Chapter 2 Chapter 2

Volumetric gas Reservoir EngineeringVolumetric gas Reservoir Engineering

11

Gas PVTGas PVT

Gas is one of a few substances whose state, as Gas is one of a few substances whose state, as defined by pressure, volume and temperature defined by pressure, volume and temperature (PVT)(PVT)

One other such substance is saturated steam.One other such substance is saturated steam.

22

The equation of state for gasThe equation of state for gas Classical ideal gas lawClassical ideal gas law ClassicalClassical non-ideal gas law non-ideal gas law

Cubic Equations of State Cubic Equations of State van der Waals equation of state van der Waals equation of state Redlich-Kwong equation of state Redlich-Kwong equation of state Soave modification of Redlich-Kwong Soave modification of Redlich-Kwong Peng-Robinson equation of statePeng-Robinson equation of state Elliott, Suresh, Donohue equation of state Elliott, Suresh, Donohue equation of state Non-cubic Equations of State Non-cubic Equations of State Dieterici equation of state Dieterici equation of state

Virial Equations of State Virial Equations of State Virial Equations of StateVirial Equations of State The BWRS equation of state The BWRS equation of state

Other Equations of State of Interest Other Equations of State of Interest Stiffened equation of state Stiffened equation of state Ultrarelativistic equation of state Ultrarelativistic equation of state Ideal Bose equation of stateIdeal Bose equation of state

33

The equation of state for an ideal gasThe equation of state for an ideal gas

(Field units used in the industry)(Field units used in the industry) p [=] psia; V[=] ftp [=] psia; V[=] ft33; T [=] ; T [=] OOR absolute temperatureR absolute temperature

n [=] lbn [=] lbmm moles; n=the number of lb moles; n=the number of lbmm moles, one lb moles, one lbmm mole is mole is

the molecular weight of the gas expressed in pounds.the molecular weight of the gas expressed in pounds. R = the universal gas constant R = the universal gas constant

[=] 10.732 psia∙ ft[=] 10.732 psia∙ ft33 / (lb / (lbmm mole∙ mole∙00R)R)

Eq (1.13) results from the combined efforts of Boyle, Charles, Eq (1.13) results from the combined efforts of Boyle, Charles, Avogadro and Gay Lussac.Avogadro and Gay Lussac.

)13.1(nRTpV

44

2m

2m

In other book1 Darcy = ０ .986927×10-8 cm2

1 Darcy = ０ .986927×10-

12 m2

Note: In this text 1 Darcy = 1.0133×10-8 cm2

or 1 Darcy 10-8 cm2

or 1 Darcy

10-12 m2 = 1

1 Darcy

1

Non-ideal gas lawNon-ideal gas law

Where z = z-factor =gas deviation factorWhere z = z-factor =gas deviation factor =supercompressibility factor=supercompressibility factor = compressibility factor= compressibility factor

)15.1(nzRTpV

PandTatgasofmolesnofvolumeIdeal

PandTatgasofmolesnofvolumeActual

V

Vz

i

a

),,( ncompositioTPfz )1( airgravityspecificncompositio g

77

Determination of z-factorDetermination of z-factor

There are three ways to determine z-factor :There are three ways to determine z-factor :

(a)Experimental determination(a)Experimental determination

(b)The z-factor correlation of standing and(b)The z-factor correlation of standing and

katz katz

(c)Direct calculation of z-factor(c)Direct calculation of z-factor

88

(a) Experimental determination(a) Experimental determination n moles of gas n moles of gas

p=1atm; T=reservoir temperature; => V=Vp=1atm; T=reservoir temperature; => V=V00

pV=nzRTpV=nzRT z=1 for p=1 atm z=1 for p=1 atm =>14.7 V=>14.7 V00=nRT=nRT

n moles of gas n moles of gas

p>1atm; T=reservoir temperature; => V=Vp>1atm; T=reservoir temperature; => V=V pV=nzRTpV=nzRT pV=z(14.7 VpV=z(14.7 V00))

By varying p and measuring V, the isothermal z(p) function can beBy varying p and measuring V, the isothermal z(p) function can be readily by obtained.readily by obtained.

