Thermodynamics 3

Petroleum Engineering Applications of Phase Equilibria , Colombia, Summer 2000-

Author: Dr. Maria Barrufet - Summer, 2000 /163

Instructional Objectives After seeing this module the student should be able to:

• Evaluate volume roots from a cubic equation of state when two-phases coexist.

• Derive and evaluate fugacity coefficients from cubic EOS.

• Evaluate phase boundaries (dew and bubble points) using EOS.

• Evaluate flash separations using an EOS.

• Understand the production mechanism for a gas condensate or a volatile oil (Constant Volume Depletion, CVD).

• Interpret the results of a simulated CVD test from a commercial simulator.

• Determine oil and gas in place and recoveries using CVD compositional data.

• Determine oil and gas in place from recombination.

• Understand vertical compositional variation due to gravity.

• Understand the effects of a porous media upon phase equilibria.

Petroleum Engineering Applications of Phase Equilibria High Pressure Phase Equilibria Applications (Reservoir). Equations of State Models (EOS). Cubic EOS. Root Selection. Evaluation of Fugacity Coefficients from Equations of State. Evaluation of Phase Boundaries (Dew and Bubble Points) and Flash Equilibrium with EOS. Tuning of Equations of State (EOS). Miscible Gas Injection. Swelling Tests. Constant Volume Depletion Studies (Gas Condensates and Volatile Oils). Determination of Oil and Gas in Place by Recombination. Areal and Vertical Composition Variation in Hydrocarbon Reservoirs. Estimation of Compositional Gradients due to Gravity. Phase Equilibria in Porous Media. The Effect of Curved Surfaces. Suggested Reading: EL, WM, MAB



Equations of State (EOS) Single Component Systems Equations of State (EOS) are mathematical relations between pressure (P) temperature (T), and molar volume (V). For a pure component in a single phase (i.e. gas or liquid) given any pair of these variables the third can be evaluated. If the component exhibits phase equilibrium and additional constraining equation is placed and only one variable, either P or T are needed to specify the STATE of the system. For VLE the constraining equation is the equality of the Gibb’s energy of the GAS and LIQUID phases. A model is required to evaluate this ENERGY and a CUBIC EOS is one of the many different models available. The ideal gas EOS was the first mathematical expression used to describe the PVT behavior of gases but it’s use is limited to low pressures (near atmospheric) additionally, since it only provides a single volume for a given pressure and temperature, it cannot describe phase transitions

Multicomponent Systems For multicomponent mixtures in addition to these variables (P, T & V) , the overall molar composition and a set of mixing rules are needed.

Modern EOS’s are versatile tools for reservoir engineering applications. They can be used for all states (gas, liquid, and solid), and they can describe phase transition conditions and properties of the coexisting phases. Some of the EOS uses include: evaluation of gas injection processes (miscible and immiscible), evaluation of properties of a reservoir oil (liquid) coexisting with a gas cap (gas), simulation of volatile and gas condensate production through constant volume depletion evaluations, recombination tests using separator oil and gas streams, evaluation of paraffin deposition in the wellbore, etc.

Types of EOS There are many families of EOS’, suitable for different purposes and substances i.e. hydrocarbons, electrolytes, solids, gas-liquid-solid equilibria, etc. In petroleum engineering the most commonly used EOS are cubic polynomials in volume. More



complicated EOS’s exist but these are not practical (unaffordable in terms of computer time) for compositional simulations. Cubic equations are EXPLICIT in pressure and can be written as the sum of a term indicating repulsion forces and a term indicating attraction forces

attrrep PPP += (1)

One of the most used EOS’ in petroleum engineering is the Peng-Robinson EOS (1975).

)()( bVbbVVa

bVRTP

−++α−

−= (2)

The critical point conditions are that the first and second derivatives of pressure with respect to molar volume evaluated at the critical point are equal to zero.

0

0

2

2

=

∂∂

=

∂∂

c

c

T

T

VP

VP

(3)

Solving these two equations simultaneously provides different values for the parameters a and b depending upon the EOS used. For the Peng-Robinson EOS these are,

c

ca P

TRa22

Ω= (4)

and

c

cb P

RTb Ω= (5)



Where

07780.045724.0

=Ω=Ω

b

a (6)

and

( )( )211 rTm −+=α (7)

with

22699.054226.137464.0 ω−ω+=m (8)

The Peng Robinson EOS is a three-parameter corresponding states model.

In practical petroleum engineering applications we are concerned with the phase behavior of the hydrocarbon liquid mixture, which at a specified pressure and temperature, is in equilibrium with a hydrocarbon gas mixture at the same pressure and temperature (the production path sketched in previous lectures).

Equation (2) can be expressed as a cubic polynomial in (V) or (Z). In terms of compressibility factor this becomes

0231 32223 =−−−−−+−+ )BBAB(Z)BBA(Z)B(Z (9)

with



( )

RTbPB

RTPaA

=

α= 2

(10)

Cubic EOS’s are the simplest polynomials that can provide an adequate description of both: liquid and gas properties. They can describe the state of pure fluids and mixtures (single or multiphase) and their properties.

When working with mixtures the same expressions apply except that (aα) and (b) are evaluated for a mixture using a set of mixing rules. The most commonly used mixing rules (MR) are:

Quadratic MR for a

( ) ( ) ( )∑∑= =

−αα=αNc

iji

.Nc

jjijijim kaaxxa

1

50

1

1 (11)

Linear MR for b

∑=

=Nc

iiim bxb

1

(12)

where the kij’s are called interaction parameters and by definition

0=

=

ii

jiij

kkk

(13)

These MR’s are called the van der Waals mixing rules. Many more MR’s have been proposed in the literature, but these are the most popular.

Example - For a three-component mixture (Nc = 3) expand Eqs. (11) and (12) to

evaluate the attraction (a) and the repulsion constant (b). This gives:



( ) ( ) ( )( ) ( ) ( )

( )3323

222211

213

5.0313131

235.0

323232125.0

212121

)1(2

)1(2)1(2

1

α+

α+α+−αα+

−αα+−αα=α

ax

axaxkaaxx

kaaxxkaaxxa m

(14)

bxbxbxbm 332211 ++= (15)

A system may not exist as a SINGLE PHASE at the specified P, T, and overall composition zi. Energy requirements may cause the system to split into two (or more) phases such that the Gibb’s ENERGY of the system (i.e VAPOR + LIQUID) is at a MINIMUM. These two phases will have different compositions from the original system (yi & xi).

To find these gas and liquid compositions such that the Gibb’s energy is at a MINIMUM you need to solve a set of non-linear equations simultaneously. The overall mixture composition will be used to evaluate a unique set of liquid and gas compositions such that material balance is satisfied. The overall composition zi is normally provided to you, this composition may coincide with the composition of a phase boundary , or you may need to evaluate gas and liquid compositions (yi, xi, i = 1, 2…Nc) which will normally be evaluated from

Therefore the constants a and b can be evaluated using

• Overall compositions zi i = 1, 2…Nc

• Liquid compositions xi i = 1, 2…Nc

• Vapor compositions yi i = 1, 2…Nc

The cubic expression for a mixture is then evaluated using



( )( )

RTPbB

RTPaA

mm

mm

=

α= 2

(16)

In most of the current literature the subscript “m” is omitted for simplicity, and whether it refers to the mixture or to a single component is understood from the context.

Analytical Solution of Cubic Equations The cubic EOS can be arranged into a polynomial as Eq. (9) and be solved analytically as follows. (Solution taken from standard manuals on mathematical formulas)

Let’s write the polynomial in the following way

032

21

3 =+++ axaxax (17)

Note: “x” could be either the molar volume, or the density, or the z-factor, when the equation is expressed in terms of the z factor as in Eq. (9) the coefficients a1 to a3 are:

)BBAB(a)BBA(a

)B(a

323

22

1

23

1

−−−=

−−=

−=

(18)

Analytical solution of a cubic polynomial The procedure to evaluate the roots of a cubic equation analytically follows. This has been extracted from a mathematical handbook (Spiegel, Schaum series).



Let

3 23

3 23

31321

212

542279

93

RQRT

RQRS

aaaaR

aaQ

+−=

++=

−−=

−=

(19)

The solutions are,

( ) ( )

( ) ( )TSiaTSx

TSiaTSx

aTSx

−−−+−=

−+−+−=

−+=

321

31

21

321

31

21

31

13

12

11

(20)

If a1, a2 and a3 are real and if D = Q3 + R2 is the discriminant, then

1. One root is real and two complex conjugate if D > 0;

2. All roots are real and at least two are equal if D = 0;

3. All roots are real and unequal if D < 0.

If D < 0, computation is simplified by use of trigonometry, as follows,

13

12

11

31240

31cos2

31120

31cos2

31

31cos2

0 If

aQx

aQx

aQx

D

−

+θ−=

−

+θ−=

−

θ−=

⇒<

o

o (21)



where

3cos

QR

−=θ (22)

3321

2133221

1321

axxxaxxxxxx

axxx

−==++

−=++ (23)

where x1, x2 and x3 are the three roots.

The range of solutions that are used for the engineer are those for positive volumes and pressures, we are not concerned about imaginary numbers.

Basically, from the general shape of the polynomial we are interested in the first quadrant of Figure 1.



Figure 1 - Solutions of a cubic polynomial.

The following web site contains Fortran codes to solve the roots of polynomials up to fifth degree.

http://www.uni-koeln.de/math-nat-fak/phchem/deiters/quartic/quartic.html

http://www.uni-koeln.de/math-nat-fak/phchem/deiters/quartic/quartic.html



Figure 2 - Web site to download Fortran source codes to solve polynomials up to fifth degree.

Phase equilibrium for a single component at a given temperature can be graphically determined by selecting the saturation pressure such that the areas above and below the loop are equal, these are known as the van der Waals loops (see Figure 3 Maxwell Equal area rule).



-100

0

100

200

300

400

500

600

700

2 4 6 8 10

12

14

A1

A2

Pres

sure

Molar Volume

Tc

T2

T1P1

v

L

2 - Phases

CP

V

LV

1

2

34

7 6

5

0>

∂∂

TV~P

-100

0

100

200

300

400

500

600

700

2 4 6 8 10

12

14

A1

A2

Pres

sure

-100

0

100

200

300

400

500

600

700

2 4 6 8 10

12

14

A1

A2

Pres

sure

Molar Volume

Tc

T2

T1P1

v

L

2 - Phases

CP

V

LV

1

2

34

7 6

5

0>

∂∂

TV~P

Figure 3 - van der Waals loops showing the Maxwell Equal Area rule (A1 = A2).

Although the EOS does not provide viscosities (a transport property), it gives phase compositions that are used in the evaluation of viscosities.

In practical petroleum engineering applications we are concerned with the phase behavior of the hydrocarbon liquid mixture, which at a specified pressure and temperature, is in equilibrium with a hydrocarbon gas mixture at the same pressure and temperature (production path).

Along this production path we will need to evaluate:

(1) Bubble point pressure (Pb) at given reservoir T and overall zi.

(2) Properties of gas & liquid, below (Pb), through equilibrium computations in the two-phase region (flash computations).



For a retrograde fluid, along the production path we will need to evaluate:

(1) Dew point pressure (Pd) at given reservoir T and overall zi.

(2) Properties of gas & liquid, below (Pd), through equilibrium computations in the two-phase region (flash computations).

To accomplish this we make use of a set of criteria for thermodynamic equilibrium. To solve the phase equilibrium problem requires time consuming iterations, often the bottle-neck in all reservoir compositional simulation package.

To evaluate phase equilibrium requires the simultaneous solution of a set of non-linear equations.

If two (or more) phases exist in equilibrium, then the equal fugacity criteria must be satisfied.

The number of equilibrium equations to be solved is equal to the number of components that make up the mixture times the number of phases minus one. (i.e. 10 components and three-phases (gas / liquid 1/liquid 2 ! 20 equations).

! Physically, the difference of the fugacities of one component in one phase with respect to another phase gives a measure of the potential for transfer of that component between these phases. Equal fugacities of a component in the two (or more phases) results in zero net mass transfer across the phases, or equilibrium."

At equilibrium pressure, temperature and compositions within the phases remain constant.

Two-phase VLE For two-phase vapor-liquid-equilibrium VLE these equations are expressed as:

li

vi ff ˆˆ = (i = 1, 2, 3, …Nc) (24)



When an EOS is used to evaluate the fugacities an alternative expression is using the

fugacity coefficients ( )li

vi

ˆ,ˆ φφ . The relationship between fugacity and fugacity

coefficients is

Pˆxf lii

li φ= (25)

lii

vii xy φ=φ ˆˆ (26)

Fugacity coefficients will be evaluated from an EOS as will be seen later.

Typical phase equilibrium problems to solve are indicated in Table 1.

Given Variables

(independent)

Unknown Variables

(dependent)

Problem Type Example Application

P, zi = xi T, yi Bubble Point Distillation

T, zi = xi P,yi Bubble Point Gas injection,

Production

P, zi = yi T,xi Dew Point Separations

T, zi = yi P,yi Dew Point Gas Condensates,

Production

P, T, zi xi, yi, fv Flash Production

Separation

Table 1 - Dependent and independent variables used in typical phase equilibrium problems.



We will present the algorithms for the: DEW – BUBBLE & FLASH equilibrium calculations which are general regardless of the EOS and mixing rule used. We will solve phase equilibrium for binary mixtures and we will compare our results from those evaluated with a commercial package.

The first objective is to understand the type of calculations involved, next we will know how to use and interpret the results from software packages that will do the job for us.

The phase equilibria equations are expressed in terms of the equilibrium ratios, or more commonly called the “K-values”. The K-value of component “i” is defined as:

vi

li

i

ii ˆ

ˆ

xyK

φφ== (27)

Dew Point Calculations

Equilibrium is always stated as:

PˆyPˆx vii

lii φ=φ (i = 1, 2, 3, …Nc) (28)

with the following material balance constrains

11

=∑=

Nc

iix (29)

11

=∑=

Nc

iiy (30)

11

=∑=

Nc

iiz (31)

At the dew-point the overall composition is identical to the vapor (or gas composition). Therefore making use of the K-value definition:



vii

lii

ˆzˆx φ=φ

iii zKx = (i = 1, 2, 3 ,…Nc) (32)

Rearranging Equation (32) and adding over all components we obtain the Dew-Point objective function. For a Dew-Point equilibrium calculation the objective is to find a unique set of K-values that satisfies the following equation,

011

=−∑=

Nc

i i

i

Kz (33)

Bubble Point Equilibrium Calculations

For a Bubble-point equilibrium calculation, the objective function is derived following the same reasoning as:

011

=−∑=

Nc

iiiKz (34)

The objective is to find a unique set of K-values that satisfies the above equation. In searching for these K-values, we will have to find the correct equilibrium compositions and bubble point pressure ( when the temperature is specified), or the equilibrium compositions and the bubble point temperature (when the pressure is specified).

Flash Equilibrium Calculations

Flash calculations are the work-horse of any compositional reservoir simulation package. In a flash separation calculation the objective is to find the fractional amount of vapor in a Vapor-Liquid mixture at a specified temperature and pressure such that the following equation is satisfied.

011

11

=−+

−∑=

cN

i iv

ii

)K(f)K(z (35)



This is one of the alternative mathematical expressions that can be used and is chosen because of its good behavior in a Newton-Raphson algorithm. This is also known as the Rachford-Rice function and it is derived from:

∑∑==

=−Nc

ii

Nc

ii xy

11

0 (36)

Notice that the first derivative of Eq. (35) is monotonic (always has the same sign and it is negative). Figure 4 shows three different flash functions, notice that the three have the same solution but only the Rachford Rice function is monotonic and therefore it is well behaved from a numerical stand point.

Numerical Behavior of Flash Functions

-6.00

-4.00

-2.00

0.00

2.00

4.00

6.00

0.00 0.20 0.40 0.60 0.80 1.00

Molar Fraction of vapor (fv)

F(fv

)

Sum XiSum YiRachford Rice

Figure 4 - Numerical behavior of flash functions.



The two-phase equilibria criteria is that both functions (dew and bubble) must be greater than one.

11

>∑=

Nc

iiiKz (37)

11

>∑=

Nc

i i

i

Kz (38)

The equilibrium GAS compositions are evaluated as:

)K(fKzy

iv

iii 11 −+= (39)

with the liquid compositions evaluated as,

i

ii K

yx = (40)

As an example here zi may indicate the composition of the reservoir oil fluid above the Bubble-Point (point t1). Later in production the pressure in the reservoir drops and this initial single-phase fluid separates into two phases. Both, gas and liquid, move within the reservoir and up to the surface according to their relative permeabilities. As fluids are produced the composition of the reservoir fluids keep changing and consequently properties such as: density, viscosity, solution gas-oil-ratios, etc. Figure 5 illustrates the compositional changes that may occur due to production or injection.



Temperature

t1

Composition Changes Due to Production

and Gas InjectionPr

essu

re

t3

t2

GasInjection

Production

Temperature

t1

Composition Changes Due to Production

and Gas InjectionPr

essu

re

t3

t2

GasInjection

Production

Figure 5 - A sequence of phase envelopes resulting from compositional changes during production or injection.

The following figures indicate the mathematical behavior of the DEW, BUBBLE, and FLASH functions when evaluated at different locations of the phase envelopes.

Note that this behavior is for the converged K-values.

The following figures illustrate the behavior of the DEW and BUBBLE functions projected on a temperature composition diagram at a fixed pressure and on pressure-composition diagram at a fixed temperature.

The behavior of these functions for multicomponents is the same.



TD

T B

Ta

z1 y1x1 10

11

>∑=

Nc

i i

i

Kz

11

<∑=

Nc

i i

i

Kz

11

=∑=

Nc

i i

i

Kz P = Pa

Tem

pera

ture

Figure 6 - Behavior of the Dew Point function at the boundaries and in the two phase region in a TX diagram.

