1968), ~, 3~

10. Spivak, A.: “MathematicalSimulation of Condensate-GasReservoirs,” Ph.D. Dissertation,The University of Texas, Austin,(May 1971).

APPENDIX A - SOLUTION OF THEFINITE DIFFERENCE APPROXIMATIONS-----TO THE FLOW EQUATIONS —

The finite difference equatj.ons 7 aresolved on a time step basis. Thesolution each time step is an iterativeone. The equations are first expressedin terms of gas potential @g byeliminating @w and ‘$~using equations4 and 5. The right sides are thenexpanded in terms of time differencesin the dependent variables. Thisexpansion is different according towhether a grid block is Case 1(equilibrium) or Case 2 (non-equilibrium) . The three flow equationsare then added together after beingmultiplied by appropriate factors sothat the addition will eliminateterms in AtS and Atrs. The resultantequation is then solved for @ thegas potential at the new timeg~evelin an iterative process. When thesolution for gas potential has beenobtained, the remaining dependentvariables (either Sw and S for Case 1or S and rs for Case 2) age solvedfor ?rom the original flow equations 7.This completes a time step.

—— ---- A on-I- llAv .-A T N II TYOM 13

The Flow Equations in for Case 1 and

Terms of Gas Potential

From the definition of potential, the A @ At Sw, and At rst g’water and liquid hydrocarbon poten-tials are expressed in terms of gaspotential as follows:

for Case 2. The expansions are done

- (Y~- @ has follows:

‘t= ‘~ -PC9L Case 1

= % - PCQ, - A ‘u h. . (A-1)

Ifor the water equation:

$W = @g- PCgW-(YW - Yg) h~ At (b. Sw) =

I‘g - Pcgw - A ywgh . . (A-2)

[ 1H (bw%)..+~-(b‘W)*=where

. EY[ 1Swn ~ AtPw + %n+l At SwAt

P = Pcgt + ~clwcgw .

where

Therefore, in terms of gas potentials, b - bwnequations 7 become w n+l

b~ =l?n+~ - Pn

ATw A(4g-pcm-Aywgh) -b’ is not a derivative but is a chordY If it is assumed that

CIW =& At(~Sw) .,, .,,.s ope.

(A-3a) ‘t ‘Cwl = At PcgL = O, then

AtPw =Atpl=At Pg

ATtA($g-pcg4-Ay~g h)+

ATgr~A$g -q&=

‘AAt t (b% ‘L ‘bgrs ‘g) . . . . (A-3b)

AT8A$g-qg= ~ At (bg ‘g)

● ✎☛☛✎ ✎✌☛☛☛ ● ☛✎✎✎ ● (A-3c)

Expansion of the Right Sides

The right sides of equations A-3 areexpanded in terms of

This assumption is made for mostmethods of solving the equations offluid flow through porous media andis consistent with dating saturationdependent terms on the left side ofthe flow equations at the previoustime level. If, in addition it isassumed that At y = O, then

g

At Pg = At $g

This assumption is acceptable if timesteps are such that density changesover the time step are small.

A @ At Sw, and At St g’g ~–

.4 SIMULATION OF GAS-C

Thus~At(bW Sw) =

c~ Swn 1~’ At $g +bwn+l At G

=ciOAt$g +cil At Sw +C12 At ‘g

where

Clo = ~ S~dnb~

Cll = fibw n+,

C12 = o

for the liquid hydrocarbon equation:

Pv~(AEb4S1 ‘At bgr~ Sg) =

[-~ S}nb; At P+

b~n+l At SL + Sgn(bg r~)’ At P +

1(bg%)n+.&‘g=c20bt 4g+ c21 At Sw +

c22 At sg

(since At st = -At sg - At Sw)

where

C20 = &[sin b~ 1+Sgn(bgr.)‘

C21 = H (-% n+l)

C22 = ~

[7(bg‘s)n+l - b{n+~

DENSATE RESERVOIRS SPE 427:

and

b lan+l - bLn

b; ‘P & n+l - ‘/,n

(bg ‘~)n+l - (b~ ‘.s~n(bgr~) ‘pgn+, ‘Pgn

for the gas equation:

~At (bg sg) =

~At (Sgn b~ Atp + bgn+~ At sg)

= c30 At +g + c31 At sw + c32 At ‘g

where

C30 = ~ Sgn b;

C31 = o

c32 = ~ bg n+l

and

bg n+1 - bgnb~=

Pgn+l ‘Pgn

Case 2

The expansion of the right side forthe water equation in Case 2 is thesame as for Case 1. The right sideof the liquid hydrocarbon equationfor Case 2 is:

~A (bg r~ Sg) =

[~ (r. ‘gL ‘t bg +

bg n+l 1At(r,Sg)

“n?- 1,.5’71 A. SPTVAK and T. N. DIXON 15ar~ -tLIA . .. ---- ----—--——

[

.=

=~ (r. Sg). (b~btp+b~ At%)+ At [Sgn (b: Atp+b; At r.) +

1 1gn+l Atsgb

bgn+l (r~n At Sg + Sg n+l At ‘s)

= c20 At P +c21At Sw +C22 At rs= c30 At @g + c31 At sw + c32 At r~

where

and

bg (Pn.l, rsn) - bg (PnI r~n )b_! =..

~ Pn+ ~ - Pn

where

C30 = =At (Sgn b:)

C31 = - H bg n+l

c32 = ~ (Sgnb’:)

Substituting the right side expansionsin equations A-3 yields

for Case 1

ATW A (Sg - PCg,, -A Ywg h) - qV,=

bg (Pn+l~‘Sn+l) - bg (Pn+l f ‘S*) c1O At $g A cll At St!+ C!12At Sgb; =

‘s n+l - ‘sn

I . ..*. . . . . . ,. .0. . (.4-4a

It should be noted that the aboveexpansion of At b

Fwhere b = bg

(rs, p) is consis ent withgrespect tothe definition of the time differenceoperator.

Thus At bg ‘b& Atp+b’’Atrg s

= bg (Pn+l~r.n)-bg(pn, r.n)’

bg (pn+l~ rs n+l) - bg (%+1~‘sri)

AT/, A(Qg-~cgL-Ay4g h)+

ATgr~ACg -q&=

c20 At ~g + c21 At Sw + C22 AC S’g

(A-4b. . . . . .. *-* ● *”*” ‘

ATg A$g-qg=c30&+g+

= b~ (Pn+1~ ‘s n+l) - bg (pn’ ‘sri) c31 At SW + c32 At Sg . . . . (A-4c

for the gas equation:for Case 2——

~At (bg Sg) =

[ 1~ Sgn At bg + bgn+l At ‘g

ATw A(4g-pCgw-Aywg@- qw=

CIO At @g + Cll At Sw + C12 ~ rs

. ...0. . . . . . . ● *** (A-5t

; SIMULATION OF GAS-CONDENSATE RESERVOIRS SPE 427

ATgr~A$g - qt = C20 At *g +

C21At Sw+C22Atr~. . . . (A-5b)

ATg A@g-~ =C30AC @g+

c31At Sw+C32Atr~ . . . (A-be)

Terms of the type At Sw, At S forCase 1 and A SW,

f“At r~ for C~se 2

can now be e lmlnated. The elimina-tion of these terms will be demonstra-ted for Case 1. The procedure forCase 2 is identical.

Equaticns A-4a and A-4b are multipliedby al and 22, respectively, and theresultant equations are added toequation A-4c. This yields

alATWA@g + aa A TLA @g +

H = (al * CIO + a2 “ c20 + C30)

At $g + (al * cll + a2 * C21 +C3L)

At Sw +(al * C12 + a~ ● C22 “32)

At Sg. . . . . ● ...* ● (A-6)

where

H= al A TW A (p~gw - A Y,qgh) +

aa A T1 A (Peg% - AY4gh),

To eliminate terms in At SY

andAt S , a2 and al are calcu ated asfoll&.7s:

a2 ..=C22 ‘ “ ● ‘ ‘“” ““ (A-7)

Cql + a2 * C21al.-Cll ● .*.O (A-8)