07.14 V

pVz

0

0

Vp

pVz

zT

pV

Tz

Vp

scsc

sc

99

(b)The z-factor correlation of standing and katz(b)The z-factor correlation of standing and katz Requirement: Requirement: Knowledge of gas composition or gas gravityKnowledge of gas composition or gas gravity Naturally occurring hydrocarbons: primarily Naturally occurring hydrocarbons: primarily

paraffin series Cparaffin series CnnHH2n+22n+2

Non-hydrocarbon impurities: CONon-hydrocarbon impurities: CO22, N, N22 and H and H22SS

Gas reservoir: lighter members of the paraffin series, CGas reservoir: lighter members of the paraffin series, C11

and Cand C22 > 90% of the volume. > 90% of the volume.

1010

The Standing-Katz CorrelationThe Standing-Katz Correlation

knowing Gas composition (nknowing Gas composition (nii)) Critical pressure (PCritical pressure (Pcici)) Critical temperature (TCritical temperature (Tcici) of each component ) of each component ( Table (1.1) and P.16 ) ( Table (1.1) and P.16 ) Pseudo critical pressure (PPseudo critical pressure (Ppcpc) ) Pseudo critical temperature (TPseudo critical temperature (Tpcpc) for the mixture) for the mixture

Pseudo reduced pressure (PPseudo reduced pressure (Pprpr) ) Pseudo reduced temperature (TPseudo reduced temperature (Tprpr) )

Fig.1.6; p.17 Fig.1.6; p.17 z-factor z-factor

iciipc

iciipc

TnT

PnP

pcpr P

PP

).(IsothermalconstT

TT

pcpr

1111

1212

(b’)The z-factor correlation of standing and katz(b’)The z-factor correlation of standing and katz

For the gas composition is not available and the gas gravity For the gas composition is not available and the gas gravity (air=1) is available.(air=1) is available.

The gas gravity (air=1)The gas gravity (air=1) ( ) ( ) fig.1.7 , p18fig.1.7 , p18

Pseudo critical pressure (PPseudo critical pressure (Ppcpc))

Pseudo critical temperature (TPseudo critical temperature (Tpcpc))

g

1313

(b’)The z-factor correlation of standing and katz(b’)The z-factor correlation of standing and katz

Pseudo reduced pressure (PPseudo reduced pressure (Pprpr))

Pseudo reduced temperature (TPseudo reduced temperature (Tprpr))

Fig1.6 p.17Fig1.6 p.17

z-factorz-factor The above procedure is valided only if impunity (COThe above procedure is valided only if impunity (CO22,N,N22 and and

HH22S) is less then 5% volume.S) is less then 5% volume.

pcpr P

PP

).(IsothermalconstT

TT

pcpr

1414

(c) Direct calculation of z-factor(c) Direct calculation of z-factor The Hall-Yarborough equations, developed using the Starling-Carnahan The Hall-Yarborough equations, developed using the Starling-Carnahan

equation of state, areequation of state, are

where Pwhere Pprpr= the pseudo reduced pressure= the pseudo reduced pressure

t=1/Tt=1/Tprpr ; T ; Tprpr=the pseudo reduced temperature =the pseudo reduced temperature y=the “reduced” density which can be obtained as the y=the “reduced” density which can be obtained as the

solution of the equation as followed:solution of the equation as followed:

This non-linear equation can be conveniently solved for y using the simple This non-linear equation can be conveniently solved for y using the simple Newton-Raphson iterative technique.Newton-Raphson iterative technique.

)20.1(06125.0

2)1(2.1

y

tePz

tpr

2323

432)1(2.1 )58.476.976.14(

)1(06125.0

2

yttty

yyyyteP t

pr

)21.1(0)4.422.2427.90( )82.218.2(32 tyttt

1515

(c) Direct calculation of z-factor(c) Direct calculation of z-factor The steps involved in applying thus are:The steps involved in applying thus are:

making an initial estimate of ymaking an initial estimate of ykk, where k is an iteration counter (which in , where k is an iteration counter (which in this case is unity, e.q. ythis case is unity, e.q. y11=0.001=0.001

substitute this value in Eq. (1.21);unless the correct value of y has been substitute this value in Eq. (1.21);unless the correct value of y has been initially selected, Eq. (1.21) will have some small, non-zero value Finitially selected, Eq. (1.21) will have some small, non-zero value Fkk. .