If the objective is to find the DEW POINT TEMPERATURE at a fixed pressure (Pa) and overall composition.

If 11

<∑=

Nc

i i

i

Kz ! Superheated fluid ! lower T (41)

One way of updating the temperature is

∑=

+ ×=Nc

i i

ikk

KzTT

1

1 (42)

If 11

>∑=

Nc

i i

i

Kz ! Subcooled fluid ! increase T (43)



The same scheme works for updating the temperature that is

∑=

+ ×=Nc

i i

ikk

KzTT

1

1 (44)

The schemes indicated are secure but slow, there are other well documented numerical accelerating schemes, but numerical methods is not the main objective of this course.

! At the dew point 11

=∑=

Nc

i i

i

Kz (45)

11

>∑=

Nc

i i

iKz

11

=∑=

Nc

i i

iKz

11

<∑=

Nc

i i

iKzPr

essu

re

PD

PB

Pa

z1 y1x1 10

T= Ta

-

Figure 7 - Behavior of the Dew point function in a PX diagram.

If the objective is to find the DEW POINT PRESSURE at a fixed temperature (Ta) and overall composition.

If 11

<∑=

Nc

i i

i

Kz ! Under pressurized fluid ! increase P (46)



∑=

+ ×=Nc

i i

ikk

KzPP

1

1 (47)

If 11

>∑=

Nc

i i

i

Kz ! Over pressurized fluid ! lower P (48)

∑=

+ ×=Nc

i i

ikk

KzPP

1

1 (49)

! At the dew point 11

=∑=

Nc

i i

i

Kz (50)

To evaluate a bubble-point pressure or bubble-point temperature we use the same functions and correction schemes.

These correction schemes can be accelerated by using a larger correction step, but the directionality is the same provided that the fluid does not exhibit a double dew-point (Gas Condensates).

For the upper DEW POINT of GAS CONDENSATES we must use other numerical approaches. When the initial guess is in the two-phase region the approach just explained will converge to the lower dew point.

For a BUBBLE-POINT calculation



11

<∑=

Nc

iiiKz

11

>∑=

Nc

iiiKz

11

=∑=

Nc

iiiKz

Pres

sure

PD

PB

Pa

z1 y1x1 10

T= Ta

Figure 8 - Behavior of the Bubble Point function in a PX diagram.

If the objective is to find the BUBBLE POINT PRESSURE at a fixed temperature (Ta) and overall composition, then

If 11

<∑=

Nc

iiiKz ! Over pressurized fluid ! lower P (51)

∑=

×=+

Nc

iii

k KzPP k

1

1 (52)

If 11

>∑=

Nc

iiiKz ! Under pressurized fluid ! increase P (53)

∑=

×=+

Nc

iii

k KzPP k

1

1 (54)

! At the bubble point 11

=∑=

Nc

iiiKz (55)



To ensure the existence of two-phases both functions (DEW & BUBBLE) must be:

11

>∑=

Nc

i i

i

Kz 1

1>∑

=

Nc

iiiKz (56)

The following graphs complete the illustration of the behavior of these functions.

11

<∑=

Nc

iiiKz

11

>∑=

Nc

iiiKz

11

=∑=

Nc

iiiKz

Pres

sure

PD

PB

Pa

z1 y1x1 10

T= Ta

Tem

pera

ture

TD

TB

Ta

z1y1x1 10

11

<∑=

Nc

iiiKz 1

1=∑

=

Nc

iiiKz

11

>∑=

Nc

iiiKz

P= Pa

Figure 9 - Behavior of the Dew and Bubble functions on a TX and a PX diagram.



.

P

b

Pd

Pres

sure

Temperature

CP

Bubble-Curve

Dew-C

urve

1

1

>∑=

Nc

i i

iKz

1

1

<∑=

Nc

i i

iKz

11

=∑=

Nc

i i

iKz

11

=∑=

Nc

iiiKz

11

>∑=

Nc

iiiKz

11

<∑=

Nc

iiiKz

2-phases

Tr

A

B

Figure 10 - Behavior of the Dew and Bubble functions on a PT diagram.

As pressure keeps dropping gas and liquid compositions are continuously changing in the reservoir cells. Describing these changes is an essential part of reservoir engineering that requires the solution of flash equations million of times (a) at different locations of the reservoir for a fixed time, and (b) at different times and the same location within the reservoir. The material balance equation plus the diffusivity equation pressure (Darcy’s equation) are solved simultaneously.

Evaluation of Fugacity Coefficients and K-values from an Equation of State (EOS) The general expression to evaluate the fugacity coefficient for component “i” in any phase (vapor, liquid-I, liquid-II, etc.) from any model is given by the following expression.



fixedT

P

ivi dP

PRTVˆlnRT

=

−=φ ∫

0

(57)

This is not a suitable expression to evaluate fugacity coefficients using a pressure explicit equation of state. Therefore, the goal is to make a transformation of variables to transform the integral and use volume as the dependent variable.

We need to follow some algebraic steps involving rules of differentiation and the thermodynamic identities seen in Module 2.

The partial molar volume of component “i” is defined as:

in,P,T

v

vv

vj

i

t

i nV

V≠

∂∂

= (58)

for simplicity, many books omit the superscripts that indicate the phase.

Following the (-1) rule of thermodynamics and rules of differentiation

in,T

vn,T

v

in,P,T

v

v

vj

ivt

t

vj

i

t

nP

PV

nV

≠≠

∂∂×

∂∂

−=

∂∂

(59)

vi

n,T

v

n,T

v

in,P,T

v

v

in,T

v VVP

VP

nV

nP

vt

tvt

tvj

i

t

vj

i

×

∂∂−=

∂∂×

∂∂

−=

∂∂

≠≠

(60)

and

vt

t

vj

i n,T

v

in,T

vv

i PV

nPV

∂∂

×

∂∂−=

≠

(61)

(Note that first order partial derivatives can be inverted)



Replacing the partial molar volume into the general expression for the fugacity coefficient

fixedT

P

n,T

v

in,T

vvi dP

PRT

PV

nPˆlnRT

vt

t

vj

i=

≠

+

∂∂

×

∂∂−=φ ∫

0

(62)

dPP

RTdPP

VnPˆlnRT

P

n,T

v

in,T

v

Pvi

vt

t

vj

i

∫∫ −

∂∂

×

∂∂−=φ

≠ 00

(63)

dPP

RTdVnPˆlnRT

Pv

t

in,T

v

Vvi

vj

i

vt

∫∫ −

∂∂−=φ

≠∞ 0

(64)

Note that we have changed the limits of the first integral accordingly.

The second integral in the right hand side can be expanded using the following identity:

VZRTP = at constant T

)V/Z(RT)V/Z(RTd

PdP

t

t= (65)

or

tVlndZlndPlnd −= (66)

t

t

VdV

ZdZ

PdP −= (67)

Replacing Eqn.(67) into Eqn.(64) the final expression to evaluate the fugacity coefficient using an EOS is.

vv

tv

in,T

v

Vvi ZlnRTdV

VRT

nPˆlnRT

tvj

i

vt

−

−

∂∂−=φ

≠∞∫ (68)



Alternative expressions can be obtained using the compressibility factor as follows

v

V

nnVTi

vi Z

VdV

nnZv

j

ln1ˆln,,

−

−

∂∂−=φ ∫

∞

(69)

ZVdV

nnZl

j

V

nnVTi

li ln1ˆln

,,

−

−

∂∂−=φ ∫

∞

(70)

When using cubic EOS, and linear, or quadratic mixing rules, this expression can be evaluated analytically.

Equilibrium calculations are expressed in terms of K-values. We will illustrate how to evaluate these K-values from an EOS using the definitions outlined previously.

Physically, the difference of fugacities of one component in one phase with respect to another phase gives a measure of the potential for transfer of the component between the phases. Equal fugacities of a component in the two (or more phases) results in zero net mass transfer across the phases, or equilibrium. Pressure, temperature and compositions within the phases remain constant.

For two-phase vapor-liquid-equilibrium VLE these equations are expressed as:

li

vi ff = (i = 1, 2, 3, …Nc) (71)

When an EOS is used to evaluate the fugacities an alternative expression is using the

fugacity coefficients ( )li

vi φφ ˆ,ˆ

lii


As seen previously the K-value of component “i” is defined as:



vi

li

i

ii ˆ

ˆ

xyK

φφ== (73)

vi

liv

i

li

iˆlnˆlnˆ

ˆlnKln φ−φ=

φφ= (74)

dPP

RTdVnPdP

PRTdV

nPKlnRT

P

t

in,T

VP

t

in,T

V

i

ji

vt

ji

lt

∫∫∫∫ −

∂∂−−

∂∂−=

≠∞≠∞ 00

(75)

t

in,T

V

Vi dV

nPKlnRT

ji

vt

lt ≠

∂∂= ∫ (76)

Note that to evaluate the K-values we need to evaluate the derivative within the integral and at the limits we need to evaluate the corresponding molar volumes of gas and liquid. To do so we will require different compositions for the liquid and the gas which are evaluated iteratively.

The following example shows the steps to evaluate the fugacity coefficient from a cubic EOS using the expressions just presented.

Example – Evaluation of the fugacity coefficient for van der Waals EOS

We will illustrate the evaluation of a K-value for species “i” in a multicomponent mixture using the Van der Waals EOS. The idea can be easily generalized to other EOS.

Relations to keep in mind:

Total number of moles: lv nnn +=

Number of liquid moles - binary: lll nnn 21 +=



- multicomponent: li

Nc

il nn ∑

=

=1

Number of vapor moles - binary: vvv nnn 21 +=

- multicomponent: vi

Nc

iv nn ∑

=

=1

Mole fraction of component “i” in liquid: l

li

i nnx =

Mole fraction of component “i” in vapor: v

vi

i nny =

Overall mole fraction n

nnz

vi

li

i+

=

Partial derivatives:

1=

∂∂

≠in

l

l

lj

inn , 1=

∂∂

≠ in

v

v

vj

inn

0=

∂∂

≠in

l

lj

vj

inn

, 0=

∂∂

≠in

v

vj

vj

inn

Recall that the volume used in the EOS is a molar volume:

nVV t=

The mixture parameters used in the EOS if linear mixing rules are used are:

Liquid



∑∑==

==Nc

iil

lNc

iii

l ann

axa i

11

∑∑==

==Nc

iil

lNc

iii

l bnn

bxb i

11

Vapor

∑ ∑= =

==Nc

i

Nc

iiv

v

iiv a

nn

aya i

1 1

∑∑==

==Nc

iiv

vNc

iii

v bnn

byb i

11

The van der Waals EOS is:

2Va

bVRTP −−

= = 2

2

tt Van

bnVRTn −−

To evaluate the K-values we need to evaluate the following partial derivative

( ) ( )

∂∂+−

∂∂+

−+

−=

∂∂

≠≠≠ initinittin,V,Tijjj

nana

Vn

nbnb

bnVRTn

bnVRT

nP

222

These partial derivatives are

n

aana i

inij

−=

∂∂

≠

n

bbnb i

inij

−=

∂∂

≠

Replacing these expressions into

( ) ( )

−+−

−+

−+

−=

∂∂

≠ naana

Vn

nbbnb

bnVRTn

bnVRT

nP i

t

i

ttin,V,Tij

222



Now put the expression back using the molar volume nVV t= and replace

ndVdVt =

( ) ( ) ndVn

aaVn

nVnna

nbVRTnb

nbVRTKlnRT i

tt

iV

Vi

v

l

−−−

−+

−= ∫ 22

2

22222

After cancellations and integration between the molar liquid and vapor volumes the final expression for the K-value of species “i” is

+−

++

−−

−−

−−= l

il

vi

v

llvvill

vv

i Vaa

Vaa

bVbVRTb

bVbVRTKRT 11lnln

The same algebraic procedure is used for all EOS. The reason that non-cubic EOS are not popular in Reservoir Engineering is because of their complexity without gaining much more accuracy and at the expense or much more CPU time.

Fugacity Coefficients from Cubic EOS The following expressions, taken from Stanley & Walas, give the results obtained for the fugacity coefficients using the most popular cubic EOS.

The K-value for component “i” is obtained just by the ratio of the liquid to the vapor fugacity coefficient of species “i”.



Van der Waals

( )

RTVaa

Vbz

bVb

byb

ayaVa

bVRTP

iii

ii

ii

21lnˆln

2

2

−

−−

−=φ

=

=

−−

=

∑∑

(77)

Note

For the liquid we have an analogous expression.

! The a and b constants are evaluated using the liquid compositions (xi’s)

! The molar volume for the liquid Vl is evaluated from the EOS at the same P, and T as for the gas, but using liquid compositions.

Equilibrium compositions are the unknown variables in equilibrium problem. This is an iterative process therefore volumes and fugacity coefficients are evaluated many times before they converge.

Redlich-Kwong

( )( )

( )

( ) [ ]

+

−+−−−=φ

+

−+

−−−=φ

=

=

+−

−=

∑∑

zB

AA

BB

BABzz

BB

Vbaa

bab

bRTVbzz

bb

byB

ayA

bVVTa

bVRTP

iiii

iii

i

ii

ii

1ln2ln1ˆln

1ln211ln1ˆln

)

5.1

2

(78)



Soave

( ) ( ) ( ) ( )

( )

( )( )

( ) ( )

( ) [ ] ( )

+

α

α−+−−−=φ

+

α

α−α+

−−−=φ

=

=

α=

=

α=α=

αα−=α

∑

∑

∑∑∑

zBay

aBB

BABzz

BB

Vbay

abb

bRTa

Vbzz

bb

RTPbB

RTbPB

RTPaA

byb

ayyak

aaka

jijj

iii

jijj

iii

ii

ii

ijji

ii

jjiiijij

1ln2ln1ˆln

1ln21ln1ˆln

0

1

2

(79)

When using cubic EOS, and linear, or quadratic mixing rules, this expression can be evaluated analytically.

Peng-Robinson

( ) ( ) ( ) ( )

( )

RTPbB

RTPbB

RTPaA

BzBzay

aBB

BABzz

BB

ii

vv

vv

vv

vv

jijj

vv

i

v

vvvv

v

ivi

=

=

α=

−−++

α

α−+−−−=φ ∑

2

)12()12(ln2

22ln1ˆln

(80)

You will have a homework were you will have to proof Eq. (80) (that is a good algebraic exercise!).



To evaluate the K-values we just need to ratio the liquid and gas fugacity coefficients.

Tuning of Equations of State (EOS) For two-phase vapor-liquid-equilibrium VLE, the equilibrium equations were expressed as:

lii


These fugacity coefficients are evaluated from an EOS, in this case we will chose the Redlich-Kwong EOS (RK).

The phase equilibria equations are expressed in terms of the equilibrium ratios, or more commonly called the “K-values”. The K-value of component “i” is defined as:

vi

li

i

ii ˆ

ˆ

xyK

φφ== (82)

Example - Tuning of an EOS to VLE data

Given the experimental VLE data on C3-CO2 reported in Table 2 we wish to determine kij in the Redlich-Kwong EOS, such that predicted values from the EOS are as close as possible to the experimental values.

These values chosen for prediction will be saturation pressures and either the liquid or the gas compositions. If the liquid compositions are selected as independent the type of calculation will be a bubble point and the vapor compositions will be evaluated. Likewise if the gas compositions are selected as independent variable the type of calculation will be a dew point and the liquid compositions will be evaluated. This choice is arbitrary, but it is normally



constrained by selecting as “independent” variables those for which one has the highest precision.

Even though this matching is done for just one set of isothermal data EOS have the power to be extrapolated with accuracy to other conditions!!

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 0.4676

13.61 0.1602 0.6036

17.01 0.2402 0.6864

20.41 0.3316 0.7431

23.81 0.4361 0.7876

27.22 0.5532 0.8309

30.62 0.6714 0.8688

34.02 0.7956 0.9102

37.43 0.9401 0.9702

Table 2 - Experimental VLE data for CO2/C3.

A graph of the data is



0

10

20

30

40Pr

essu

re (A

tm)

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

Figure 11 - Experimental PX diagram for the system Propane/CO2 at 4.4 oC.

Mathematically this minimization problem is expressed such that the following objective function is minimized.

( ) MinPPFM

j

cali

datai =−= ∑

=

2

1 (83)

Where;

M = number of data points (here is 10) data

iP = experimental saturation pressures

caliP = saturation pressures calculated from the RK EOS.



The Redlich-Kwong EOS is

)bV(VTa

bVRTP / +

−−

= 21 (84)

Step 1.

Evaluate pure component EOS constants (a & b) for all components. The expressions for these from the RK EOS are,

ci

.ci

i PTR.a

522

427480= and ci

cii P

RT.b 086640=

The critical properties for CO2 and C3 are tabulated and are,

Component Tc (K) Pc (Atm)

CO2 304.19 72.85

C3 Fill up Fill up

The gas constant to be used in these calculations has the following value and units

K molliter Atm.R 082050=

Therefore the attraction constants are:

Let (1) be CO2

211 COaa =



( )

××=

AtmK

K molliter Atm

....a

2.5. 2522

11 857219304082050427480

2

1.52

molK Atm l.a 75376311 =

Likewise for propane :

2

1.52

C molK Atm l.aa 48180223

==

The interaction constant of attraction is

( )12221112 1 kaaa −=

And the “repulsion constants” are:

××=

AtmK

K moll Atm

....b8572193040820500866401

moll.b 0296801 =

and for propane

mollb 0062682 =

Equilibrium is always stated as:



PˆyPˆx vii

lii φ=φ (i = 1, 2, 3, …Nc)

with the following material balance constrains

11

=∑=

Nc

iix

11

=∑=

Nc

iiy

11

=∑=

Nc

iiz

Step 2.