For Case 2, al and a2 are identicallydefined, however, C21, C31, etc., aredifferent. The resultant singleequation in @

gcan be written as

A~A@g-G=cAt% “*-s (A-9)

where, for Case 1

5’ al Tw +a2T + a2 T r~ + TgL g

G= al A ~ A (pcgw -AYwgh)+

a2A T% A (pcgL - A y~g ‘) +

al*~+a2*q4+qg

and, for Case 2

T= al Tw + a2 Tg r~ + Tg

G= al A TW A (PCgW - A Ywg h) +

al o qw + aa ‘ q~ +clg

and

c= al ● CIO + a2 ● C20 + C30

In both cases the coefficients aland a2 are treated as constants whenthe_difference operator is appliedto T.

The combined difference equation A-9must be solved iteratively due toits non-linear nature. The systemof equations can be solved usingeither SIP (Strongly ImplicitProcedure), ADIP (AlternatingDirectiofi Implicit Procedure), orLSOR (line successive overrelaxation).The coefficients al, a2, C1O, Cll, etcare functions of the dependentvariables at the new time level andmust therefore be updated after eachiteration during the solution. After

a ‘O1utiOn ‘or ‘g n+l ‘as been ‘btaine’

sP17 11771 A. SPTVAK and T. N. DIXON 1

‘~-------..-

from equation A-9, the other dependentvariables are solved for fromequations A-1 and A-2 (for @w and @l)and equations A-3,

For Case 1, the remaining dependentvariables are S and S and theseare solved for ~rom eq~ations A-3aand A-3c. Actually, any two ofequations A-3 could be used to solvefor Sw and S but A-3a and A-3cinvolve the feast amount of computa-tional work.

For Case 2, the remaining dependentvariables are SV7and rs and theseare solved for using equations A-3aand A-3b. For this case, sinces~ = O equation A-3b is

Switching of Blocks from Case 1to Case 2 and Vice Versa

Since the method depends on treatingequilibrium and non-equilibrium blocksdifferently, a suitable method ofswitching from Case 1 to Case 2 andvice versa is essential. The switch-ing methods employed are based on thephysical significance of the twocases.

Case 2 to Case 1 ICase 2 implies that the gas in a gridblock is holding less liquid at theblock pressure than the equilibriumliquid content. When a Case 2 blockis ready to drop out liquid, thecalculated r~ will exceed theequilibrium r . Thus the switch fromCase 2 to Cas&! 1 is made when,

r~2Fs

Case 1 to Case 2

Case 1 implies that there is someliquid hydrocarbon present whileCase 2 implies no liquid hydrocarbonpresent. When all of the liquid hasbeen vaporized from a Case 1 block,the calculated liquid saturation willbecome negative. Therefore, theswitch from Case 1 to Case 2 is madewhen,

L--- “-——

s <0A

The procedure is to check each gridblock during the iterations to seeif a switch should be made.

APPENDIX B - DETERMINATION OFFLUID PROPERTIES FOR ANEXAMPLE CONDENSATE-GAS —

Compositions of the pseudo-componentsand fluid properties were determinedfor an example condensate-gas froman actual condensate-gas reservoir.The initial reservoir fluid composi-tion is given in Table 6. The specifigravity and molecular weight of theC7+ fraction are also given in thistable. The reservoir fluid had alaboratory-measured dew point of 3632psig at the reservoir temperature of175°F.

It was assumed that the reservoirwould be produced through a 100%efficient gas plant which strips outall of the Cj+ components . Therefore;the two pseu o-components were chosento be a dry gas containing the C2.components in the same relativeproportions as in the initial reser-voir fluid and a hydrocarbon liquidcontaining the C3+ components in thesame relative proportions as in theinitial fluid. The pseudo-componentdry gas and liquid hydrocarboncompositions are given in Table 7.The molecular weights and basedensities for the two pseudo-componentare also shown in this table.