(3) using the first order Taylor series expansion, a better (3) using the first order Taylor series expansion, a better estimate of y can be determined asestimate of y can be determined as

wherewhere

(4) iterating, using eq. (1.21) and eq. (1.22), until satisfactory (4) iterating, using eq. (1.21) and eq. (1.22), until satisfactory convergence is obtained(5) substitution of the correct value of y inconvergence is obtained(5) substitution of the correct value of y in eq.(1.20)will give the z-factor. eq.(1.20)will give the z-factor. (5) substituting the correct value of y in eq.(1.20)will give the z-factor. (5) substituting the correct value of y in eq.(1.20)will give the z-factor.

)22.1(1

dydF

Fyy

k

kkk

yttty

yyyy

dy

dF k

)16.952.1952.29()1(

4441 324

432

)23.1()4.422.2427.90)(82.218.2( )82.218.1(32 tytttt

1616

1717

The equation of state for real gasThe equation of state for real gas

The equation of Van der WaalsThe equation of Van der Waals (for one lb mole of gas(for one lb mole of gas

where a and b are dependent on the nature of the gas.where a and b are dependent on the nature of the gas. The principal drawback in attempting to use eq. The principal drawback in attempting to use eq.

(1.14) to describe the behavior of real gases (1.14) to describe the behavior of real gases encountered in reservoirs is that the maximum encountered in reservoirs is that the maximum pressure for which the equation is applicable is still pressure for which the equation is applicable is still far below the normal range of reservoir pressures far below the normal range of reservoir pressures

)14.1())((2

RTbVV

ap

1818

Peng-Robinson equation of Peng-Robinson equation of state state

where, ω is the acentric factor of the species R is the universal gas constant.

1919

2020

Peng-Robinson equation of Peng-Robinson equation of statestate

The Peng-Robinson equation was developed in 1976 in order to satisfy the following goals:[3]

1. The parameters should be expressible in terms of the critical properties and the acentric factor.

2. The model should provide reasonable accuracy near the critical point, particularly for calculations of the Compressibility factor and liquid density.

3. The mixing rules should not employ more than a single binary interaction parameter, which should be independent of temperature pressure and composition.

4. The equation should be applicable to all calculations of all fluid properties in natural gas processes.

For the most part the Peng-Robinson equation exhibits performance similar to the Soave equation, although it is generally superior in predicting the liquid densities of many materials, especially nonpolar ones. The departure functions of the Peng-Robinson equation are given on a separate article.

Application of the real gas equation of stateApplication of the real gas equation of state Equation of state of a real gasEquation of state of a real gas This is a PVT relationship to relate surface to reservoir volumes of This is a PVT relationship to relate surface to reservoir volumes of

hydrocarbon.hydrocarbon.(1) the gas expansion factor E,

Real gas equation for n moles of gas at standard conditionsReal gas equation for n moles of gas at standard conditions

Real gas equation for n moles of gas at reservoir conditions Real gas equation for n moles of gas at reservoir conditions

>>

> surface volume/reservoir volume > surface volume/reservoir volume [=] SCF/ft[=] SCF/ft3 3 or STB/bblor STB/bbl

)15.1(nzRTpV

conditionsreservoiratgasofmolesnofvolume

conditionsdardsatgasofmolesnofvolume

V

VE sc tan

scscscsc RTnzVp sc

scscsc p

RTnzV

nzRTpV p

nzRTV

)1:(7.14

6.519

scsc

sc

sc

scscsc

scsc

sc znotezTzTp

pT

nzRTp

pRTnz

pnzRT

pRTnz

V

VE

][35.35 zT

pE

2121

ExampleExample

Reservoir condition: Reservoir condition: P=2000psia; T=180P=2000psia; T=1800F=(180+459.6)=639.60R; z=0.865F=(180+459.6)=639.60R; z=0.865 >>

surface volume/reservoirsurface volume/reservoir or SCF/ftor SCF/ft33 or STB/bbl or STB/bbl