Evaluate the fugacity coefficients using the RK EOS. These expressions are,

+−

+

+

−

+−

−+

−=φ

∑=

bVb

VbVln

bRTab

RTPVln

VbVln

bRT

az

bVb

bVVlnˆln

.

.i

ii

2511

51

2

11

11

2

For a vapor phase: vv ˆˆ ,VV ,yz1111 φ=φ==

For the liquid phase: ll ˆˆ ,VV ,xz1111 φ=φ==

To demonstrate the procedure, we will choose one data point:

( )T,P,y,x 11

For this example:



K ...TAtm .P

.y.x

55277152734461136036016020

1

1

=+====

To begin the iterations we will assume the interaction parameter:

06012 .k =

And we must select what will be our dependent and our independent variables:

Given Variables

Independent

Unknown Variables

To evaluate Problem Type

T, zi = xi

T=277.55 K

x1=0.1602

P,yi Bubble Point

The selection of dependent and independent variables is completely arbitrary. It will be dictated by the weight or confidence given to the experimental points. Usually, the Temperature and the liquid compositions are more easily and accurately measured than the pressure and gas compositions.

To evaluate the fugacity coefficients we need:

Assume an initial guess for the gas composition and pressure. We can use the experimental data as the initial guess.



Evaluate the interaction constant a12, and the mixture parameters for both gas and liquid phases. We will use quadratic mixing rules for the a parameter and linear for the b parameter.

Once these are evaluated the EOS must be solved twice to obtain the gas and liquid molar volumes.

! For the liquid phase using xi , al , bl, Vl

! For the vapor phase using yi , av, bv, Vv

( ) ( )∑∑= =

−=2

1

502

1

1i

ji

.

jjijil kaaxxa ( ) ( )∑∑

= =

−=2

1

5.02

1

1i

jij

jijiv kaayya

∑=

=2

1iiil bxb ∑

=

=2

1iiiv byb

The interaction constant of attraction using our initial guess for k12 = 0.06 is

( )12221112 1 kaaa −=

( ) 2

0.52

molK l Atm....a 831000601481807546312 =−×=

1221222111

21 2 axxaxaxal ++=

( )( )

2

0.52

l

l

molK l Atm.a

.......a

05156

831008398016020248180839807546316020 22

=

+×+×=

and



moll.b

....b

l

l

057390

062608398002968016020

=

×+×=

With these constants, we now must solve the cubic EOS for Vl,

)bV(VTa

bVRTP

lll/

l

ll

data

+−

−= 211

( ) ( )0573905527705156

057390552770820506113 50 .VV.

..V

...ll

.l +

−−

×=

which is solved for Vl, and if three real roots are obtained the smallest root is chosen. If not the real root of the equation. This gives;

moll.Vl 087850=

Now we can evaluate the fugacity coefficient for CO2 (component 1) of the liquid.

+

−

+

+

−

+−−

+

−

=φ∑

=

ll

l

l

ll

l.

l

l

l

ll

l.

iii

llll

ll

bVb

VbVln

bRTba

RTPVln

VbVln

bRT

ax

bVb

bVVlnˆln

2511

51

2

11

11

2

Replacing the results obtained, we obtain:

43228083870

7116299750

22

11

.ˆ .ˆln

.ˆ .ˆlnll

ll

=φ⇒−=φ

=φ⇒=φ

Now, repeat the same procedure for the vapor phase.

1221222111

21 2 ayyayayav ++=



( )( )

2

0.52

v

v

molK l Atm.a

.......a

8499

831003964060360248180396407546360360 22

=

+×+×=

2211 bybybv +=

moll.b

....b

v

v

042760

0626803964002968060360

=

×+×=

Now use the RK EOS with bv, av, and Pidata and solve for the vapor volume. Again, if

three real roots are obtained we must select the largest. This provides;

moll.Vv 42511=

Calculate the fugacity coefficients for both components in the vapor phase. This provides.

784830

930930

2

1

.ˆ.ˆ

v

v

=φ

=φ

Next we evaluate the K-values and a new set of vapor compositions

v

l

ˆˆ

xyK

1

1

1

11 φ

φ==

912829309071162

1

11 .

.

.ˆˆ

K v

l

==φφ=

55080784830432280

2

22 .

.

.ˆˆ

K v

l

==φφ=



The new vapor compositions are:

462605508083980466609128216020

222

111

...Kxy...Kxy

=×===×==

the sum is not = 1, Whoops…!

The material balance constraint must be satisfied. Therefore we must normalize the new vapor compositions such that its sum is equal to one.

92920462604666021 ...yy =+=+

502209292046660

2

1

11 .

.

.

y

yy

ii

k, ===∑

=

497809292046260

2

1

22 .

.

.

y

yy

ii

k, ===∑

=

where “k” is iteration index.

The New pressure is evaluated as

∑=

− ×=2

11

iiikk KxPP

Atm...Pk 6512929206113 =×=

Recall the phase envelope and the bubble point criteria to satisfy.



12

1

=∑=i

ii Kz ! Bubble point - Solution

12

1>∑

=iii Kz ! Superheated ! lower T or increase P.

12

1<∑

=iii Kz ! Subcooled ! Increase T, or lower P.

Now return to Step 1. And using these new values for P and vapor compositions repeat the procedure until the bubble point criteria is satisfied.

This is your first answer for kij = 0.06. We have only calculated the saturation pressure for one composition.

! You must calculate VLE for all the data points provided still using kij = 0.06

! Next you must repeat all of the above for different values of kij. A Newton

Raphson procedure can be followed to select successively better values for kij.

This is a minimization problem. The final answer to this exercise gives an optimum kij of 0.07605.

A flow chart summarizing this procedure is as follows.



Given data:

Temperature

Vapor Compositions

Liquid Compositions

Pressure composition data

M = # of data sets

),...,...,,( 21 Nci yyyyy =r

),...,...,,( 21 Nci xxxxx =r

( ) MjyxP j ,...1 ,, =rr

Pure fluid propeties

Hydrocarbon Characterization Schemes

...),,,,( MwTTP bcc ω

Mixing Rules - EOS

Select dependent and independent variables

(i.e., Dependent: - Independent )yP r, xT r,

For all data sets: k = 1,…M do

Set initial guesses and interaction parameters (j = 1)( )

kyy exp1 rr = ( )kPP exp1 = 1ijij kk =

1



Evaluate for (i = 1,…Nc)k( )

kli

vilv VV φφ ˆ,ˆ,,

Evaluate for (i = 1,…Nc)k

kvi

li

iK

φφ

= ˆˆ

Re-evaluate for (i = 1,…Nc)k

k

Nc

jjj

iii

Kx

Kxy

=∑

=1

Is

for (i = 1,…Nc)k ?k

Nc

jjjKx

=−∑

=01

1N

Update

k

Nc

iiijj KxPP

= ∑=

+1

1

Y

Print

( ) ijk kxyTP ,,,, rr

Is

optimal?

( ) ε<−∑=

M

k

calc PP1

2exp N

Change interaction parameters

Y

Print

( ) ijk kxyTPrr

End

1



Evaluation of Reservoir Engineering PVT Properties from an Equation of State

The material balance equation is one of the most powerful tools for estimating oil and gas in place by using production data and PVT fluid properties. Accurate oil and gas properties are needed for making any meaningful estimate of OOIP/OGIP for any type of reservoir fluid.

Oil and gas formation volume factors (Bo’s and Bg’s), bubblepoint pressure, and solution gas oil ratios (Rs’s) can be obtained from flash vaporization, differential liberation, and separator separation tests. These are standard PVT experimental procedures for the oil industry, although they may not always be economically affordable.

In the flash vaporization experiment, a variable-volume cell is filled with a reservoir liquid sample and brought to reservoir temperature. The pressure in this cell is changed by changing the sample volume through a movable piston. Pressure volume pairs are recorded and the bubblepoint pressure and bubblepoint volume are obtained by intersecting two lines with different slopes, one very steep within the liquid phase region, and the other less steep, corresponding to the two-phase region. The cell contents remain constant throughout the experiment.

Differential liberation experiments are conducted to determine gas and oil formation volume factors and solution gas oil ratios. The difference with the flash vaporization experiment is that all the gas formed in the PVT cell is expelled after every pressure depletion stage, then all these gas volumes are brought to standard conditions and added. The total volume of gas removed during the entire process is the amount of gas in solution at the bubblepoint. The solution gas/oil ratio at any given pressure lower than the bubblepoint pressure is obtained by adding all the volumes of gas released at lower pressures down to atmospheric conditions. These are converted to the volume at standard conditions and divided by the volume of residual oil left at the very last expansion stage.

Properties obtained from this test are indicated as differential vaporization values. These are time-consuming experiments, prone to errors that accumulate in every expansion stage. To make matters worse, the Rs obtained from a differential depletion experiment will vary depending upon the pressure decrements taken, and if the



expansion from reservoir conditions (above the bubblepoint) to standard conditions is done in just one or two stages – such as a separation process – the Rs obtained at the bubblepoint will differ from the Rs obtained from a liberation experiment.

In the Separator tests, the reservoir sample brought to reservoir temperature and bubblepoint pressure is expanded through two separation stages; the second is in the stock tank itself. The pressure and temperature of the first separator are selected such that the stock tank oil has a maximum API gravity, which means that the gas/oil ratio at the bubblepoint will be at a minimum, and the stock oil will have a maximum volume, which will yield a minimum oil formation volume factor at the bubblepoint.

For volatile oils, three or more separation stages are usually used to maximize the oil volume in the stock tank. To optimize this would require various combinations of pressure-temperature pairs for all the separators used, plus economic considerations.

The underlying assumption for having all these tests is that the reservoir process can be simulated from a differential vaporization, while the process from the bottom of the well to the stock tank can be simulated with a separator test. Under these assumptions, fluid properties at pressures below the bubblepoint pressure can be obtained by some sort of empirical averaging between the two tests.

( )oDb

oSbsDsDbsSbs B

BRRRR −−= (85)

and

oDb

oSboDo B

BBB = (86)

The well-known averaging formula for Rs may provide some unrealistic negative values, particularly at low pressures. McCain devised a new empirical formula, analogous to the formula for the oil formation volume factor, that prevents these negative results.

sDb

sSbsDs R

RRR = (87)



There are a variety of correlations for these standard PVT properties, although many of them have been developed for specific data sets, and extrapolation to other fields may be quite dangerous. Additionally, most of these correlations require as input variables some experimental information such as bubblepoint pressures, API and/or gas gravities, or Rs’s.

Equations of State provide an alternative method of estimating these properties provided that the reservoir fluid composition is known. Simulated PVT experiments provide simultaneously all liquid and gas properties in a manner that is self consistent.

The Phase Equilibrium Problem Many petroleum engineering applications deal with the phase behavior of the hydrocarbon oil liquid mixture, which at a specified pressure and temperature coexists with a hydrocarbon gas mixture at the same pressure and temperature (i.e. in the production path). To be able to determine the properties of these two phases, thermodynamic phase equilibrium is assumed.

To solve phase equilibrium problems, a set of thermodynamic criteria must be satisfied. The set of equilibrium equations is mathematically presented in terms of compositions and fugacity coefficients, or in terms of the K-values. If a vapor and a liquid phase exist in equilibrium, then the equal fugacity criterion for every component in each phase must be satisfied. For vapor liquid equilibrium, this is equivalent to

PxPy lii

vii φ=φ ˆˆ (i = 1, 2, 3, …Nc) (88)

where by definition yi denotes the mole fraction of component i in the gas phase, while xi denotes the molar composition of i in the liquid phase, and Nc is the total number of components in the mixture. The composition of the overall mixture (gas + liquid) is usually denoted by zi.



In terms of K-values

vi

li

i

ii x

yKφφ== ˆˆ

, iii xKy = (i = 1, 2, 3, …Nc) (89)

The following material balance constraints must also be satisfied:

11

=∑=

Nc

iix 1

1

=∑=

Nc

iiy 1

1

=∑=

Nc

iiz (90)

Physically, the difference between the fugacities of a component in one phase with respect to the other phase gives a measure of the potential, or driving force, for transfer of the component between these phases. Equal fugacities in both phases results in zero net mass transfer across the phases, or equilibrium. Pressure, temperature, and compositions within the phases must remain constant.

Table 3 indicates the governing equations for the most common VLE (vapor-liquid-equilibrium) computations performed by the petroleum engineer along with examples. In all these problems the objective is to find the dependent variables that will give a unique set of K-values that satisfies the governing equation. Once convergence is achieved, the equilibrium compositions (i.e., gas for a bubblepoint and liquid for a dewpoint) are evaluated.

In a flash problem, a set of K-values is proposed which is usually evaluated from correlations as a function of pressure and temperature. The objective is then to find the fractional molar amount of vapor (fv) in a vapor/liquid mixture at the specified temperature and pressure such that the flash equation is satisfied. At convergence the equilibrium gas and liquid compositions are evaluated from the equations listed in Table 3

When an EOS is used, K-values are also dependent upon compositions thus the iteration procedure does not stop here. These equilibrium compositions will provide a new set of K-values when used with the EOS, and this new set of K-values must be



used to restart the iterations in the flash function. Iterations proceed until the K-values and all compositions don’t change.

Flash calculations are the work-horse of any compositional reservoir simulation package.

Given Variables

(independent)

Unknown Variables

(dependent)

Problem Type

Governing Equations Example Application

P , zi = xi T, yi 011

=−∑=

Nc

iiiKz Separator-

Distillation tests

T, zi = xi P,yi

Bubble Point

iii Kzy = Gas injection

Production

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 of

Bo, Rs, Rv, Bg

Table 3 - Governing equations for typical VLE problems.

Simulated PVT Experiments



As seen from Table 3, an EOS can be used to evaluate bubblepoint pressures and flashes provided the overall mixture composition and a choice of independent variables is given. Use of an EOS to evaluate Bo and Rs below the bubblepoint is accomplished by proper accounting of the fractions of gas and liquid obtained in a sequence of depletion stages. The liquid compositions from each stage are stored and this mixture is expanded (flashed) to successively lower pressures. Table 4 indicates the procedure followed when a fixed amount of mixture (say one mole) is subjected to these depletion stages. As a result of the flash computations, not only are gas and liquid compositions obtained, but also liquid and gas volumes which form part of the evaluation of the K-values.

Depletion Stage

Starting Moles

fv fl Moles of Gas Produced (nv)

Moles of Liquid in PVT

cell (nl)

0 1 0 1 0 1

1 1 fv1 fl1 fv1*1 fl1*1

2 fl1*1 fv2 fl2 fv2*fl1*1 fl2*fl1*1

3 fl2*fl1*1 fv3 fl3 fv3*fl2*fl1*1 fl3*fl2*fl1*1

… … … … … …

Nd 11

1×∏

−

=

dN

ilif

fvNd flNd 11

1×∏

−

=li

N

ivN ff

d

d 1

1×∏

=

dN

ilif

Table 4 - Accounting of fractional molar amounts of gas and liquid produced from a simulated differential flash vaporization process.

Accounting for the gas produced and liquid remaining after each depletion stage, as indicated in Table 4, the formation volume factor of the oil from a differential vaporization test is obtained as:

Bo at P > Pb



∏=

×=

dN

iilSCSCo

Roo

fPTV

PTVB

1,),(~

),(~ (91)

where ),(~ PTV Ro is the molar volume of the oil evaluated at reservoir temperature (TR)

and P. Nd is the number of depletion stages and fl,i is the liquid molar fraction obtained in depletion stage i.

Bo at P< Pb

∏

∏

=

=

×

×=

dN

ii,lSCSCo

j

ii,ljRo

o

f)P,T(V~

f)P,T(V~

B

1

1 (92)

it can bee seen that the depletion stages up to pressure j cancel out leaving the following expression

∏+=

×=

dN

jii,lSCSCo

jRoo

f)P,T(V~)P,T(V~

B

1

(93)

The initial solution gas/oil ratio is evaluated as:

∏

∑ ∏

=

=

−

=

×

=d

d

N

iilSCSCo

N

j

j

kkljvSCSCg

si

fPTV

ffPTVR

1,

1

1

1,,

),(~

),(~

(94)

at a pressure stage ‘E’ below the bubblepoint pressure



∏

∑ ∏

=

=

−

−=

×

=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 (95)

where ),(~SCSCg PTV is the molar volume of the gas evaluated from the ideal gas law at

the standard conditions (TSC and PSC).

Simulated PVT Properties for a Black Oil and a Volatile Oil Two reservoir fluids were selected using fluid compositions representative of a black oil and a volatile oil system. These two fluids were subjected to a differential depletion test at T =180 oF using the Soave-Redlich-Kwong EOS. The volatile system was subjected to various separator tests using one and two separators in addition to the stock tank itself, with the purpose of maximizing the liquid produced.

Figure 12 indicates the optimization of the pressure in Separator 2 (the low pressure separator), provided the conditions of the high pressure separator have been fixed at 900 psia and 100 oF. It can be easily seen that the optimal conditions found for this case will vary depending upon the choice of temperature and conditions in Separator 1.



3160

3180

3200

3220

3240

3260

3280

3300

Solu

tion

Gas

Oil

Rat

io a

t Pb

(Rsb

)

2.76

2.78

2.80

2.82

2.84

2.86

Form

atio

n Vo

lum

e Fa

ctor

at P

b (B

ob)

50 100 150 200 250 300Separator 2 Pressure (psia)

High Pressure Separator @ 900 psia and

TSep2 = 75oF

100oF

Figure 12 - Optimization of separator 2 conditions for a volatile oil.

Figure 13 shows the effect of pressure in Separator 1 upon the Rsb and Bob, when the conditions in Separator 2 and the temperature have been fixed. This figure also illustrates the advantage of having two separators instead of one when producing a volatile fluid. Again, these two separators are in addition to the stock tank oil.