Table 8 gives the properties of theequilibrium vapor during a constant-volume depletion analysis performedon a sample of this fluid by acommercial laboratory,

Liquid Densities

Figure 6 shows ‘Ehehydrocarbon liquiddensity vs pressure. These datawere determined using the Alani-Kenneccorrelation . The hydrocarbon liquidformation volume factor at any pressuzis simply the density at that pressur~divided by the liquid density atstandard pressure and temperature.The liquid density at stock tarlkconditions was also determined usingthe Alani-Kenrledy correlation.

Gas Densities

The dry gas and equilibrium gasdensities vs pressure are shown onFigure 7. The dry gas densitieswere obtained from the dry gas composi.tion and the compres ibility factors

5of Standing and Katz . The equilibria]gas density vs pressure curve wasdetermined using the laboratory-measured compressibility factorsobtained during the constant-volumedepletion test. The dew-pointcomposition and the compressibilityfactors of Standing and Katz wereused to obtain the equilibrium gasdensities at pressures above the dewpoint.

Gas Viscosities

Figure 8 shows the dry gas andequilibrium gas viscosities vs pressurThe dry gas viscosities were obtainedfrom the dry gas composition and thecorrelation of Carr, Kobayashi, andBurrows7. The equilibrium gasviscosity vs pressure curve wasobtained using the equilibrium gascompositions measured during thelaboratory constant-volume depletiontest (Table 8) and the correlation ofCarr, et al.

Equilibrium Liauid Conter,t

The equilibrium liquid content curve vpressure is given in Figure 9, Thedew-po~nt liquid content of .174 STB/MCF was calculated such that a mixturecontaining this ratio of liquid hydro-carbon to dry gas has a compositionexactly equal to the dew-point gascomposition. The remainder cf the rvs pressure curve was determined sucRthat when this curve was used in azero-dimensional material balancecalculation simulating the laboratoryconstant-volume depletion test, itclosely matched the measured retrogradliquid volume vs pressure curve. Thesolid line in Figure 10 is thelaboratory retrograde liquid vclumecurve. The dots represent the retro-grade liquid volumes calculated usingthe final rs curve and the zero-dimens:onal mater.al balance calcula-tions .

!DENSATE RESERVOIRS SPE 427:

TABLE 1

Dependence of Variables i:)the Flow Equations

dependence

variable Case 1 Case 2

k k (X,l’, z) k (X, Y,Z)

k k,,-

rw t-w \~w ) k I-W(%)

k r9.k r), (s; , Sg) kr; (S;, Sg)

k rg krg (Sg) rg (ss)k

bg bg (Pg) bg (Pg , rs )

L .,: ‘)L,w (p!,> .,A,(Pw )

N f. ‘ ; (P?) ‘P, ):.>.,

U.g Ug (Pg) ;-g (Pg. rs)

!Yg Dg (Pg) o t:-nF g’ rc )

dependentvariable

TABLE 2.

Three-Dimensional Example, Gas Pressures at 1500 Days

I-

J Layer I&top)

I 1 ? 3 4 51 3219.3 +213~.7 3326.1 3347*3 3357*5? 3269.7 3316*93

333$1.63326.1

3355.9.333P.6

3365*33354.9

3347.33369.9 3379.1

6 3355.9 3370.0 3385.15

3798.53357.5 3356.3 3379.1 3s98.5 3427*’?

Layer 2

1 ? 3 6 51 3%2H.9 3294,~ 3330.5 3351,7 3361.~.? 3296.1 3319.3 3343.0 3360.3 336~.R

333n.53351.7336109

1

3236.1

3298.53334.933%6.23366*4

3343.0 335+.33360.3 3374.4336’3.8 33n3*7

Layer 3

? 3?298.5 3334.93323.7 3347.53347.5 3363.83366.8 3378.93376.3 33!18.Z

337h.43389.63402.2

4

3356.2

3364.83378.93394.23406. q

3383*73402,23b28. Q

53366.4

3374933388.234015.93431*4

Layer 4

1 2 3 4 51 3243.1 -+303.1 3339.4 3360.7?