8.1276.639865.0

200035.35

E

iwi ESVOGIP )1(

2222

2323

2424

(2) Real gas density(2) Real gas density

where n=moles; M=molecular weight)where n=moles; M=molecular weight)

at any p and Tat any p and T

For gasFor gas

For airFor air

Vm V

nM

V

m

zRT

MP

pnzRT

nM

RTz

PM

gas

gasgas

RTz

PM

gas

gasgas

RTz

pM

air

airair

air

air

gas

gas

air

gas

gas

gas

gair

gas

ZM

ZM

RTzpM

RTzpM

air

gas

g

ZM

ZM

)(

)(

2525

(2) Real gas density(2) Real gas density

At standard conditions zAt standard conditions zairair = z = zgasgas = 1 = 1

in generalin general

(a) If is known, then or , (a) If is known, then or ,

(b) If the gas composition is known, then (b) If the gas composition is known, then

where where

air

gas

g

ZM

ZM

)(

)(

)28.1(97.28

gas

air

gasg

air

gas M

M

M

8.0~6.0g

g 97.28 ggasM airggas

i

iigas MnM

97.28gas

g

M

airggas

30763.0)(ft

lbmscair

2626

(3)Isothermal compressibility of a real gas(3)Isothermal compressibility of a real gas

nzRTpV ))(:(1 pfznotenRTzpp

nzRTV

p

znRTppnRTz

p

V

12 ][

p

z

p

nRT

p

nzRT

p

V

2

)11

()11

(p

z

zpV

p

z

zpp

nzRT

p

V

)]11

([11

p

z

zpV

Vp

V

VCg

p

z

zpCg

11

pCg

1

since p

z

zp

11

p.24, fig.1.9

2727

2828

Exercise 1.1Exercise 1.1 - Problem - Problem

Exercise1.1 Gas pressure gradient in theExercise1.1 Gas pressure gradient in the

reservoirreservoir (1) Calculate the density of the gas, at (1) Calculate the density of the gas, at

standard conditions, whose standard conditions, whose

composition is listed in the table 1-1.composition is listed in the table 1-1. (2) what is the gas pressure gradient in(2) what is the gas pressure gradient in

the reservoir at 2000psia andthe reservoir at 2000psia and

1800F(z=0.865)1800F(z=0.865)

2929

3030

3131

Exercise 1.1 -- solutionExercise 1.1 -- solution -1 -1 (1) Molecular weight of the gas(1) Molecular weight of the gas

sincesince

or from or from

At standard conditionAt standard condition

i

iigas MnM 15.19 661.097.28

15.19

97.28 gas

g

M

airggasair

gasg

)(0504.0)(0763.0661.0 33 ftlbmftlbmgas

nzRTpV nMzRTpVM mzRT

zRT

pM

V

m

)(0505.06.51973.101

15.197.14 3ftlbmRTz

MP

scsc

scgas

3232

Exercise 1.1 -- solutionExercise 1.1 -- solution -2 -2

(2) gas in the reservoir conditions(2) gas in the reservoir conditions

nzRTpV nMzRTpVM mzRT

)(451.6)1806.459(73.10865.0

15.192000 3ftlbmzRT

pM

V

m

3333

Exercise 1.1 -- solutionExercise 1.1 -- solution -3 -3

gDp gdDdp

gdD

dp 232.32)

2.32

1451.6(

sft

lbm

slug

ft

lbm

23451.6

s

ft

ft

slug

3451.6

ft

lb f

2

2

2 144

11451.6

in

ft

ftft

lbf

ftin

lb f 10448.0

2

ftpsi0448.0

3434

Fluid Pressure RegimesFluid Pressure Regimes

The total pressure at any depth The total pressure at any depth

= weight of the formation rock = weight of the formation rock

+ weight of fluids (oil, gas or water)+ weight of fluids (oil, gas or water)

~ 1 psi/ft * depth(ft)~ 1 psi/ft * depth(ft)