3360

3380

3400

3420

3440

3460

Solu

tion

Gas

Oil

Ratio

at P

b (R

sb)

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

Figure 13 - Comparison between one and two separator stages for a volatile oil.

This is a multiple variable optimization problem.

Figure 14 shows the oil formation volume factor obtained for the black oil and the volatile system. Additionally, the values obtained from one of the most modern correlations for the black oil (Kartoadmodjo, T. and Schmidt) correlation is indicated. This correlation is only indicated for reference purposes, and it requires the API gravity of the residual oil, the specific gravity of the gas, and the total gas produced from a separator test. In return it will give bubblepoint pressures, solution gas oil ratios, and formation volume factors.



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 5000

Pressure (psia)

Black OilVolatile OilBlack Oil Correlation

TR =180 oF

PbPb

Figure 14 - EOS simulated oil formation volume factors obtained from a differential depletion test at T = 180 oF for a black and a volatile oil.

Figure 15 shows the solution gas/oil ratios obtained for the black oil and the volatile system. Notice that knowledge of the bubblepoint is crucial for this type of evaluation.



0

1000

2000

3000

4000

Solu

tion

Gas

Oil

Rat

io (R

s)

0 1000 2000 3000 4000 5000

Pressure (psia)

Black OilVolatile OilBlack Oil Correlation

TR =180 oF

PbPb

Figure 15 - EOS simulated solution gas oil ratios obtained from a differential depletion test at T = 180 oF for a black and a volatile oil.

Figure 16 shows the gas specific gravity as a function of pressure for the two reservoir systems. It can be seen clearly that the gas produced from the volatile oil is heavier, and it will likely produce some additional liquid when expanded to standard conditions (Rv). Those calculations are not indicated here, though.



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

Figure 16 - Gas specific gravities obtained from a differential depletion test at T = 180 oF of a black and a volatile oil.

=

scfft cu0282.0.0

PZTBg (96)



0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18

Bg (c

u ft

/ SC

F)

0 1000 2000 3000 4000Pressure (psia)

Black OilVolatile Oil

T = 180oF

Figure 17 - Gas formation volume factors from a differential depletion test at T = 180 oF for a black and a volatile oil.

Figure 17 shows the gas formation volume factors for both fluids evaluated using the standard formula which assumes that the mass of gas is constant, or that no oil is being produced from the gas when expanded to standard conditions. The compressibility factor is obtained from the EOS at the indicated pressure and temperature, using the gas compositions in equilibrium with the remaining liquid. If the gas releases oil at the surface, Bg could be affected by as much as 30%.

The Less Popular Volatized Oil or Solution Oil-Gas Ratio, Rv Up to this point a systematic evaluation of the standard PVT properties used in black oil and gas material balance equations was presented. To evaluate Bg only the Z factor for the gas phase obtained after each depletion stage is needed. The main assumption



here is that the stock-tank oil (also called surface oil) cannot be volatilized into the gas phase.

As pressure declines below the dewpoint, gas condensate reservoirs release liquid within the reservoir pore space, which can be as high as 30 to 35% of the pore volume. Condensate and volatile reservoirs also produce distillate oil from the vapor phase produced at surface conditions.

These facts indicate that surface oil is also dissolved in the gas phase. To deal with all types of reservoir fluids one must take into account the so called “volatilized oil,” i.e. oil dissolved in the gas phase.

This not-so-popular PVT property that takes into account the volume of surface volatilized oil dissolved per unit volume of surface gas is called Rv. The Rv of a reservoir gas at P1,T1 (reservoir pressure and temperature) is obtained by expanding a fixed volume of this gas from to PSC, TSC. The volumes of surface gas and stock tank oil obtained are recorded, and Rv is calculated as the number of STB of surface oil obtained per cubic feet of surface gas at standard conditions. This oil that appeared at surface was volatilized at reservoir conditions. For pressures above the saturation pressure, Rv is just the inverse of Rs. But at pressures P < Psaturation, it is not the same because we are considering the oil in the gas phase.

MMSCFSTBRv ==

conditions standardat gas surface offeet cubic ofmillion conditions standardat oil of barrels (97)

Figure 18 and Figure 19 sketch how this volatilized oil can be obtained from the gas phase of a volatile oil and from a retrograde gas respectively. The volatized oil (Rv) is used in a generalized compositional material balance formulation that can be used for volatile and condensate systems. This subject will be described in greater detail in future lectures.



separator oil

stock tank oil

P > Pb P, T P = PSC

Reservoir Separator SurfaceConditions Conditions Conditions

TR

T= TSC

Figure 18 - Schematics of the additional distillate oil produced from a volatile oil.

surface oil

P > Pd P = PSC

ReservoirSurface

Conditions Conditions

TR

T = TSC

Figure 19 - Schematics of the distillate oil produced from a gas condensate.



Example - Evaluation of Reservoir Engineering Fluid Properties using an Equation of State (EOS)

The following example illustrates a set of fluid property calculations using the Soave-Redlich-Kwong EOS. The fluid is a hypothetical reservoir oil (BigBucks) with 4 components.

Initial Reservoir Composition

Component

Name & Short Cut Identification

Mole

Fraction Tc (K) Pc (atm) ωωωω Mw

Methane

Ethane

Pentane

Nonane

C1

C2

C5

C9

0.3

0.2

0.2

0.3

190.6

305.4

469.6

594.6

48.5

48.2

33.3

22.8

0.008

0.098

0.251

0.440

16.042

30.068

72.146

128.259

PVT Properties of Reservoir Fluid at given reservoir conditions

Given initial reservoir pressure and temperature are:

T = 200.0 F & P = 1800.0 psi

Assume we start with 1 lb-mole of fluid ! upscaling easily done

Properties evaluated from the EOS are

Volume (ft3) (one lb-mole)

Z-Factor (liquid)

Density (lb/ft3)

Molecular Weight

1.982

0.5041

32.151

63.73



Single phase (above bubble-point pressure).

The gas constant in these units is:

Rmole lbft psi732.10 o

3

=R

The bubble point pressure corresponding to T = 200.0 oF is P = 1507.5 psi.

Temperature

Pres

sure

P = 1507.5 psi

T=200oF

CP

Temperature

Pres

sure

P = 1507.5 psi

T=200oF

CP

Fluid properties at the bubble-point are:

Volume (ft3/ lbmol)

Z-Factor

Density (lb/ft3)

Viscosity (cP) Molecular Weight

2.018

.4298

31.585

.1273

63.73



Simulated Constant Mass Expansion

(Also called flash liberation - No gas is being produced).

This experiment is used to obtain isothermal compressibilities.

First Expansion – T = 200.0 oF, P = 1500.0 psia

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xI

C1 30.000 72.017 29.848

C2 20.000 22.507 19.991

C5 20.000 4.499 20.056

C9 30.000 .977 30.105

Property Total

Mixture Vapor Liquid

Mole Fraction 1.000 .004 .996

Volume (ft3) 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 .4281

Viscosity (cP) .0163 .1278

Molecular Weight 63.73 22.82 63.88

Surface Tension (dyne/cm)

2.667



Observe the differences between gas and liquid properties (compositions, density, viscosity, molecular weight).

Verify the evaluation of the liquid and gas molar volumes using the following formula:

PRTZV v

v =

And use a similar equation for the liquid volume.

Second expansion at T = 200.0 oF, P = 1200.0 psia

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xI

C1 30.000 70.211 23.663

C2 20.000 24.399 19.307

C5 20.000 4.597 22.427

C9 30.000 .793 34.603



Property Total



Volume (ft3) 2.524 .694 1.830

Volume (ft3/lbmol)

Volume% 100.0 27.5 72.5

Density (lb/ft3) 25.249 4.499 33.118

Z-Factor .4280 .8643 .3592




4.008

Verify the evaluation of gas and liquid densities and molecular weights using the following equations:

∑=

=Nc

iii MwyvaporMw

1

)(

vv V

vMw )(=ρ

Similarly for the liquid phase use liquid compositions (xi) and liquid volume.



Another expansion - T = 200.0 oF, P = 800.0 psia

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xI

C1 30.000 66.387 15.284

C2 20.000 27.699 16.886

C5 20.000 5.246 25.967

C9 30.000 .669 41.863

Property Total



Volume (ft3) 3.897 2.276 1.621


Volume % 100.0 58.4 41.6

Density (lb/ft3) 16.352 2.988 35.115

Z-Factor .4405 .8934 .2574




6.408



Summary Constant Mass Expansion at 200.0 oF

Pressure

psi

Relative

Volume

V/Vb

Compressibility

(1/psi) Y-Factor

1800.0 .9824 .517E-04

Pb = 1507.5 1.0000 .589E-04

1500.0 1.0047 1.07

1200.0 1.2509 1.02

800.0 1.9315 .95

Simulated Differential Depletion Test

Let’s suppose that as the pressure drops following the same decrements as in the previous case the gas that is being generated is produced and the liquid remains in the reservoir.

This test allows the determination of the oil and gas formation volume factors (Bo & Bg) and solution gas-oil ratios (Rs)

Here we assume that ALL the gas generated from a flash expansion is being produced.



Depletion Stage # 1 - T = 200.0 oF, P = 1500.0 psia

Properties as this stage are still the same (no production took place yet).

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xI

C1 30.000 72.017 29.848

C2 20.000 22.507 19.991

C5 20.000 4.499 20.056

C9 30.000 .977 30.105

Property Total



Volume (ft3) 2.027 .014 2.013


Volume% 100.0 .7 99.3

Density (lb/ft3) 31.438 5.713 31.622

Z-Factor .4296 .8465 .4281




2.667

In the next depletion stage note that the composition using for flashing are identical to the liquid composition from the previous stage (i.e., ALL gas has been produced).



Depletion Stage # 2 - T = 200.0 oF, P = 1200.0 psia

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xi

C1 29.848 70.202 23.661

C2 19.991 24.407 19.314

C5 20.056 4.597 22.426

C9 30.105 .794 34.599

Property Total Mixture

Vapor Liquid


Volume (ft3) 2.515 .678 1.837

Volume% 100.0 27.0 73.0

Density (lb/ft3) 25.404 4.499 33.117

Z-Factor .4263 .8643 .3592




4.008



Depletion Stage # 3 - T = 200.0 oF , P = 800.0 psia

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xi

C1 23.661 65.493 15.112

C2 19.314 28.574 17.421

C5 22.426 5.263 25.934

C9 34.599 .670 41.533


Vapor Liquid


Volume (ft3) 3.225 1.339 1.887

Volume% 100.0 41.5 58.5

Density (lb/ft3) 21.752 3.011 35.048

Z-Factor .3646 .8916 .2568




6.346

In a differential depletion test a last stage is added at standard conditions to determine the residual oil volume that will be used to determine the oil formation volume factor.



Last Depletion Stage - T = 60.0 F, P = 14.7 psia

Component

Name

Total

Moles %

zi

Vapor %

yi

Liquid %

xi

C1 15.112 40.581 .247

C2 17.421 44.305 1.731

C5 25.934 14.901 32.373

C9 41.533 .212 65.649


Vapor Liquid


Volume (ft3) 139.930 138.367 1.564

Volume% 100.0 98.9 1.1

Density (lb/ft3) .569 .082 43.661

Z-Factor .3695 .9915 .0065




21.432



Summary of Results from Differential Depletion at 200.0 oF

Pressure (psi)

Bod

(Bbl/STB)

Rsd

(Scf/bbl) Bg

ρρρρo

(lb/ft3) Z γγγγg

1800.0 1.767 32.15

Pb = 1507.5 1.799 1030.1 31.58

1500.0 1.795 1023.3 .010595 31.62 .846 .788

1200.0 1.632 773.3 .013524 33.12 .864 .792

800.0 1.453 496.8 .020934 35.05 .892 .820

14.7/60 1.000 43.66

EXERCISE - Verify the calculation of formation volume factors and solution gas-oil ratios using the formulas derived in class.



Production Mechanism for Gas Condensates/Volatile Oils: Constant Volume Depletion To understand condensate production we must explain what kind of physical processes take place in the reservoir and in the surface.

None of the routine experimental PVT analysis (flash liberation, constant, differential liberation, or separator separation tests), mimics the depletion behavior of a condensate reservoir.

A more realistic (and more expensive) PVT analysis for studying the depletion of a gas condensate is a constant volume depletion (CVD) test. This test provides not only volumetric information but also compositional measurements on the gas and the liquid at equilibrium.

During the laboratory CVD studies, only a certain amount of gas phase is removed from the PVT cell, and it is subsequently flashed to standard conditions. This gas produces liquid at standard conditions, the gas and the liquid are separated and quantified for their mass, volumetric ratios, and compositions. The composition of liquid phase in the cell is usually calculated from mass balance equations. This method of determination of liquid phase composition is associated with considerable uncertainties on the quantity and composition of fluid initially in the cell and quantification of the flashed fluid.

We assume the reservoir to be a cell with a constant pore space filled initially with the reservoir gas condensate. For simplicity we will not consider water in the analysis, but this could be easily incorporated. As production begins, the reservoir pressure will drop causing the reservoir gas to expand. Unlike the differential liberation test in which all produced gas is taken out from the cell, in the reservoir only a portion of the gas is produced. This amount is determined by keeping the pore volume constant.

The gas produced is the gas contained in this excess volume with respect to the original pore volume.

We normally assume that the gas is the mobile phase and that the liquid remains in the reservoir. This is the actual process in the reservoir since the oil saturation rarely reaches the critical saturation.

For volatile oils, both the liquid and the gas are mobile. Moving according the relative permeability saturation relationships. But the rest of the process may be considered identical.



The following figures (Figure 20 and Figure 21) sketch the CVD production mechanism.

After every depletion stage, we must perform an accounting of:

• moles and volume of gas to be produced,

• moles and volume of liquid and gas left in the reservoir,

• composition of the produced (surface) and the remaining phases inside the reservoir, and

• moles and volume of condensate liquid produced at surface from surface gas.



Constant Volume Depletion (CVD)

Pi = Initial Reservoir PressureT = Reservoir TemperatureVi = Reservoir Pore Volumezi = Initial Reservoir Fluid Composition

(Gas Condensate)

(1)

1. Mass in place (either in moles or pounds)

Given

Pi =3,000 psia

P1 = 2,800 P1= 2,800 P2 = 2,500

np1 np2Vp1 Vp2

( zi )o ( zi )1 ( yi )2

( xi )2

( yi )1

( x i ) 1

Vi V1 = Vi + Vp1 Vi V2= Vi + V p2V1 = VL1 + VG1 V2 = VL2 + VG2

ni = P iVi/zRT

Figure 20 - Sketch of a production mechanism for a gas condensate.



• • •

V

( y i ) n

( x i ) n

( z i ) n-1

pnnpn

Target for gascycling

End of Depletion

• • •

Constant Volume Depletion of a Condensate Reservoir (cont.)

At any depletion stage need:

• Moles and composition of liquid phase (nl , xi's)• Moles of gas phase formed (nv, yi's)• Compressibility factor of the liquid phase, z l• Compressibility factor of the gas phase, zv

n zi = yinv+ xinl From flash calculations

VG = nvzv RTP

VL = nlzl RTP

From EOS

Figure 21 - Sketch of a production mechanism for a gas condensate.



Simulation of a Constant Volume Depletion (CVD) Test for a Gas Condensate Using an EOS The following steps illustrate how a CVD test is simulated using an EOS.

System = gas condensate/ ac-ft reservoir bulk volume

Most of the nomenclature used here has been seen previously, however we must use several subscripts to identify depletion stage, fluid produced or remaining in the reservoir, and type of fluid (gas or liquid).

The following nomenclature will be used:

tn = initial # of moles of condensate gas [lb moles/ac-ft]

G = initial gas in place [SCF/ac-ft]

N = initial oil in place [STB/ac-ft]

Gp = gas produced [SCF/ac-ft]

Np = oil produced [STB/ac-ft]

tlv VVV ~ , ~ , ~ = vapor, liquid, and total mixture molar volumes

[ft3 / lb mole]

Subscripts:

i = component number i = 1, 2, 3,…Nc

j = depletion stage j = 1, 2, 3, … E

d = property at dew point (d)

R = at reservoir pressure and temperature.

SC = at standard conditions

p = produced

r = remaining



v, l, t = vapor (gas), liquid (oil), and total, respectively

2φ = two-phase property

Subscripts are used combined as follows,

jljvjt nnn ,,, ,, = total, gas, and liquid (oil) number of moles at

`the end of depletion stage “j”

jpG , = gas produced in depletion stage “j” at Pj.

np,j = moles of gas produced in depletion stage “j”

The gas volume expressed in SCF is usually taken as:

G = ) (

4.379 molelbnmolelb

SCFt×

(98)

Where 379.4 is the volume in cubic feet occupied by 1 lb-mole of ideal gas at standard conditions T = 60°F, P =14.7 psia -

(Use ideal gas law to derive this conversion factor).

Computational Procedure for Depleting a GAS CONDENSATE reservoir. Step 1: Determine the size your reservoir from reservoir chararacterization data (volumetrics). Or express results in terms of the pore space occupied by the hydrocarbon in cubic feet per acre -foot.

)1(560,43 wiHC SV −×φ×= [RCF/ac-ft] (99)

This will remain fixed throughout the process. Upscaling can be easily accomplished. Recall EOS works with intensive properties!!



Step 2: Given the composition of the reservoir fluid (zi's) i = 1, 2, 3 ... Nc (Nc = total number of components), and the reservoir temperature TR. Estimate the upper dew point using any EOS with a set of mixing rules.

For the dew point pressure the objective is to find Pd ( using the initial fluid composition zi’s at TR ) that satisfies:

ε≤−∑=

11

Nc

i i

i

Kz ( )6101 −×=ε (100)

with the equilibrium ratios (Ki) evaluated from any EOS/MCR.

vi

li

i

ii x

yKφφ== ˆˆ

(101)

Recall that one can evaluate the derivative within this integral regardless of the number of moles the system has.