3371*O33(J3. I 332R.2

2335?.0

3339.43369.4 3378.8

33s7.0 3368.4 3383.4 3392.83300.7 3369.4 3343.4 3398.8

;3412.1

3371. r7 337Q.8 3392.7 3412.1 3438. (I

Layer 5 (bottom)

1 ? 3 .1

53259.5 3317,4 3353.7 3375. . 33R5.2

? 3317.4 334?.4 3366.2 3383.63

3392,93253.7 3366.2 338?.5

4 3375*O3397*3 3606.6

3383.65 33e5*?

33’?7.3 3412.4 3424.7339?,9 3406.6 3424.7 344483

OmoooQt-ooo

UlmlmoooCooco

. ..**000

m-mooF-- f-N00

U-!lwm*ooCclaoo. . . . .

00

meutvo

a,aaoo4mmmo0

00000*..**

0

.erloob.o -loo

U!mml-noo00c’oo

● . . . .00

~. . . . .Li

!-m Coo*mm-lo

li7NWNN000000● .*.*

0

. ...*

. . . . .

0000000000

InoooooOoDOO–*. .

22000

. . .O“: 000

+-..4. .-H .+-. #-..-Oooco

** UC?--- Cecoc00000 00000 OD WON 00000

lnooooo In Ooooo U! OooooA.-4 . . .

Lflcaoumm.W .+..

U)ooooo&. -t.-

.* ***+dmfnm COoc=a

. . . . . .. **. . ..00 . . . . .

d. . . .

A“.-l. .Ooooc

.iCcoac.-I A.-.

.* *.*

.+.+.-l..00000

3COOC0.+ . . . .

.* ***

M. A. .-4 g~luu$ 00000GOOOO 00000

d.=o=oo aOODDm Ul=ooco--t . . . *.t.s r!@- Ccooc

. . . . . . ...0 . . . . ..-! -.4..

.

.-1 -..

l-v++.+- tv-l .-4.- W. .-f-. O.+r-+c. n Ccooc

3CJ=OC Ooo, co O=JG=O Ai mooa 00000dc Gcc O escocca -C OOO.3 -In ccc-c.+ Co Coo

. ..4*dd dd #c-t. “-.-t. ..* **.

.*e*4e 00000● 9... .. 0.. ● . . . . . . . .

&* Add

TABT.’F5

Three-Dimensional Example, r~ Liquid Content at Uoo Days

T-

J Layer 1 (top)

. 3 ‘4 5

1 .1:47 .1<18 .1354 ,1375 .1385

? ,131n ●1343 .13~7 ,1384 .1393

3 .1 354 .1347 .]3QJ .1398 .13f+5

4 .] 375 .13~4 ●139~ ●1230 .0519=. .] 3~5 .1.?.93 .1344 ●0519 .0001

Layer 2

1 ? ?, 4 5

1 .1? ’37 .13?2 .l~%H ,13H0 ,1390

7 .1327 .1747 ● 1371 .13U8 .13993 •135~ .i 771 ~13R7 .1402 ●1612

4 .]3’30 .13JI13 .]402 .14]8 .0765c! .]3~o .l?QH c14]2 .07b5 .0026

1hycr 1

? 3 4 5

1 .]264 .13?7 ,1363 .1384 .]394

? 913.27 .1352 .1375 ● 1393-4 ●1363 913?5

.1402.13’32 .i 407 ●1416

~ ●]384 * 1393 .1407 .16?.2 .11905 ● ]394 .1402 .1416 01189 ,0336

L_iyer i

1 ? 3 4 51 .1271 .13?1 .1367 .13 fl’4 .1399

.? .1331 .1356 e13H~ ,1397 .~407

3 .1367 .l~no .1396 .1411 .] 421

4 ,l~~o .1397 .1611 91-.27 .146ty

5 .1399 .14~7 .1421 .1440 .1279

I.aycr 5 (bottom)

1 ? 3 4 5

1 0.0000 9.0000 0.0000 0.0000 0.00002 0.0000 U.000o 0.0000 0,0000 0,00003 0.0000 0,0000 o.oono 0,0000 0,00004 0.00G0 U.oono 0.0000

50.0000

0.0000(?.0000

O.onoo 0.7JOO0 0.0000 0.0000

TABLE ~ TABLE 8

Compositions of the Liquid Hydrocarbfm andDry Gas Pseudo-Components

—.