3535

Fluid Pressure RegimesFluid Pressure Regimes

Density of sandstoneDensity of sandstone

3

3

3 )1(

)1003048.0(

1000

2.27.2

ft

cm

gm

lbm

cm

gm

lbm

slug

ft

lbm

7.32

1202.168

3

322.5

ft

slug

3636

Pressure gradient for sandstonePressure gradient for sandstone

Pressure gradient for sandstonePressure gradient for sandstone

gD

p

gDp

3084.1682.3222.5

ft

lbf

)/(16.1144

1084.168

22

2

2ftpsi

ftin

lbf

in

ft

ftft

lbf

3737

Overburden pressureOverburden pressure

Overburden pressure (OP)Overburden pressure (OP) = Fluid pressure (FP) + Grain or matrix pressure (GP)= Fluid pressure (FP) + Grain or matrix pressure (GP)

OP=FP + GPOP=FP + GP

In non-isolated reservoir In non-isolated reservoir PW (wellbore pressure) = FPPW (wellbore pressure) = FP

In isolated reservoir In isolated reservoir PW (wellbore pressure) = FP + GP’PW (wellbore pressure) = FP + GP’ where GP’<=GPwhere GP’<=GP

3838

Normal hydrostatic pressureNormal hydrostatic pressure

In a perfectly normal case , the water pressure at any depthIn a perfectly normal case , the water pressure at any depth Assume :(1) Continuity of water pressure to the surfaceAssume :(1) Continuity of water pressure to the surface (2) Salinity of water does not vary with depth.(2) Salinity of water does not vary with depth. [=] psia [=] psia

psi/ft for pure waterpsi/ft for pure water psi/ft for saline waterpsi/ft for saline water

7.14)( DdD

dPP water

4335.0)( waterdD

dP

4335.0)( waterdD

dP

3939

Abnormal hydrostatic pressure Abnormal hydrostatic pressure ( No continuity of water to the surface)( No continuity of water to the surface)

[=] psia[=] psia

Normal hydrostatic pressureNormal hydrostatic pressure

c = 0 c = 0

Abnormal (hydrostatic) pressureAbnormal (hydrostatic) pressure c > 0 → Overpressure (Abnormal high pressure)c > 0 → Overpressure (Abnormal high pressure) c < 0 → Underpressure (Abnormal low pressure)c < 0 → Underpressure (Abnormal low pressure)

CDdD

dPP water 7.14)(

4040

Conditions causing abnormal fluid pressuresConditions causing abnormal fluid pressures

Conditions causing abnormal fluid pressures in enclosed water Conditions causing abnormal fluid pressures in enclosed water bearing sands includebearing sands include

Temperature change ΔT = +1 → ΔP = +125 psi in a ℉Temperature change ΔT = +1 → ΔP = +125 psi in a ℉sealed fresh water systemsealed fresh water system

Geological changes – uplifting; surface erosionGeological changes – uplifting; surface erosion Osmosis between waters having different salinity, the Osmosis between waters having different salinity, the

sealing shale acting as the semi permeable membrane in sealing shale acting as the semi permeable membrane in this ionic exchange; if the water within the seal is more this ionic exchange; if the water within the seal is more saline than the surrounding water the osmosis will cause saline than the surrounding water the osmosis will cause the abnormal high pressure and vice versa.the abnormal high pressure and vice versa.

4141

Are the water bearing sands abnormally Are the water bearing sands abnormally pressured ?pressured ?

If so, what effect does this have on the extent of any If so, what effect does this have on the extent of any

hydrocarbon accumulations?hydrocarbon accumulations?