Step 3: Calculate the gas compressibility factor at the dew point (Pd, Td) (reservoir temperature = TR = Td ) and the number of moles in the reservoir.

Solve the cubic EOS for the gas compressibility factor.

023 =γ+β+α+ ddd zzz (102)

Where the polynomial coefficients ( α, β, γ ) depend upon pressure, temperature, fluid composition, and the EOS with mixing rules used.

What if the polynomial gives 3 real roots? → choose the largest for gas.

Therefore the initial number of moles of gas per acre-foot in the reservoir is,

dd

HCdt RTz

VPn = (103)

(Td = TR)

Step 4: Evaluate the volume that these nt moles will occupy when expanded to a pressure less that Pd. For example for depletion stage j = 1, we lower the pressure to a predetermined value, P1, and solve the flash equations as outlined in previous lectures.



For simplicity sometimes we may drop the index “j” with the understanding that properties are evaluated at each depletion stage.

P1, TR, zi => equation for Ki = > flash equations.

0)1(1

)1(1

=−+

−∑=

Nc

i iv

ii

KfKz (104)

The equilibrium GAS compositions are evaluated as:

)1(1 −+=

iv

iii Kf

Kzy (105)

and the equilibrium LIQUID (oil) compositions are evaluated as,

i

ii K

yx = (106)

The molar volumes for the gas and the liquid ( lv VV ~,~ ) had been already evaluated when

solving the flash problem. They both participate in the evaluation of the equilibrium ratios (Ki ) from the EOS (see previous notes for the final expression of the Ki values using the Peng-Robinson EOS).

For the liquid lV~

Real root of the following polynomial or the smallest root if 3 real roots exist.

0)',,(~)',,(~)',,(~1312

211

3 =+++ sxTPaVsxTPaVsxTPaV iRliRliRl (107)

For the gas vV~

Real root of the following polynomial or the largest root if 3 real roots exist.

0)',,(~)',,(~)',,(~1312

211

3 =+++ syTPaVsyTPaVsyTPaV iRviRviRv (108)

Calculate total molar volume of mixture as

( ) vlllt VfVfV ~1~~ −+= (109)



And the total volume at the end of the depletion stage “j” is obtained by multiplying Eq. (109) by original number of moles.

( ) vltllttt VfnVfnVn ~1~~ −+= (110)

vt

vtl

t

lttt V

nnnV

nnnVn ~~~ += (111)

vlt VVV += (112)

The z-factors for the gas and the liquid at any depletion stage are,

d

jvjjv RT

VPz ,

,

~= and

d

jljjl RT

VPz ,

,

~= (113)

Note that the number of moles at the end of stage j=1 still is the same as initially. Production has not started yet.

Step 5: Evaluate oil saturation in the reservoir as

1001,1, ×=

HC

ll V

VS (114)

At any stage “j”

100100~

,,, ×=×=

HC

jl

HC

ljljl V

VV

VnS (115)

This value indicates the “current” fraction of the pore space filled with oil (liquid dropout).

Step 6: Determine the amount of gas to be produced in depletion stage “j”.

Since the total volume will be larger because of pressure reduction we must produce the excess gas to restore original volume (VHC = Vt,0).

The produced gas volume at stage “j” is

0,,, tjtjvp VVV −= (116)

Calculate # of moles of gas produced in Vvp,j



Rjv

jvpjjp RTz

VPn

,

,, = (117)

(Note zv,j is an intensive property does not depend on amounts)

Step 7: Evaluate cumulative gas produced as a % of the initial gas in place.

100% 1,

×=∑

=

t

J

jjp

p n

nG (118)

We have used the property of an ideal gas that “volume fraction = mole fraction”.

Here J = current number of depletion stages (maximum = E).

Step 8: Evaluate the two-phase gas deviation factor. For stage j =1 is,

Rpt

HC

RTnnVPz

)( 1,

12 −

=φ j = 1 (119)

the denominator has (initial # of moles) – (number of gas moles produced after 1st expansion)

After 2 expansions

Rppt

HC

RTnnnVPz

)( 2,1,

22,2 −−

=φ (120)

and after the Eth expansion.

R

E

jjpt

HCEE

RTnn

VPz

−

=

∑=

φ

1,

,2 (121)

This equation shows that a plot of (P/z) is proportional to the mass, or volume or gas production. The only difference is that the z-factor to be used is the two-phase z-factor.



This type of plots using production data is the basis of estimating the original gas in place as seen in a previous chapter.

Step 9: Evaluate the remaining moles of gas after every depletion stage (nvr,j) from mole balance.

jpjvjvr nnn ,,, −= (122)

Step 10: Calculate a NEW # of total moles and new overall composition from the mole balance.

jljvjt nnn ,,, += (123)

and the new overall composition in the reservoir is

jt

jvjijljiji n

nynxz

,

,,,,,

+= (124)

Step 11: The moles of gas produced after each depletion stage are flashed to standard conditions to determine the amounts of gas and condensate produced at surface. This step will be seen in greater detail in a subsequent example.

Step 12: Drop the pressure another step and repeat whole procedure outlined in steps 2-11.

Note that a plot of gas produced versus the ratio of pressure over the two-phase z-factor gives a straight line and the ultimate production at abandonment pressure.

Numerical Example of a CVD The steps outlined previously were followed using an EOS with a gas condensate from the Simpson’s Field. Figure 22 shows pressure over three different z-factors obtained versus the gas produced. Note that the prediction of original gas in place could have substantial errors if the wrong z-factors are used.

The data required for the EOS for this calculation are,



• initial fluid compositions

• reservoir temperature

• initial pressure and pressure depletion stage

The production data required are

• Pressures and Gp.

This gas condensate has a dew point pressure of 4,055 psi at 205oF. The plot shows three different values of “P over z” versus the gas produced in MSCF.

Shows a flow chart indicating the steps followed in a simulated CVD test.



z (liquid)z(vapor)z(2 phase)

P/ z = 4055 -1.0578e-05 Gp (Correct two-phase z)

P/ z = 4055 -8.1097e-06 Gp (z = z liquid)

P/ z = 4055 -1.6219e-05

What's the "two-phase" z-factor?

From the EOS (Equation of state ) will have three roots. The largest Vcorresponds to the vapor and the smallest to the liquid.

VL= nlzl RTP

Vv = nvzv RTP

PVvRT

= zv= f (P, T, yi's)

PVLRT

= zl = f (P, T, xi's)

z2φ= nvn zv + nln zl z2φ= PVini-npRTor

5e+84e+83e+82e+81e+80e+00

1000

2000

3000

4000

5000

G p

P/z2 #

Gp (z = z vapor)

Estimation of Reserves: Simpson's Condensate Field



P/ z = 4055 -8.1097e-06 Gp (z = z liquid)

P/ z = 4055 -1.6219e-05



VL= nlzl RTP

Vv = nvzv RTP

PVvRT


PVLRT

= zl = f (P, T, xi's)


5e+84e+83e+82e+81e+80e+00

1000

2000

3000

4000

5000



P/ z = 4055 -8.1097e-06 Gp (z = z liquid)

P/ z = 4055 -1.6219e-05



VL= nlzl RTP

Vv = nvzv RTP

PVvRT


PVLRT

= zl = f (P, T, xi's)


5e+84e+83e+82e+81e+80e+00

1000

2000

3000

4000

5000

G p

P/z2 #

Gp (z = z vapor)

Estimation of Reserves: Simpson's Condensate Field

Figure 22 - Differences in the estimation of reserves using three different compressibility factors.



Given Reservoir Fluid CompositionTemperature, Dew Point PressureControl parameters

Determine Moles ofGas to be produced

Recombine RemainingPhases

Calculate initialnumber of moles

Drop pressure below Pd

Select Hydrocarbon PoreVolume

Flow Diagram for simulating a CVD

P = P - ∆P

Calculate TotalVolume

Evaluate Flash

Calculate Volume ofCondensate

Calculate Volume ofGas

Flash to SC

Figure 23 - Flow chart for a simulated CVD test.

Exercise - Simulation of a CVD Using an EOS

This example will calculate the volumetric depletion performance of a retrograde gas condensate reservoir using the Soave-Redlich-Kwong EOS. Consider 500 psi as the abandonment pressure. The reservoir temperature is 195oF, and the initial reservoir pressure is equal to the dew-point pressure. The composition of the reservoir fluid required for the EOS is given in Table 5.



Software packages have a data base built-in with the critical properties, molecular weights, and acentric factors of all pure substances required for EOS computations.

Component Mole %

C1 75.2

C2 7.7

C3 4.4

C4 3.1

C5 2.2

C6 2.3

C7 2.099

C8 1.235

C9 0.727

C10+ 1.039

Table 5 - Initial gas condensate composition “Good Money”.

Additional information is

φ = 0.2

Swi = 0.25

TR = 195 oF

Tsc = 60 oF

Psc = 1.7 psi

Solution



Given the reservoir fluid composition and the reservoir temperature we select a SATURATION POINT calculation and the SRK EOS gives us the following information.

DEW-POINT pressure at 195oF, molar volume at the dew point, and the z-factor. This pressure is 2,960 psia, and the z-factor is 0.779.

We select a CVD option using any commercial software (PVTSIM from Calsep, PVTi from Geoquest, DeskTop PVT from Landmark) software selecting 500 psia decrements.

The following tables present the results from a CVD experiment simulated with the SRK (Soave-Redlich-Kwong EOS) EOS, as explained previously.

CVD Results Z - Factor Gas Viscosity

P(psi) 100×

d

l

VV 100×

t

p

nn 100×

∑t

p

nn zg z2φ

)(cPgµ

2,960 0 0 0 0.779 0.030

2,500 14.02 14.4 14.4 0.799 0.769 0.023

2,000 14.42 16.6 31.0 0.825 0.763 0.019

1,500 12.96 16.7 47.7 0.855 0.755 0.016

1,000 11.09 16.5 64.2 0.891 0.735 0.015

500 8.93 16.0 80.2 0.933 0.665 0.013

Table 6 - Summary results from a CVD study on condensate “Good Money” at T = 195 oF.

100×

d

l

VV → Percent of liquid dropout left in reservoir. Referred to dew-point

volume (Vd)



100×

t

p

nn → Percent of moles produced

100×

∑t

p

nn → Cumulative mole percent production

For each depletion stage the gas properties are given in Table 7

P (psi) 2,960 2,500 2,000 1,500 1,000 500

Mole % 100 85.197 83.176 82.484 81.745 80.627

~Vv (ft³/mol) 1.848 2.246 2.897 4.005 6.259 13.103

VV

v

t

% 100 88.1 88.6 90.5 92.9 95.8

ρg (lb/ft³) 14.881 10.924 7.949 5.574 3.557 1.783

Zg 0.7787 0.7994 0.8249 0.8553 0.891 0.9327

µ g (cP) 0.0303 0.0226 0.0186 0.0162 0.0146 0.0135

Mwg 27.50 24.54 23.03 22.33 22.26 23.36

Table 7 - Properties of the gas produced at 195 oF (Condensate Good Money).

The gas composition after each depletion stage is stored to perform the flash calculations at standard conditions and is given in Table 8



P (psi) 2,960 2,500 2,000 1,500 1,000 500

C1 75.2 78.84 80.695 81.403 81.021 78.595

C2 7.7 7.65 7.697 7.834 8.071 8.406

C3 4.4 4.153 4.062 4.105 4.322 4.857

C4 3.1 2.743 2.56 2.518 2.664 3.244

C5 2.2 1.806 1.576 1.471 1.516 1.943

C6 2.3 1.731 1.386 1.194 1.159 1.485

C7 2.099 1.45 1.057 0.832 0.745 0.914

C8 1.235 0.782 0.521 0.376 0.313 0.364

C9 0.727 0.41 0.242 0.157 0.118 0.126

C10+ 1.039 0.435 0.203 0.11 0.071 0.067

MwC10+ 153.8 147.7 144.7 142.9 141.6 140.7

Table 8 - Composition of produced gas with pressure at 195oF (Condensate Good Money).

From the EOS information we need to calculate the following terms:

• Initial Gross Gas in Place (MSCF/ac-ft)

• Initial Residue Gas in Place (MSCF/ac-ft)

• Initial Oil in Place (STB/ac-ft)

• Increments of Gross Gas Production (MSCF/ac-ft)

• Cumulative Gross Gas Production (MSCF/ac-ft)

• Residue Gas in Each Increment (MSCF/ac-ft)

• Cumulative Residue Gas in Each Pressure Stage (MSCF/ac-ft)

• Liquid Produced in Each Pressure Stage (STB/ac-ft)

• Cumulative Liquid Produced in Each Pressure Stage(STB/ac-ft)



• Average Gross Gas-Oil-Ratio in Each Pressure Stage (SCF/bbl)

• Average Residue Gas-Oil-Ratio in Pressure Stage(SCF/bbl)

• Cumulative Gross Gas Recovery Percentage

• Cumulative Residue Gas Recovery Percentage

• Cumulative Liquid Recovered Percentage

• These terms will be described as we move along.

• Quantities will be evaluated per acre foot of net bulk reservoir rock.

The hydrocarbon pore volume per acre foot is

)1(560,43 wiHC SV −×φ×= (125)

ftacftVHC −

=−××=3

534,6)25.01(2.0560,43

Step 1:

Calculate the initial number of moles (nt) based upon the hydrocarbon pore space per acre foot. Then, evaluate the number of moles and cumulative number of moles produced at each pressure stage.

Using VHC as a base volume.

R

RRtHC P

RTZnV =

The number of moles contained in this hydrocarbon pore volume is



RR

RHCt RTZ

PVn =

ftacmolelb

RTZPVn

RR

RHCt −

=××

×== 5829.532,365573.10779.0

960,2534,6

The number of moles produced at P =2,500 psia is

ftacmolelbnfn tpp −

=×== 6919.5085829.532,3144.0

Step 2:

Calculate the Initial Gross Gas in Place per ac-ft, then evaluate the increments of Gross Gas Production (MSCF/ac-ft) and Cumulative Gross Gas Production (MSCF/ac-ft) produced after each pressure stage.

The Initial Gross Gas in Place expressed in MMSCF/ac-ft is

7.1452073.105829.532,3 ××==

SC

SCt

PRTnG

Note that the term

mollb/SCF ..

.P

RTV~SC

SCsc −=×== 5643379

7145207310



is the standard molar volume of an ideal gas. The standard temperature and pressure have been taken as 520 R (60 F) and 14.7 psia.

ftacMSCF

feetacSCFG

−≅

−= 341,1843,340,1

The Gross Gas Production at each pressure stage is evaluated as

GfG pp ×=

where fp is the fraction of moles produced (see Table 6).

At P = 2,500 psia

ftacMSCFGfG pp −

=×=×= 08.193341,1144.0

Proceed with all the other pressures and fill the corresponding columns in the following table:



P(psi) np ∑ np

Gp ∑ Gp

GRP ∑ GRP

N p N p∑

GOR

2960 0 0 0

2500 508.69 508.69 193.08 193.08

2000

1500

1000

500

Table 9 - Calculate results from depletion test.

Step 3:

Evaluate the Initial Residue Gas in place (MSCF/ac-ft) and the Initial Oil in Place (STB/ac-ft).

NGG R +=

Note that these terms are usually expressed in different units!

To evaluate these quantities we need to flash the initial number of moles with initial gas composition to standard conditions.



Property Total Vapor Liquid

Mole % 100 94.983 5.017

V~ (ft³/mol) 358.755 377.573 2.503

ρ (lb/ft³) 0.077 0.061 43.706

Z 0.9458 0.9954 0.0066

µ (cP) 0.0103 0.5083

Mw 27.5 23.18 109.4

Table 10 - Flash results of initial composition of gas condensate Good Money to 14.7 psia and 60oF.

The compositions of the gas and liquid phases expressed in moles % are given in the following table.

Component Initial Vapor Liquid

C1 75.2 79.147 0.47

C2 7.7 8.09 0.31

C3 4.4 4.596 0.688

C4 3.1 3.168 1.812

C5 2.2 2.081 4.446

C6 2.3 1.728 13.123

C7 2.099 0.86 25.56

C8 1.235 0.258 19.741

C9 0.727 0.056 13.428

C10+ 1.039 0.015 20.423

MwC10+ 153.8 137.5 154

Table 11 - Compositions of the gas and liquid phases as a result of flashing the initial mixture to 14.7 psia and 60oF.



From Table 10 we can see that about 5% is transformed into liquid thus the Initial Residue Gas per acre-feet is:

ftacMSCFGfG vR −

=×=×= 72.273,1341,194983.0

and the Initial Liquid in Place is

ftacSCFN

VnfN otl

−=

=××=××=

6059.443

503.25829.532,305017.0~

In standard barrels

ftacSTBN

−== 0178.79

614.56059.443

Step 4:

Evaluate the following properties

Residue Gas produced (GRP ) in each pressure stage (MSCF/ac-ft)

Cumulative Residue Gas in each pressure stage(MSCF/ac-ft)

Liquid produced (Np ) in each pressure stage (STB/ac-ft)

Cumulative Liquid Produced in each pressure stage (STB/ac-ft)



This requires flashing the number of moles of gas produced in each depletion stage to standard conditions. Use compositions from Table 8.

For example at 2,500 psi,


Mole % 100 97.534 2.466

V~ (ft³/mol) 368.436 377.687 2.553

ρ (lb/ft³) 0.067 0.059 44.209

Z 0.9714 0.9957 0.0067

µ (cP) 0.0104 0.5396

Mw 24.68 22.45 112.88

Table 12 - Flash results to 14.7 psia and 60 oF from the gas produced at P = 2,500 psia (Condensate Good Money).