MoLe Fractions

Component Liquid Dry Gas

co* 0.0074

N2 .0060

c1 .8833

c= 0.1033

c~ 0.3324

i-C ~ .0550

n-C .12314

i-c = .0499

n-C .05565

c= .0670

c 0.31707+

Liquid Molecular Weight = 80.47

Liquid Density at 14.65 psia and 60° F = 40.86 lb. /ea. ft.

Dry Gas MoIecular Weight = 17.72

D:y Gas Density at 14.65 psia and 60”F = 46.7 lb.fMCF

Properties of the Equilibrium Gas During Constant-VolumeOepIetion at 175°F

pressure (psig)

Component 3692 3200 2600 1900 1200 700

mole fractions

co ~ .0062 .0059 .0061 .,0061 .0063 .0065

ii .0050 .00532

.0055 .0059 .0054 .0052

~1 .7L33 .7602 .7835 .7978 .8004 .7883

C* .0871 .0875 .0862 .0872 .0898 .0939

C3.0527 .0526 .0L98 .0L95 .0503 .0543

i-C4 .0087 .0080 .0076 .0076 .007L .0081

n-C4 .0195 .0187 .0166 .0153 .0158 .0179

i-C .00795

.0072 .0065 .0055 .0053 .0057

n-C .0088 .00865

.0072 .0060 .0053 .0060

c .0106 .01026

.0082 .0062 .0052 .0054

c7+ .0502 .0360 .0228 .0129 .0088 .0087

MoL. wt.of C7. 133 126 118 109 104 104

S. G. of CT+ .775 .768 .760 .751 .745 .745

Compress-ibility Z .8U .797 .791 .326 .880 .927

L

II

1.0

0.8

0.2

0,0

0.0 0.2 0,4 0.6 (J.8 1.0

Watm s,lt,tratio”

FIGURE 3

Ex.mplc W.t., -Oil Rvlativc Per-m ~billty OataPressure (o.sia)

FIGURE 1

Tx

z

K= 1

.2

J

5

2

3

4

5

_ /-—— .— ——

——- .— -.—,II,LL.]1 ,;d:.-wat,r

.— ---- —---- ---- ct!,)rsct

~ j~~o’ —_—— — .—— +

* Produc t idn

Tl>rcc- lli,.ensic.nal ExmnPlc

‘\

1,0 0.8 0.6 0.4 0.2 0.0Gas s.3turaclon

FIGURE U

“Exmplc Gas. Oi! RS%I. CIVC Permeability Data

35

/=

/

,>00 10[)0 1$[!11It,.c (,1,,.. )

t’l(s, Rt 5

Sx.mplr, Lsqul<! I’roc!ucti’,n IL, tr v.. Ti, w

o 1000 2000 3000 4000 5000

Pressure (psi)

FIGORE 6

Hydrocarbon Liquid Density vs. Pressure for Example C.andensate. Cas

Equil.

Dry

1000 2000 3000 4000 5000

Pressure (psi)

FIGURE ‘1

Dry and Equilibrium Gas Oensities vs. Pressurefor Example Cmdensat?. Gas

0.01

O.oh

0.03

Fquil

ory

o 1000 2000 3000 Looo 5!300

, , t..- . . . . .—u lUUV Zuuo 3000 hooo 5000

pressure (psi)

FIGURE 9

Equilibr ia.* Liquid Content vs. Preswre forExample Ccmdemate-Cas

20

5

Prrswrr (psi)

FIGURE 8

Dry and Equll, t,rir, m Gss Vlsr.. itlc.s VS.. Prvss. rc

for E.mnple Cmdm$ate -[,,.

— I.%!vr.,t,v, -wzv,rcd

● L.ll. !: I.!’,.!

●

>.

●

I ,1 Ouo 2000 3000 5000

Prcss:rc (psia)

Retrograde L,quid Vol..re vs. Prc%. ur. for

Exmple Cmd.msa:c.-Cas