4242

Hydrocarbon pressure regimesHydrocarbon pressure regimes

In hydrocarbon pressure regimesIn hydrocarbon pressure regimes

psi/ftpsi/ft

psi/ftpsi/ft

psi/ft psi/ft

45.0)( waterdD

dP

35.0)( oildD

dP

08.0)( gasdD

dP

4343

4444

Pressure Kick – Oil and WaterP(psia)

5000

5200

55005600

oil

water

OWCD=5500ft

Pw=2265 Po=2315

Pw=2355 Po=2385

Pw=Po=2490Pw=2535

Depth(ft)

psia 2265 155000*45.0)5000 (psia 2315 5655000*35.0)5000 (

psia 2355 155200*45.0)5200 (psia 23856555200*35.0)5200 (

zone oilin 565D*0.35P5655500*0.35-2490Cor

*35.0 2490) 5500 (

psia 2490 155500*45.0) 5500 (psia 2535 155600*45.0)5600 (

][ 15*45.0

o

o

ftDatPftDatPftDatPftDatP

CDOWCatorftDatPOWCatorftDatP

ftDatPzonewaterinpsiaDP

w

o

w

o

o

o

w

w

w

4545

pressure kick-gas and water

psiaftDatP

psiaftDatP

psiaftDatP

psiftDatP

zonegasinDP

Cor

CD

GWCftDatP

psiaGWCftDatP

psiaftDatP

zonewaterinDP

w

g

w

g

g

g

g

g

w

w

w

2265)5000(

245020505000*08.0)5000(

2355)5200(

246620505200*08.0)5000(

2050*08.0

20505500*08.02490

*08.0

2490)5500(

2490)5500(

2535)5600(

15*45.0

Gas

P(psia)5000

5200

55005600

Pw=2265 Pg=2450

Pg=2466Pw=2355

Pw=Pg=2490Pw=2535

Depth(ft)

GWC

water

D=5500ft

4646

pressure kick-gas, oil and water

psiaftDatp

psiaftDatp

zoneoilinDp

Cor

CD

psiaOWCftDatp

psiaOWCftDatp

psiftDatp

zonewaterinDp

w

o

o

o

o

o

w

w

w

2445155400*45.0)5400(

24555655400*35.0)5400(

565*35.0

565

*35.0

2490)5500(

2490)5500(

2535)5600(

15*45.0

psiaftDatp

psiaftDatp

psiaftDatp

psiaftDatp

Dp

Cor

CD

psiaGOCftDatpGOCftDatp

psiaGOCftDatp

psiaGOCftDatp

w

g

w

g

g

g

g

go

w

o

2265)5000(

239619965000*08.0)5000(

2355)5200(

241219965200*08.0)5200(

1996*08.0

1996

*08.0

2420)5300()5300(

2400155300*45.0)5300(

24205655300*35.0)5300(

P(psia)5000

52005300540055005600

Pw=2265 Pg=2396

Pw=2355 Pg=2412Pw=2400 Po =Pg=2420

Pw=2445 Po=2455Pw= Po=2490

Pw=2535

Depth(ft)

Gas

GOC

oilOWC

water

D=5500ft

D=5300ft

4747

Pressure KickPressure Kick

Assumes a normal hydrostatic pressure regime Pω= 0.45 × D + 15Assumes a normal hydrostatic pressure regime Pω= 0.45 × D + 15 In water zoneIn water zone at 5000 ft Pω(at5000) = 5000 × 0.45 + 15 = 2265 psiaat 5000 ft Pω(at5000) = 5000 × 0.45 + 15 = 2265 psia at OWC (5500 ft) Pω(at OWC) = 5500 × 0.45 + 15 = 2490 psiaat OWC (5500 ft) Pω(at OWC) = 5500 × 0.45 + 15 = 2490 psia

4848

Pressure KickPressure Kick

In oil zone Po = 0.35 x D + C In oil zone Po = 0.35 x D + C at D = 5500 ft , Po = 2490 psi at D = 5500 ft , Po = 2490 psi → → C = 2490 – 0.35 × 5500 = 565 psiaC = 2490 – 0.35 × 5500 = 565 psia → → Po = 0.35 × D + 565Po = 0.35 × D + 565 at GOC (5200 ft) Po (at GOC) = 0.35 × 5200 + 565 = 2385 psiaat GOC (5200 ft) Po (at GOC) = 0.35 × 5200 + 565 = 2385 psia

4949

Pressure KickPressure Kick

In gas zone Pg = 0.08 D + 1969 (psia)In gas zone Pg = 0.08 D + 1969 (psia) at 5000 ft Pg = 0.08 × 5000 + 1969 = 2369 psia at 5000 ft Pg = 0.08 × 5000 + 1969 = 2369 psia