The compositions of the gas and the liquid obtained are given in Table 13



Component Total Vapor Liquid

C1 78.84 80.821 0.477

C2 7.65 7.836 0.3

C3 4.153 4.242 0.634

C4 2.743 2.772 1.584

C5 1.806 1.757 3.753

C6 1.731 1.489 11.306

C7 1.266 0.741 22.039

C8 0.745 0.26 19.936

C9 0.439 0.064 15.27

C10+ 0.627 0.018 24.701

MwC10+ 153.8 137.5 154.3

Table 13 - Compositions of the gas and the liquid obtained from flashing the gas produced at 2,500 psia (Condensate Good Money).

The residue gas and the liquid produced are evaluated as

pvRP GfG = oplp VnfN ~=

at P = 2,500 psia is

ftacMSCFGRP −

=×= 3196.188081.19397534.0

and

ftacSCFN p −

=××= 0256.32553.269.50802466.0



in barrels

ftacSTBN p −

== 7046.5614.50256.32

Cumulative values are obtained just by adding the results obtained from each stage (left as an exercise).

Step 5:

Evaluate the following

Average Gross Gas-Oil-Ratio in Each Pressure Stage (SCF/STB)

Average Residue Gas-Oil-Ratio in Pressure Stage(SCF/STB)

Gross Gas and Cumulative Gross Gas Recovery Percentage

Residue Gas and Cumulative Residue Gas Recovery Percentage

Liquid Recovery and Cumulative Liquid Recovery Percentage

The Average Gross Gas-Oil-Ratio, and the Average Residue Gas-Oil-Ratio in Each Pressure Stage (SCF/STB) are evaluated as

p

p

NG

GOR = p

RPR N

GGOR =

The Gross Gas and Residue Gas Recovery Percentages are



100×=GGpGross 100×=

R

RP

GGGresid

The liquid recovery in each stage is

100×=NN

Nrec p

for example at P = 2,500 psia

STBSCFGOR 55.846,33

7046.50.081,193 ==

STBSCFGORR 9.011,33

7046.56.319,188 ==

4.14100341,1

081.193 =×=Gross

78.1410072.273,1

3196.188 =×=Gresid

22.71000178.79

7046.5 =×=Nrec

Steps 1 to 5 are repeated for all the depletion pressures, and the cumulative values are calculated. This will be left as an exercise.



From a CVD simulated test properties of the liquid remaining in the reservoir are also reported. These can be used for gas cycling projects.

These properties are in the following tables.

P (psi) 2,960 2,500 2,000 1,500 1,000 500

Mole % 100 14.803 16.824 17.516 18.255 19.373

~Vo (ft³/mol) 1.848 1.75 1.852 1.982 2.149 2.381

VV

l

t

% 100 11.9 11.4 9.5 7.1 4.2

ρo (lb/ft³) 14.881 25.476 28.402 31.069 33.657 36.364

Zo 0.7787 0.6229 0.5273 0.4232 0.3059 0.1695

µo (cP) 0.0303 0.0718 0.092 0.1158 0.1462 0.1891

Mwo 27.50 44.59 52.6 61.57 72.32 86.57

Table 14 - Properties of the liquid (oil) remaining in the reservoir at 195 oF after every depletion stage (Condensate Good Money).

Table 15 contains the overall composition of the fluid remaining in the reservoir after every depletion stage. This composition is obtained by recombination of the liquid and gas remaining in the reservoir after every depletion stage from Equation (124).



P (psi) 2,960 2,500 2,000 1,500 1,000 500

C1 75.2 75.2 74.586 73.118 70.465 65.611

C2 7.7 7.7 7.709 7.711 7.672 7.488

C3 4.4 4.4 4.442 4.533 4.67 4.83

C4 3.1 3.1 3.16 3.304 3.556 3.967

C5 2.2 2.2 2.267 2.432 2.74 3.303

C6 2.3 2.3 2.396 2.639 3.101 3.994

C7 2.099 2.099 2.208 2.485 3.014 4.058

C8 1.235 1.235 1.312 1.501 1.862 2.574

C9 0.727 0.727 0.78 0.909 1.151 1.625

C10+ 1.039 1.039 1.141 1.366 1.769 2.55

MwC10+ 153.8 153.8 154.2 154.5 154.7 154.9

Table 15 - Overall Composition changes with pressure at 195oF.



P (psi) 2,960 2,500 2,000 1,500 1,000 500

C1 75.2 54.252 44.382 34.107 23.195 11.576

C2 7.7 7.99 7.766 7.133 5.883 3.671

C3 4.4 5.823 6.322 6.547 6.228 4.717

C4 3.1 5.153 6.128 7.007 7.553 6.973

C5 2.2 4.47 5.679 6.961 8.222 8.963

C6 2.3 5.573 7.389 9.439 11.798 14.435

C7 2.099 5.836 7.899 10.268 13.175 17.143

C8 1.235 3.842 5.22 6.799 8.797 11.772

C9 0.727 2.548 3.44 4.454 5.775 7.865

C10+ 1.039 4.515 5.777 7.285 9.374 12.885

MwC10+ 153.8 157.2 155.8 155.3 155.2 155.2

Table 16 - Reservoir liquid composition changes with pressure at 195 oF.

Suggested Exercises - The following exercises may be assigned in the exam or in a homework.

Suggested Exercise 1.

Complete Table 9 from this example using the information from the example plus the following EOS results from flashing the produced gas at the different depletion stages to standard conditions.


Plot Np and GOR versus pressure.




Mole % 100 98.705 1.295

V~ (ft³/mol) 372.906 377.763 2.608

ρ (lb/ft³) 0.062 0.058 44.788

Z 0.9831 0.9959 0.0069

µ (cP) 0.0104 0.5763

Mw 23.18 21.95 116.83

Table 17 Flash Results from gas produced at P =2,000 psi to 15 psia and 60 oF


C1 80.695 81.747 0.479

C2 7.697 7.794 0.297

C3 4.062 4.107 0.612

C4 2.56 2.574 1.47

C5 1.576 1.553 3.317

C6 1.386 1.277 9.693

C7 0.833 0.607 18.048

C8 0.49 0.248 18.988

C9 0.288 0.07 16.917

C10+ 0.412 0.022 30.179

MwC10+ 153.8 137.6 154.7

Table 18 Gas and liquid compositions from gas produced at P =2,000 psi and flashed to 15 psia and 60 oF




Mole % 100 99.228 0.772

V~ (ft³/mol) 374.903 377.801 2.657

ρ (lb/ft³) 0.06 0.057 45.318

Z 0.9884 0.996 0.007

µ (cP) 0.0104 0.6125

Mw 22.46 21.7 120.42

Table 19 -Flash Results from Gas Produced at P =1,500 psia to 15 psia and 60oF


C1 81.403 82.033 0.478

C2 7.834 7.893 0.3

C3 4.105 4.132 0.615

C4 2.518 2.526 1.441

C5 1.471 1.458 3.113

C6 1.194 1.136 8.618

C7 0.607 0.497 14.757

C8 0.357 0.225 17.287

C9 0.21 0.074 17.74

C10+ 0.301 0.025 35.654

MwC10+ 153.8 137.7 155.2

Table 20 - Gas and liquid compositions from gas produced at P =1,500 psi and flashed to 15 psia and 60 oF




Mole % 100 99.411 0.589

V~ (ft³/mol) 375.577 377.785 2.679

ρ (lb/ft³) 0.06 0.058 45.589

Z 0.9902 0.996 0.0071

µ (cP) 0.0104 0.6313

Mw 22.39 21.8 122.14

Table 21 Flash Results from Gas Produced at P =1,000 psia to 15 psia and 60oF


C1 81.021 81.498 0.473

C2 8.071 8.117 0.308

C3 4.322 4.344 0.645

C4 2.664 2.671 1.522

C5 1.516 1.506 3.213

C6 1.159 1.116 8.457

C7 0.513 0.439 13.029

C8 0.302 0.209 16.014

C9 0.178 0.074 17.733

C10+ 0.254 0.027 38.607

MwC10+ 153.8 137.8 155.7

Table 22 Gas and liquid compositions from gas produced at P =1,000 psi and flashed to 15 psia and 60 oF.




Mole % 100 99.587 0.413

V~ (ft³/mol) 376.045 377.594 2.317

ρ (lb/ft³) 0.062 0.061 43.372

Z 0.9914 0.9955 0.0061

µ (cP) 0.0103 0.4492

Mw 23.39 23.07 100.48

Table 23 Flash Results from Gas Produced at P = 500 psia to 15 psia and 60oF


C1 78.595 78.919 0.45

C2 8.406 8.44 0.317

C3 4.857 4.874 0.719

C4 3.244 3.25 1.844

C5 1.943 1.934 4.112

C6 1.485 1.446 10.906

C7 0.267 0.238 7.339

C8 1.2 0.9 74.3

C9 1.2 0.9 74.3

C10+ 1.2 0.9 74.3

MwC10+ 1.2 0.9 74.3

Table 24 Gas and liquid compositions from gas produced at P = 500 psi and flashed to 15 psia and 60 oF




Prove that the following expression is equivalent to Equation (121).

( )ElvvvE zfzfz )1(,2 −+=φ

The identity is valid for any depletion stage.


The following information has been obtained from a simulated CVD test on Jeffs’ gas condensate reservoir. The average reservoir temperature is 240oF. From geological data the estimated hydrocarbon pore volume is:

RBVHC7101×=

The standard molar volume in Jeff’s state is

molelbSCFVg −

= 00.375~

Properties at Jeff’s Condensate at the Dew Point:

2,714 psia and 240.0 oF

Molar Volume 2.073 ft³/lb-mol

Density 15.859 lb/ft³

Z-Factor 0.7495



Component Pressure zi% yi% xi%

C1 75.000 80.090 66.562

C5 15.000 12.551 19.060

C7+

2,600

10.0 7.4 14.4

C1 74.790 84.634 54.792

C5 15.101 10.378 24.696

C7+

2,200

10.1 5.0 20.5

C1 73.191 86.467 44.952

C5 15.868 9.559 29.288

C7+

1,800

10.9 4.0 25.8

C1 70.640 87.109 35.295

C5 17.080 9.415 33.533

C7+

1,400

12.3 3.5 31.2

C1 66.749 86.523 25.379

C5 18.892 10.073 37.342

C7+

1,000

14.4 3.4 37.3

C1 60.642 83.427 14.992

C5 21.615 12.478 39.923

C7+

600

17.7 4.1 45.1

Table 25 Constant Volume Depletion at 240.0 oF

Determine:

The number of moles and SCF of gas produced at each pressure stage.

The cumulative number of moles and SCF produced at each pressure stage.

The number of moles of liquid remaining in the reservoir at each pressure stage.

The fractional gas recovery at each pressure depletion stage.




The following information has been obtained from a simulated CVD test on Jeffs’ gas condensate reservoir. The average reservoir temperature is 240 oF and the initial number of gas moles in this reservoir is,

9101×=tn lb moles

Assuming that production starts at the Dew-Point pressure and that the abandonment pressure is 400 psia.

Determine: the fraction of moles recovered at that pressure.

Pressure (psi)

z-factor z-factor (2phase)

3,000 0.777

2,800 0.757

2,714 0.750 0.750

2,600 0.789 0.748

2,200 0.831 0.735

1,800 0.856 0.717

1,400 0.878 0.690

1,000 0.899 0.645

Table 26 - Z-factor of the Jeff’s gas condensate.



The conventional material balance equation works well for dry gas and black oil. However, if the reservoir fluid is a volatile oil or a gas condensate, a substantial error may occur due to neglecting compositional effects.

Gas Condensates and Volatile Oils

Gas Condensate/Volatile Oil - Production Facts As pressure declines below the dew point gas condensate reservoirs release liquid within the reservoir pore space (liquid dropout can be as high as 30 - 35% of the pore volume).

Dry gas viscosity correlations cannot reproduce the viscosity variation of gas condensates due to compositional effects.

Condensate and volatile reservoirs produce distillate oil from the vapor phase produced.

As the hydrocarbon composition becomes richer in intermediate components (C4 - C8), Rs, Bg, and Bo become more and more dependent upon the composition in addition to pressure and temperature.

These facts indicate that surface oil is also dissolved in the gas phase. One of the approaches to deal with these types of reservoir fluids will take into account the so called “volatilized oil” i.e. oil dissolved in the gas. Still this approach is simplistic and it can be considered as an improved black oil formulation.

Equations of State (EOS) can handle this problem provided that the initial reservoir fluid is known.

We will define a new PVT property that takes into account the volume of surface volatilized oil dissolved per unit volume of surface gas. This property is called Rv.

The Rv of a reservoir gas at P1,T1 (reservoir pressure and temperature) is obtained by expanding a fixed volume of this gas from to PSC, TSC. The volumes of surface gas and stock tank oil obtained are measured. And Rv is calculated as the number of STB of surface oil obtained per cubic feet of surface gas at standard conditions. This oil that appeared at surface was volatilized at reservoir conditions. For pressures above the saturation pressure Rv is just the inverse of Rs. But at pressures P < Psaturation it is not the same because we are considering the oil in the gas phase.



MMSCFSTBRv ==

conditions standardat gas surface offeet cubic of millionconditions standardat oil of barrels (126)

Reservoir Fluid Rv (STB / MMSCF)

Heavy and Black Oils 0 – 10

Volatile Oils 10 – 200

Near Critical Oils 150 – 400

Gas Condensates 50 – 250

Wet Gases 20 – 100

Dry Gases 0 – 2

Table 27 - Typical values of Rv.

Gas Condensate Description Gas condensates are complex mixtures of hydrocarbons which behave more like a vapor phase at the conditions of high pressure and temperature found in petroleum reservoirs. When such gas condensate reservoirs are brought on production, the reservoir pressure falls and the hydrocarbon phase behavior changes continuously. The successful exploitation of gas condensate reservoirs depends on knowledge of the vapor-liquid equilibrium composition of the phases and their volumetric behavior.

The phase envelope of these fluids shows that the reservoir temperature is located between the critical temperature and the cricondentherm temperature.

The producing gas oil ratios of gas condensates range from 4,000 to 150,000 SCF/STB. A sharp increase in GOR’s during production indicates that the reservoir pressure fell below the dew-point pressure and liquid is accumulating in the reservoir. Recycling of gases into gas condensate reservoirs at the beginning of production or after the pressure depletion is a potential method of enhanced condensate recovery.

Heavy hydrocarbon components with carbon numbers higher than C7 have a large influence on the phase behavior of the hydrocarbon mixtures. The magnitude of liquid dropout and the dew point pressure in gas condensates are both increased by the



amount of heavy hydrocarbons. Typically these fluids have a mole percentage of C7+ <

12.4%. (For more rules of thumb identification procedures see McCain, 1990).

The way gas condensate fields are depleted gives rise to significantly different recovery factors. Simple blowdown, full or partial pressure maintenance strategies are used depending on the nature of the fluid, the reservoir temperature and initial pressure and the reservoir geology. Efficient and economic production of gas condensates is of great importance and must be considered before deciding the production method in the development of new fields.

Condensate banking (or buildup) near the wellbore acts as a mobile skin. It lowers the gas relative permeability thereby the gas flow rates.

Gas condensate fields are a significant portfolio of most major oil producing companies. However, the planning of the development of such needs reliable predictions of the physical processes of the multiphase flow of the fluids in the porous medium. The prediction of flow profiles and the development of expressions in terms of relative permeability and phase saturation are needed for reservoir simulation.

The relative permeability of condensate fluids is a function of many parameters including viscosity ratio, wettability and most importantly interfacial tension. In general the effect of low interfacial tension is to increase flow rates and lower the residual saturations, creating conditions for improved hydrocarbon recovery.

Volatile Oils Description Volatile oils are complex mixtures of hydrocarbons which behave more like a liquid phase at the conditions of high pressure and temperature found in petroleum reservoirs. Using density, or typical compositions as dividing lines between gas condensate and volatile oils is very subtle. The phase envelope of volatile oils shows that the critical temperature is located to the right of the reservoir temperature. When volatile oils are brought on production and the reservoir pressure falls below the bubble point, the gas phase may become mobile.

The producing gas oil ratios of volatile oils range from 2,000 to 4,000 SCF/STB. A sharp increase in producing GOR’s (R’s) indicates that the reservoir pressure falls below the bubble-point pressure.

Volatile oils are also called high-shrinkage oils (Bo’s can be > 4) and should be produced through three or more separator stages to minimize this shrinkage.



Field data indicates that solution GOR’s (Rs’s) generally increase with depth, therefore the probability of a hydrocarbon accumulation to contain gas condensates or volatile oils usually increases with depth.

Typically volatile oils have a mole percentage of C7+ between 12 to 30%. (For more

rules of thumb identification procedures see McCain, 1990).

PVT data is essential for gas condensate and volatile oil field evaluation and development. PVT data consists of volumetric measurements and compositional information on gas and liquid at equilibrium. Routine PVT analysis normally consists of constant mass and differential liberation studies. A more realistic (and more expensive) PVT analysis for studying the depletion of a gas condensate (or a volatile oil) is a constant volume depletion (CVD) which will be discussed later.

The following table illustrates some typical compositions of the 5 different types of reservoir fluids.

Component Black

Oil Volatile

Oil Gas Condensate Wet Gas Dry Gas

C1 48.83 64.36 87.07 95.85 86.67

C2 2.75 7.52 4.39 2.67 7.77

C3 1.93 4.74 2.29 0.34 2.95

C4 1.60 4.12 1.74 0.52 1.73

C5 1.15 3.97 0.83 0.08 0.88

C6 1.59 3.38 0.60 0.12

C7+ 42.15 11.91 3.80 0.42

MwC7+ 225 181 112 157

GOR 625 2000 18,200 105,000 ≡

Tank oAPI 34.3 50.1 60.8 54.7 -

Liquid

Color

Greenish

Black

Medium

Orange

Light

Straw

Water

White -

Table 28 Typical compositions of single-phase reservoir fluids (undersaturated).



The following graphs show property changes that take place in a condensate reservoir (Simpon’s Field) as production takes place. The data have been simulated with an EOS. The reservoir temperature is 190 oF.

The condensate field “SIMPSON” is at 190oF and exhibits the phase behavior illustrated in the following pages.

0

1000

2000

3000

4000

Pres

sure

(psi)

0 100 200 300 400Temperature (F)

Critical Point

Tc = 103.7 F

Pc=2952.5 psi

Retrograde Condensate - CVD DepletionEOS-Simulation

Figure 24 Initial phase envelope for Gas Condensate Simpson.



0

10

20

30

40

50

60

70

Mol

e %

0

10

20

30

40

50

Mol

e %

0 1000 2000 3000 4000Pressure (psi)

C7+C6

C1

Compositional Variation of The Liquid Phase PETE 420 Condensate Simpson's Field at 190 F

Figure 25 Changes taking place in the liquid remaining in the reservoir during production by depletion.



62

64

66

68

70

72

Mol

e %

0

1

2

3

4

5

6

7

Mol

e %

0 1000 2000 3000 4000Pressure (psi)

C6C7+

C1

Compositional Variation of The Gas Phase PETE 420 Condensate Simpson's Field at 190 F

Figure 26 Changes taking place in gas produced as pressure declines below the dew-point pressure for a gas condensate reservoir.

Gas condensates and volatile oils show a greater change in composition and behavior during the production when compared with black oil and dry gas.

The following plot shows the liquid dropped out in the reservoir as a function of pressure. This liquid is expressed as a percent of the volume occupied at the dew-point (Vd).



0

10

20

30

40

Liqu

id D

ropo

ut (%

) Vd

0 1000 2000 3000 4000 5000Pressure

Liquid Dropout for PETE 420 Condensate Simpson's Field at 190 F

Dew Point Pressure

Figure 27 Liquid dropout for Simpson gas condensate.

A problem associated with gas condensates is the formation of retrograde condensation when production is by pressure depletion. Therefore, it is extremely important to understand the compositional variation and phase behavior of gas condensate at wide range of temperature and pressure.

Pressure/ Z vs Moles of Gas Produced. The more common plot of P vs SCF of gas produced is obtained by converting moles to SCF, using the ideal gas law and the standard conditions applied for the state (TX, CO, LA, etc.).



0

1000

2000

3000

4000

5000

6000

Pres

sure

/ Z

0 10 20 30 40 50 60 70 80 90 100Percent Production (Moles)

P/Z (2 phases)

P/Z (L))

P/Z (2 phases)P/Z (V))

Pressure / Z vs Production PETE 420 Condensate Simpson's Field at 190 F

Figure 28 - P/Z plots for condensate Simpson.

0.0

0.2

0.4

0.6

0.8

1.0

Z fa

ctor

0 1000 2000 3000 4000 5000Pressure (psi)

Z (liquid)Z (vapor)Z (total)

Dew Point Pressure

Z Factors vs Pressure PETE 420 Condensate Simpson's Field at 190 F

Figure 29 - Different z-factors in condensate Simpson.



0

10

20

30

40

50D

ensit

y (lb

m /

ft3)

0 1000 2000 3000 4000 5000Pressure (psi)

Gas DensityLiquid Density

Liquid and Gas Densities for PETE 420 Condensate - Simpson's Field at 190 F

Figure 30 -Liquid and gas densities for condensate Simpson.

10-2.0

10-1.0

100.0

Visc

osity

(cp)

0 1000 2000 3000 4000 5000Pressure (psi)

Liquid ViscosityTwo Phase ViscosityGas Viscosity

Viscosities for PETE 420 Condensate Simpson's Field at 190 F

Figure 31 - Liquid and gas viscosities for condensate Simpson.



Note the peculiar shape of gas viscosity versus pressure. This is entirely due to compositional effects. Dry gas viscosity correlations can not reproduce this behavior.

0.01

0.02

0.03

0.04

0.05

0.06

0.07

Visc

osity

(cp)

0 1000 2000 3000 4000 5000Pressure (psi)

Viscosity vs Pressure PETE 420 Condensate Simpson's Field at 190 F

Figure 32 - Gas Viscosity for condensate Simpson.



0

1

2

3

4

5

6

7

8

9Re

lativ

e V

olum

e

0 1000 2000 3000 4000 5000Pressure (psi)

Relative Volume for PETE 420 Condensate Simpson's Field at 190 F

Dew Point Pressure

Figure 33 - Relative volume for condensate Simpson.

0

1000

2000

3000

4000

Pres

sure

(psi)

-100 0 100 200 300 400 500 600Temperature (F)

Critical Point (initial)

Critical Point (when Pr = 1500 psi)

Compositional Changes due to Production PETE 420 Condensate Simpson's Field at 190 F

Initial Reservoir Pressure = 5000 psiActual Reservoir Pressure = 1500 psi

Figure 34 Initial phase envelope and phase envelope when the average reservoir pressure dropped to 1500 psia for condensate Simpson.



The phase envelopes indicated in Figure 34 correspond to the fluid remaining in the reservoir.

Conceptual questions All the questions refer to Figure 24 through Figure 34.

Conceptual Question 1. Based upon the phase envelope indicated in Figure 24. Indicate which condensate fluid would give drop more liquid in the reservoir during production up to P = 1000 psia and why. The reservoir temperatures are,

TR = 150 F

TR = 200 F

TR = 300 F

Conceptual Question 2. What causes the compositional variation of gas and liquid phases during production.

Conceptual Question 3. What type of error would you have in the predicted performance of your reservoir if you use either the gas or the liquid z-factor?

Conceptual Question 4. What causes the changes in gas and liquid viscosity during pressure depletion?

Conceptual Question 5. Explain what happened in Figure 34.

Conceptual Question 6. What is the major difference between a volatile oil and a condensate gas?

Suggested Exercise - Simulated CVD Test

The following output corresponds to a simulated CVD test for Lee’s condensate reservoir.



Z -Factor P

(psia)

Liquid Dropout

% Vd

Moles % of Gas

Produced z-gas z-two-phase Viscosity

(cp)

4,000 0.886 0.044

3,500 0.829 0.040

3,000 0.777 0.036

2,714 0.750 0.750 0.033

2,500 34.46 7.3 0.804 0.745 0.024

2,000 28.25 24.0 0.844 0.727 0.019

1,500 23.64 40.6 0.873 0.697 0.017

1,000 19.64 57.1 0.899 0.644 0.015

500 15.10 74.2 0.925 0.536 0.014

Vd = volume at the dew point in ft3.

Table 29 - Constant Volume Depletion at 240.0 oF of Lee's condensate reservoir.

R = 10.73 psia ft3/ oR lb-mole

Tsc = 60 oF

1 barrel = 5.614 ft3

Psc = 14.7 psia

Based upon the given information.

• Plot Pressure vs. the gas z-factor, and the two-phase z-factor.

• Determine the molar volume for this condensate at the dew point (in ft3 / lb-mole).

• If Lee’s reservoir hydrocarbon pore volume is estimated to be VHC = 1x107 RB. How many moles of gas condensate fit into this volume?



• Determine the total volume of liquid dropout in the reservoir at 2,500 psi.

• How many standard cubic feet of gas would be produced at 1,750 psi?



Evaluation of Gas and Oil in Place Using Surface Separation Data and EOS. The estimates of oil and gas in place from gas condensate fields using surface separation data and an EOS requires a series of “accounting steps” consisting of:

1. Flash calculations using different fluid compositions

2. Recombination of produced gas and liquid to determine the initial oil composition

3. Determination of fluid (gas and liquid) properties at standard, reservoir, and separator conditions.

This will be best presented through an example.

Example

Evaluate Oil and Gas in place using the following separator, production, and reservoir data.

Given:

PR = 4,000 psi

TR = 195 oF

φ = 0.2

Swc = 0.3

Tsc = 60oF

Psc = 15.025 psia

Stock Tank Production (STB)2 = 30 STB/day

Separator Gas Production (SCF)1 = 179,030 SCF/day



Separator 1 yi1

(STB)2

Well Head

Reservoir

xi1 xi2

yi2?

?

(SCF)1

(SCF)2?

Separator 2

Figure 35 - Sketch illustration two separator stages.

Additional information provided is the composition of the produced gas (from the separator) and the composition of the oil from the separator.



Component yi1 xi1

N2 0.160 -

CO2 0.220 -

C1 75.310 7.780

C2 15.080 10.020

C3 6.680 15.080

i-C4 0.520 2.770

C4 1.440 11.390

i-C5 0.180 3.520

C5 0.240 6.500

C6 0.110 8.610

C7+ 0.060 34.330

molelblb

molelblbMw

C

C

/ 70.0 and

/ 103

7

7

=γ

=

+

+

molelblb

molelblbMw

C

C

/ 79.0 and

/ 143

7

7

=γ

=

+

+

Table 30 - Gas and oil compositions from Separator 1.

Note that N2 and CO2 are not part of the liquid stream.

Step 1:

Determine the number of moles of gas and oil produced in the stock tank (separator 2).

We use an EOS and flash the liquid composition from separator 1 to TSC and PSC.




Mole % 100.000 42.446 57.554

V~ (ft³/mol) 160.564 374.993 2.427

V % 100.0 99.1 0.9

ρρρρ (lb/ft³) 0.516 0.111 46.659

z-Factor 0.4217 0.9849 0.0064

µ µ µ µ (cP) 0.00 0.0086 0.6044

Table 31 - Flash results of liquid from separator 1 at T=60 oF and P = 14.7 psia.

Component Total

Mole (%)

Vapor

Mole (%)

Liquid

Mole (%)

C1 7.78 8.196 0.098

C2 10.02 22.514 0.806

C3 15.08 29.866 4.176

i-C4 2.77 4.398 1.569

C4 11.39 15.56 8.315

i-C5 3.52 2.8 4.051

C5 6.5 4.151 8.232

C6 8.61 1.953 13.52

C7+ 4.33 0.562 59.234

Table 32 - Liquid and gas compositions from flashed Separator 1 liquid at T=60 oF and P = 14.7 psia.

The daily number of moles of gas (nv2) and liquid (no2) produced in the stock tank is,



22221 oivii fxfyz +=

daymolelbn

STBSCF

daySTB

SCFmolelb

VVn

o

o

STOo

3943.69

614.5 30 427.21

~

2

22

=

××

==

Since

22

22

vo

oo nn

nf+

=

Then

−×=2

222

1

o

oov f

fnn

daymolelbnv 1782.51

57554.057554.013943.692 =

−×=

Step 2:

Using the ideal gas law convert the daily number of moles of gas produced in the stock tank (nv2) to SCF.

Convert the (SCF)1 obtained in separator 1 to moles (nv1)

The molar volume of the gas produced at the specified standard conditions is:



lb moleSCF

psiaR

Rlb mole SCFpsia

PRTV

o

osc

scg 35.371

025.15 520 73.10~ =×==

lb moleSCFVg 35.371~ =

Alternatively, we can use the volume at these standard conditions but evaluated from the EOS. Note that these are pretty close (374.994 SCF/lb mole).

And the SCF produced are

molelbSCF

daymolelbVn

daySCF

gv 35.3711782.51~)(

22 ××=×=

daySCF

daySCF 25.005,19)( 2 =

( )daymolelb

SCFmolelb

daySCF

VSCFn

gv

11.482 35.371

030,179~ 1

1 =×

==

Step 3:

Determine the recombination ratio from moles of oil and gas produced and using this determine the initial gas composition (zi).

11

1

vo

vv nn

nf+

= but 221 ovo nnn +=



8.01782.5111.4823943.69

11.482

212

1 =++

=++

=vvo

vv nnn

nf

Thus the initial gas (zi) compositions are

oivii fxfyz 11 += i= 1, 2, …Nc

Example for methane

61804.00778.02.075310.08.01 =×+×=z

Component zi (%)

N2 0.128

CO2 0.176

C1 61.804

C2 14.068

C3 8.36

i-C4 0.97

C4 3.43

i-C5 0.848

C5 1.492

C6 1.81

C7+ 6.915

Table 33 - Overall overhead compositions.



Some round off error is inevitable therefore you will have to normalize all zi’s to make sure that their sum is equal to 1.

Note:

Many commercial software codes do this step for you given that you provide

(a) the recombination ratio expressed as number of moles the gas per number of moles of liquid.

41782.513943.69

10.482

1

1 =+

==o

v

nnratio

or

(b) The producing gas oil ratio expressed as SCF/STB

Step 4:

Using the initial gas composition obtained from Step 3 use the EOS to determine the z-factor at the reservoir PR and TR. You can also verify, by calculating the phase envelope, that the initial fluid is a condensate gas.

The calculated fluid properties at PR = 4,000 psia and TR = 195 oF are



V~ (ft³/lb- mole) 1.487

ρρρρ (lb/ft³) 22.605

z-Factor 0.8468

Viscosity (cP ) 0.0531

Mw (lb / lb mole) 33.62

Step 5:

Evaluate the total number of moles per ac-ft (1 ac-ft = 43,560 ft3)

)1(560,43 wiHC SV −×φ×=

ftacmolelbn

zRTSP

zRTVPn

t

R

wiR

R

HCRt

−=

+××−×××=

−φ××==

77.099,4)460195(73.108468.0

)3.0.1(2.0560,43000,4

)1(560,43

Step 6:

Flash these initial number of moles with the initial composition obtained in Step 3 to standard conditions, and determine the number of moles and (MSCF) and (STB) of gas and oil in place per acre-foot.

The total number of moles is:

ogt nnn +=




Mole % 100.000 92.327 7.673

V~ (ft³/mol) 348.497 377.231 2.739

V % 100.0 99.9 0.1

ρρρρ (lb/ft³) 0.096 0.067 48.788

z-Factor 0.9188 0.9945 0.0072

µ µ µ µ (cP) 0.0101 0.8583

Table 34 - Flash results at T= 60 oF and P=15.025 psia.

Therefore

ftacmolelbnfn tvg −

=×=×= 27.784,377.409992327.0

and

ftacmolelbnnn gto −

=−=−= 49.31527.784,377.099,4

Step 7:

Translate the number of moles of gas to SCF and the number of moles of liquid to STB.

ftacMSCF

molelbSCF

ftacmolelbVnG gg −

=

×

−=×= 76.404,1

21.371 27.784,3~

The initial oil in place in standard stock tank barrels is evaluated using the molar volume obtained in step 6

ftacSTBN

ftSTB

molelbSCF

ftacmolelbVnN oo

−=

×

×

−

=×=

92.153

614.5 739.2 49.315

614.5

~3



Step 8:

Evaluate the daily reservoir voidage and the daily condensate production.

The daily reservoir voidage is

( )

daySCFV

PRTznnn

PRTzn

VR

RRovv

R

RRp

97.896000,4

65573.108468.0)1782.513923.6929.482(

221

=××++=∆

++==∆

daySTBV 77.159

614.597.896 ==∆

And the daily condensate production is

ppp NGG +=∆

( )daySCF

zTPPTVGRRsc

Rscp 874,223

8468.0655025.15000,452097.896 =

××××=∆=∆

Suggested Exercise

Determine the initial composition of a condensate reservoir that is producing

40 STB/day with a separator daily gas production of 200,000 SCF day. The gas and oil compositions for the separator are the same as those used in this Example.



The Effect of Gravity and Porous Media on the Phase Equilibria of Hydrocarbon Reservoirs Generally, when calculating phase equilibria of reservoir fluid systems, the effects of a gravitational field and the curvature of the phase interface are neglected. Under certain conditions, compositional variation due to gravitational field, and the effect of capillarity on phase behavior may become important. In this chapter, the criteria of phase equilibrium both in a gravitational field, and for curved surfaces, will be derived. In the first part, the theory of phase equilibria of flat surfaces in the absence of the gravity field will be reviewed.

Equilibrium Conditions Under the Influence of Gravity

Closed System The work term in the first law of thermodynamics included only the compression expansion (PdV ) contribution. Now consider a mass m (either single or multicomponent) undergoing both expansion (or compression) and a change in position (Figure 36). To raise m to some height h, a certain amount of work must be done. Therefore, the expression for dW should include both expansion/compression and gravity contributions.

( )dhmgPdVdW t −= (127)



dQ

dW

dW

dQ (mg)dh positive upwards

2

1 m

m

dQ

dW

dW


2

1

dQ

dW

dW


2

1 m

m

Figure 36 - Work to lift mass from position 1 to 2 including gravitational effects.

Substituting Eq. (127) into the first law equation provides.

( )dhmgPdVTdSdU ttt +−= (128)

( )∑=

==

∂∂ c

t

N

iii

tVS

t gMwnmgh

U1,

(129)

Equation (128) indicates that the internal energy is a function of the independent variables St, Vt, and h.

( )hVSUU tttt ,,= (130)

Similar expressions can be found for the other state properties.



Let's examine the criterion of equilibrium for a single or multi-component closed system in a gravitational field. The Gibbs free energy should be a minimum, therefore, dGt should vanish subject to the constraint of constant temperature.

( ) 0=+= dh mgdP VdG tt (131)

gdhdP ρ−= (132)

This is the equation for a hydrostatic head. Note that in a gravitational field

tt TdSdH = (133)

Open System Recall that the system can exchange matter and energy with surroundings. For an open system in the normal gravity field,

( )hnnnVSUUcNtttt ,...,,,, 21= (134)

The differential form of Eq. (134) can be written as:

dhh

UdnnUPdVTdSdU

ttjjtt

c

VSn

ti

inhVS

N

i i

tttt

,,,,,1

∂∂+

∂∂+−=

≠=∑

(135)

But,

TSU

hnnVt

t =

∂∂

,,

(136)



and

PVU

hnnSt

t −=

∂∂

,,

(137)

( )∑=

==

∂∂ cN

iii

hnVnS

t gMwnmgh

U1,,

(138)

ghMwnU

iiinhVS

N

i i

t

jtt

c

+µ=

∂∂

≠=∑ ˆ

,,,1

(139)

Combining Eqs. (135) through (139)

( ) mgdhdnghMwPdVTdSdU i

N

iiittt

c

++µ+−= ∑=1

ˆ (140)

Similar expressions for dHt, dGt, and dFt can be derived.