5050

Pressure KickPressure Kick

In gas zone Pg = 0.08 D + CIn gas zone Pg = 0.08 D + C At D = 5500 ft, Pg = Pω = 2490 psiaAt D = 5500 ft, Pg = Pω = 2490 psia 2490 = 0.08 × 5500 + C2490 = 0.08 × 5500 + C C = 2050 psiaC = 2050 psia → → Pg = 0.08 × D + 2050Pg = 0.08 × D + 2050 At D = 5000 ftAt D = 5000 ft Pg = 2450 psiaPg = 2450 psia

5151

GWC error from pressure measurementGWC error from pressure measurement

Pressure = 2500 psia Pressure = 2450 psia Pressure = 2500 psia Pressure = 2450 psia at D = 5000 ft at D = 5000 ftat D = 5000 ft at D = 5000 ft in gas-water reservoir in gas-water reservoirin gas-water reservoir in gas-water reservoir GWC = ? GWC = ?GWC = ? GWC = ? Sol. Sol.Sol. Sol. Pg = 0.08 D + C Pg = 0.08 D + CPg = 0.08 D + C Pg = 0.08 D + C C = 2500 – 0.08 × 5000 C = 2450 – 0.08 × 5000C = 2500 – 0.08 × 5000 C = 2450 – 0.08 × 5000 = 2100 psia = 2050 psia= 2100 psia = 2050 psia → → Pg = 0.08 D + 2100 → Pg = 0.08 D + 2050Pg = 0.08 D + 2100 → Pg = 0.08 D + 2050 Water pressure Pω = 0.45 D + 15 Water pressure Pω = 0.45 D + 15Water pressure Pω = 0.45 D + 15 Water pressure Pω = 0.45 D + 15 At GWC Pg = Pω At GWC Pg = PωAt GWC Pg = Pω At GWC Pg = Pω 0.08 D + 2100 = 0.45 D + 15 0.08 D + 2050 = 0.45 D + 150.08 D + 2100 = 0.45 D + 15 0.08 D + 2050 = 0.45 D + 15 D = 5635 ft (GWC) D = 5500 ft (GWC)D = 5635 ft (GWC) D = 5500 ft (GWC)

5252

Results from Errors in GWC or GOC or OWCResults from Errors in GWC or GOC or OWC

GWC or GOC or OWC location GWC or GOC or OWC location

affecting affecting

volume of hydrocarbon OOIPvolume of hydrocarbon OOIP

affectingaffecting

OOIP or OGIPOOIP or OGIP

affectingaffecting

development plansdevelopment plans

5353

Gas Material Balance: Recovery FactorGas Material Balance: Recovery Factor Material balanceMaterial balance

Production = OGIP (GIIP) - Unproduced gasProduction = OGIP (GIIP) - Unproduced gas (SC) (SC) (SC)(SC) (SC) (SC) Case 1Case 1 ：： no water influx (volumetric no water influx (volumetric depletion reservoirs)depletion reservoirs) Case 2Case 2 ：： water influx (water drive reservoirs) water influx (water drive reservoirs)

5454

Volumetric depletion reservoirs -- 1Volumetric depletion reservoirs -- 1

No water influx into the reservoir from the adjoining aquiferNo water influx into the reservoir from the adjoining aquifer Gas initially in place (GIIP) or Initial gas in placeGas initially in place (GIIP) or Initial gas in place （（ IGIPIGIP ）） ＝ ＝ G G ＝ ＝ Original gas in place Original gas in place （（ OGIPOGIP ） ） [=] Standard Condition Volume[=] Standard Condition Volume

Material Balance Material Balance （（ at standard conditionsat standard conditions ）） Production Production ＝ ＝ GIIP GIIP － － Unproduced gasUnproduced gas （（ SCSC ）） （ （ SCSC ） （） （ SCSC ））

Where G/EWhere G/Eii = GIIP in reservoir volume or reservoir volume filled with gas = GIIP in reservoir volume or reservoir volume filled with gas ＝ ＝ HCPVHCPV