The expression for dGt is:

( ) mgdhdnghMwdPVdTSdG i

N

iiittt

c

++µ++−= ∑=1

ˆ

(141)

We have already seen that in a normal gravity field the hydrostatic head for a single component single-phase fluid could be derived from the criteria of equilibrium. Now we would like to derive equilibrium criteria for a multicomponent single phase (i.e., dni = 0, i = 1, . . . nc). Let's apply the criteria of equilibrium to Eq. (140); dGt should vanish at equilibrium. Therefore following the same arguments as before, for arbitrary dni.

0ˆ =+µ ghMwii (142)

and



0=+ mgdhdPVt (143)

and the process should be isothermal. We can write Eq. (142) in a differential form and noting that the process is isothermal,

( ) 0ˆ =+µ Tii gdhMwd (144)

Both Eqs. (142) and (144) indicate that in a gravitational field the chemical potential in the same phase is not constant but varies with depth. Eq. (143) represents the hydrostatic head.

Conditions for Pronounced Compositional Variation in Hydrocarbon Reservoirs At constant temperature,

( )Ncii nnnP ,...,,,ˆˆ 21µ=µ (145)

khVSn

Nc

k k

iii dn

ndP

Pd

ttj ,,,1

ˆˆˆ ∑=

∂µ∂+

∂µ∂=µ (146)

iTn

i VP

=

∂µ∂

.

ˆ (147)

khVSn

Nc

k k

iii dn

ndPVd

ttj ,,,1

ˆˆ ∑=

∂µ∂+=µ (148)

Combining Eqs.(144) and (148) and rearranging,

( )gMwVdhdn

n iik

hVSn

Nc

k k

i

ttj

−ρ=

∂µ∂∑

= ,,,1

ˆ (149)



The above equation shows that compositional variation in a hydrocarbon reservoir could be expected when the term

( )gMwV ii −ρ (150)

is large and when the following term becomes singular

PTnk

i

jn

,,

ˆ

∂µ∂ (151)

Equation (150) for a component of a reservoir fluid system can be large if the molecular weight of one of the components is much larger than the average molecular weight of the reservoir fluid. This could be the case if asphalt materials with a molecular weight of 500 to 5000 are present in the crude residue.

The term in Eq. (151) become singular around the critical point. Therefore, we expect compositional grading and bubblepoint variation in an oil column when asphalt materials are present in the oil. Compositional variation in a gas column is enhanced when a gas condensate fluid is in the vicinity of the critical point.

Figure 37 shows schematically the variation of saturation pressure with depth in both the oil leg and the gas cap. Saturation pressure variation of Figure 37 are due to compositional grading. Kingston and Niko report considerable bubblepoint pressure variation for the Brent and Statfjord reservoirs of the North Sea.

The bubblepoint pressure gradient in the oil zone is reported to be 3.6 and 4.0 psi/ft for the Brent and Statfjord reservoirs, respectively.



Dep

th

GOCPressure

WOC

Dew Point

Bubble Point

Reservoir PressureSaturation Pressure

Figure 37 - Variation of saturation and reservoir pressure due to compositional gradients.

Next we will present some figures indicating the evaluated gravitational gradients effects on a Volatile Oil (Cuisiana Oil). Compositions of the C7

+ and C1 show the most pronounced differences.



0

10

20

30

40

50

60

70

80

Com

posit

ions

(%)

12000. 12500. 13000. 13500.Depth (feet)

C1C2-C6C7+

Compositional Gradient for Volatile Oil MAB1at T = 256 F

Figure 38 - Compositional gradient for Volatile Oil MAB1.

12000

12200

12400

12600

12800

13000

13200

13400

13600

Dep

th (f

eet)

5000. 5500. 6000.Pressure (psia)

Reservoir PressureSaturation Pressure

Reservoir Pressure and Saturation PressureGradient for Volatile Oil MAB1 at T = 256o F

Oil

Gas

Figure 39 - Reservoir pressure and saturation pressure gradient for Volatile Oil MAB1.



Figure 38 shows the variation of the C1, C2-C6, and C7+ content as a function of depth

for a Cuisiana Reservoir Fluid (Colombia). Figure 39 shows the variation of the saturation pressure (Dew or Bubble point) as a function of depth.

We will know describe a procedure to calculate compositional grading in hydrocarbon reservoirs.

Procedure for the Calculation of Compositional Grading Eq. (144) gives the condition of equilibrium under the influence of gravity in terms of chemical potential. It is more convenient to express the condition of equilibrium in terms of fugacity. The fugacity of a component " i" in a mixture is defined by,

( )Tii dPVfRTd =ˆln i=1,2, ….Nc (152)

1limˆlim0

=

=

→ Pyffi

iiP

(153)

( )Tii dPVd =µ (154)

Combining Eqs.(144), (152), and (154)

( )Tii gdhMwfRTd −=ˆln (155)

Integrating Eq. (155) from a reference depth of zero to h

−= gh

RTMwff io

ii expˆˆ i=1, 2,.., Nc (156)

This equation states that the fugacity of component "i" in a given phase is a function of position. In other words, given composition and pressure at the reference point, one should be able to calculate both the composition and pressure at any point. Variation of pressure with depth is automatically accounted for by Eq (155) Note that the phase rule in the presence of a gravity field is given by

3+−= pc NNF (157)



Given the composition and pressure at a reference point, the following equations are used to calculate pressure and composition at a desired depth.

( ) cio

iNcii ,...N, ighRTMwfyyyPTfF 21 0expˆ,...,,,,ˆ

21 ==

−−= (158)

Subject to the constraint equation

∑=

=Nc

iiy

1

1 (159)

The unknowns are pressure and composition at the desired depth. The elements of the Jacobian matrix in terms of composition need special care to be evaluated. As an example,

,, jyPTj

i

kyF

≠

∂∂ (160)

may have no physical meaning if all mole fractions with the exception of yj are held constant. One way to resolve this problem is to express mole fractions in terms of moles in the equation for fugacity Then,

( ) cio

iNcii ,...N, ighRTMwfnnnPTfF 21 0expˆ,...,,,,ˆ

21 ==

−−= (161)

In the calculation of Jacobians

nnNc

ii =∑

=1

(162)



1=

∂∂

≠in

l

l

lj

inn , 1=

∂∂

≠ in

v

v

vj

inn (163)

0=

∂∂

≠in

l

lj

vj

inn

, 0=

∂∂

≠in

v

vj

vj

inn

(164)

will be needed.

Further details will be discussed in the example problem.

Suggested Exercise – Evaluation of a Gravitational Gradient

Consider a mixture of methane (C1 ) and normal butane. (n-C4) in a gravitational

field. Use the PR-EOS to compute the composition of C1 and n-C4 system at

intervals of 1000, 2000, 3000, 4000, 5000, and 7000 ft. below the reference depth for the following cases:

a. At the reference depth, C1 and n-C4 compositions are 27.27 and 72.73 mole

percent, respectively. Pressure at the reference depth is 1300 psi. Temperature throughout the liquid column is assumed to be 220 ˚F.

b. At a given reference depth, C1 and n-C4 compositions are 88.88 and 11.12

percent, respectively. Pressure at the reference depth is 514 psi. Temperature throughout the gas column is assumed to be 160 ˚F.



Equilibrium Conditions for Curved Surfaces The interface between phases (say, gas and oil) in a porous media is not flat. Therefore, in the derivation of equilibrium conditions the influence of interface curvature should be taken into account. Our approach to account for the effect of interface curvature would be to modify the expression for the work term in Eq. (128) Consider a bubble of vapor in Figure 40 In the course of expansion process d expression for the bubble would be,

dAPdVdW t σ−= (165)

1 21 2

Figure 40 - Two stages of a growing bubble.

In the above equation, PdV is the contribution from expansion (or compression), and σdA is the work required to increase the surface area by dA.

Substituting Eq. (165) into Eq. (128) the expression for dUt for a curved interface is obtained.



dAPdVTdSdU ttt σ+−=

(166)

Therefore,

( )cNtttt nnnVSUU ...,,,, 21= (167)

( )AVSUU tttt ,,= (168)

Similar expressions for the enthalpy, Helmholtz, and Gibbs's energy and are provided below;

dAdPVTdSdH ttt σ++= (169)

dAPdVdTSdF ttt σ+−−= (170)

dAdPVdTSdG ttt σ++−= (171)

Surface Effects on Phase Equilibria In this section, the criteria of equilibrium for a multicomponent two-phase system with a curved interface will be derived. The case of a single-component two-phase system is a special case.

Consider a gas bubble surrounded by a liquid phase in a constant volume, closed system of constant temperature. In each phase nc components will be present. The pressure inside the bubble is P' and the liquid phase pressure is P" (as indicated in Figure 41). The volume of the gas bubble and liquid phase are V' and V", respectively. For the multicomponent system, the expression for the differential of total internal energy is,



P'P’

P”

P'P’

P”

Figure 41 - A bubble with pressure P’ surrounded by a liquid at P”.

"'

tt dUdUdUt += (172)

dAdndVPTdSdndVPTdSdU i

N

iitti

N

iittt

cc

σ+µ+−+µ+−= ∑∑==

''

1

'''''''''

1

'''' ˆˆ (173)

and the expression for the total Helmholtz free energy is,

( ) dAdndndVPdVPdTSSdF i

N

iii

N

iittttt

cc

σ+µ+µ+−−+−= ∑∑==

''

1

'''

1

'''''''''' ˆˆ (174)

The necessary condition for the system to be in equilibrium is that dFt should vanish subject to the constraints of constant temperature and,

constant

constant"'

"'

=+

=+

ii

tt

nn

VV (175)

Therefore,



( ) ( ) ( ) 0ˆˆ '

1

'''''''''' =σ+µ−µ+−−+−= ∑=

dAdndVPPdTSSdF i

N

iiitttt

c

(176)

Previously we found that at equilibrium the chemical potentials must be the same, then:

( )tdV

dAPP '"' σ=− (177)

If we assume the gas bubble has a spherical shape, then,

24 rA π= and rdrdA π= 8 (178)

3

34 rVt π= and drrdVt 24π= (179)

Combining Eqs. (177), (178) and (179),

( )r

PP σ=− 2"' (180)

In the case of a curved interface, in addition to equality of chemical potentials, phase pressures are related by Eq. (180). The chemical potential should be evaluated at the corresponding phase pressures. Eq. (180) is the famous equation for capillary pressure (Young-Laplace).

In the following, we will discuss reservoir fluids phase equilibrium in porous media. A flat interface has an infinite radius of curvature, therefore the capillary pressure is zero.



Effect of Porous Media on Phase Behavior Several authors have studied the effect of porous media on the phase behavior of reservoir fluid systems. Russian authors Trebin and Zadora reported a strong influence of the porous media on the Dewpoint pressure and VLE of gas condensate systems. The porous media used by these authors was a silica sand mixture (0.300 to 0.215 mm diameter) grinned by a special cutter-pulverizer. Three different packings with permeabilities of 5.6, 0.612, and 0.111 darcy and porosities of 34, 31.4, and 29.8 percent were used. The calculated surface areas for these packings were 563, 1,307, and 3,415 cm2/cm3, respectively. Fig. 10 shows the effect of the type of porous media on the liquid content of the produced fluids. This figure shows that as the surface area of the porous media increases, the Dewpoint pressure increases. Tribune and Zadora state a 10 to 15 percent increase in the Dewpoint pressure in the porous media of the type that was used in their work. When these authors increased the temperature, the effect of porous media on VLE was decreased.

Tint and Raynal measured the bubble-point pressure of two reservoir crudes in both an open space (PVT cell) and a porous media with grain sizes in the range of 16-200 microns. The bubblepoint pressure of these two crudes were higher in porous media than in a PVT cell by 7 and 4 kg-cm. As an example, the bubblepoint pressure of one of the two crudes measured at 80˚C in a PVT cell was 121 kg/cm2 and the bubblepoint pressure at the same temperature in a porous media of 160-200 microns was 128 kg/cm2. On the other hand, when these authors used a mixture of methane and n-heptane they observed no differences in the saturation pressure. Sigmund and co-authors have also investigated the effect of porous media on phase behavior of model fluids. Their measurements on dewpoint and bubblepoint pressures showed no effect of porous media. The fluid systems used by these authors were C1 /n-C4, and C1/n-C5. The smallest bead size used was 3-40 US mesh.

Since these results seem quite controversial. In the example problem to be presented next, the effect of surface curvature on dewpoint pressure and equilibrium phase composition will be discussed.

Example – Effect of Curvature Radius on Saturation Properties

Consider a binary system of C1/n-C10 with the following composition at 100˚F.



x1 = 99.894%

x10 = 0.106%

a. Compute the dewpoint pressure of the above system in a PVT cell (measured Pd # 1450 psi)

b. What would be the dewpoint pressure of the above fluid system for a mean radius of curvature of 10,1, 0.1, and 0.01 microns?

c. What are the equilibrium liquid phase composition of the above cases?

Data

σ (C1/n-C10) at 100˚F & 1500 psi = 9.76 dynes/cm.

In conclusion, we would like to state that the effect of porous media on phase behavior should be attributed to:

1) interface curvature (i.e., capillarity), and

2) adsorption on the solid surface.

The effect of surface curvature has been taken into account in the derivation of Eqs. (176) through (179). The effect of adsorption is difficult to quantify and we may not be able to separate adsorption phenomena and capillarity. If only interface curvature is taken into account, one may expect that VLE will not be affected by the presence of porous media unless the radius of curvature becomes of the order of 0.1 microns and less (which is equivalent to permeabilities less than 1 µdarcy)

The following figures show the computed results for this exercise using PR EOS.



Figure 42 - Dewpoint pressure as a function of interface curvature for C1/nC10 system.

Figure 43 - Liquid phase composition at the dewpoint as a function of interface curvature.



Note that both: bubble point & dew point will change with curvature.



Nomenclature A = surface area

Ft = Helmholtz free energy

Gt = Gibbs free energy

if = fugacity of component "i"

g = acceleration of gravity

Ht = enthalpy

m = mass of the system

Mwi = molecular weight component "i"

n = moles - with subscript is for component; without is total,

nt = number of components

P = pressure

Q = heat

R = universal gas constant

r = radius of a gas bubble

St = entropy

T = temperature

Ut = internal energy

iV = partial molar volume component "i"

V = volume

W = work

h = elevation

xi,yi = mole fraction

σ = surface tension

ρ = mass density



π = number of phases

Subscripts i,j,k = component index

Superscripts ', '' = phase index

α,β,γ..= phase index



References

• McCain, W. D. Jr. The Properties of Petroleum Fluids, 2nd Edition – Penn Well Books (1990).

• McCain, W. D. Jr. “Calculation of Fluid Properties from Black Oil PVT Reports Revisited” SPE 36017 (1996).

• Barrufet, M.A., Habiballah, W. A., Liu K., and Startzman, R., “A Warning on the Use of Composition Independent K-Value Correlations to Solve Reservoir Engineering Problems.” J. of Petroleum Science and Engineering 14, (1995), pp. 25-34.

• Kartoadmodjo, T. and Schmidt, Z., “Large Data Bank Improves Crude Physical Property Correlations”, Oil and Gas Journal (July 1994), 51-55.

• Walsh P. M., “New Improved Equation Solves for Volatile, Condensate Reserves”. Oil and Gas Journal, Aug. 22, 1994, pp. 72-76.

• Walsh P. M., “New Improved Equation Solves for Volatile, Condensate Reserves. Oil and Gas journal, Aug. 22, 1994, pp. 72-76

• Havlena, D., and Odeh A. S., “The Material Balance as an Equation for a Straight Line”, JPT, August 1963, pp. 896-900.

• Spivak, A., and Dixon, T. N. “Simulation of Gas Condensate Reservoirs”, SPE 4271, 1973, SPE Symposium of Reservoir Simulation, Houston, Jan. 10-12, 1973.

• McCain W. D., “The Properties of Petroleum Fluids”. PennWell Books. (1990).

• “Thermodynamics of Hydrocarbon Reservoirs” Abbas Firoozabadi. McGraw Hill 1. (1999).

• "The Collected Works of J. Willard Gibbs," Vol. I- Thermodynamics, Yale University Press, New Haven, 1957.

• Kingston, P.E., and Niko, H.: "Development Planning of the Brent Field," JPT (Oct. 1975) 1190-98.

• Creek, J.L., and Schrader, M.L.: "East Painter Reservoir: An Example of a Compositional Gradient from a Gravitational Field," SPE 14411, paper presented at the 60e Annual Technical Conference and Exhibition of the Society of Petroleum Engineers, Las Vegas, Nevada, Sept. 22-25, 1985.



• Tribune, F.A., and Zadora, G.I.: "Experimental Study of the Effect of a Porous Media on Phase Changes in Gas Condensate Systems," Neft'i Gaz (1968) Vol. 81, No. 8,37-41.

• Tint, R., and Raynal, M.: "Are Test-Cell Saturation Pressures Accurate Enough?" The Oil and Gas Journal (Dec. 1966) 128-139.

• Sigmund, P.M., e a.: "Retrograde Condensation in Porous Media," SPEJ (April 1973) 93-104.

• Stegemeier, G.L., e a.," Interfacial Tension of the Methane-Normal Decane System," SPEJ (Sept. 1962) 257-260.

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