3/][37.35

][)1(

ftSCFTz

pEwhere

SCFEsVG

ii

ii

iwc

)33.1(EE

GGG

ip

5555

Volumetric depletion reservoirs -- 2Volumetric depletion reservoirs -- 2

)34.1(1 i

p

E

E

G

G

3

37.35sinft

SCF

zT

pEce

.:137.35

37.351 constTTnote

z

pz

p

Tz

pzT

p

G

Gi

i

i

ii

i

p

)35.1(1

G

G

z

p

z

p p

i

i

factoreryreGas

depletionduringstageanyateryregasfractionaltheG

Gwhere p

cov

cov

pi

i

i

i GGz

p

z

p

z

p

1

5656

In Eq.In Eq. （（ 1.331.33 ））

HCPVHCPV≠≠const.const. because: because: 1. the connate water in reservoir will expand1. the connate water in reservoir will expand

2. the grain pressure increases as gas2. the grain pressure increases as gas (or fluid) pressure declines(or fluid) pressure declines

?.constE

GHCPV

i

5757

wherewhere

4.~3.)()(

)3.1(

ppGPdFPd

GPFPOP

)36.1(

)/()(

fw

i

dVdV

EGdHCPVd

HCPVinreductionato

leadswaterofansionsignnegative

volumeporeinitialV

volumewaterconnateinitialV

f

w

exp""

5858

dpVcdV

p

V

Vc

p

V

Vc

GP

V

Vc

fff

f

ff

f

ff

f

ff

1

)(

1

1

pore vol.

GP

GPGP

GP

fV

5959

dpVcdV

dp

dV

VFPd

V

Vc

www

w

w

w

ww

11

FP

FP

FP

FP

FP=gas pressure

FPFP

FP=gas pressureFP

FP

fV

wV

wV

6060

wc

fwcw

initialiti

wc

fwcw

initialiinitialiti

wcf

wc

wcw

initialitiinitiali

wcif

wc

wc

iw

i

wc

wc

iwc

wcwcw

wciwcf

ffwwi

S

pcSc

E

G

E

G

S

pcSc

E

G

E

G

E

G

pS

cS

Sc

E

G

E

G

E

G

dpSE

Gcdp

S

S

E

Gc

E

Gd

S

S

E

GS

S

HCPVSPVV

SE

G

S

HCPVPVV

Since

dpVcdpVcHCPVdE

Gd

11

1

1

1

1

11

11

11

6161

difference

E

E

G

Gwithcomputing

E

E

G

G

S

cSc

SandpsicpsicFor

E

E

S

cSc

G

G

ES

pcSc

E

GGG

EE

GGG

i

p

i

p

wc

fwcw

wcfw

iwc

fwcwp

wc

fwcw

ip

ip

%3.1

1987.01

987.0013.011

1

2.01010;103

111

11

)33.1(

1616

6262

p/z plotp/z plot

From Eq. (1.35) such asFrom Eq. (1.35) such as

A straight line in p/z v.s Gp plot means that the reservoir is A straight line in p/z v.s Gp plot means that the reservoir is

a depletion type a depletion type

pi

i

i

i

p

i

i

GGz

p

z

p

z

p

G

G

z

p

z

p

)35.1(1

In plotGpsv

z

p.

Y=a+mx

i

i

i

i

p

z

pa

Gz

pm

Gxz

py

p/z

Abandon pressure pab

0Gp G

p/z

Gp/G=RF 1.00

6363

Water drive reservoirsWater drive reservoirs If the reduction in reservoir pressure leads to an expansion of If the reduction in reservoir pressure leads to an expansion of

adjacent aquifer water, and consequent influx into the reservoir, adjacent aquifer water, and consequent influx into the reservoir, the material balance equation must then be modified as:the material balance equation must then be modified as:

Production = GIIP Production = GIIP － － Unproduced gasUnproduced gas （（ SCSC ） （） （ SCSC ） （） （ SCSC ）） GGp ＝ ＝ G G － （－ （ HCPV-WHCPV-We ）） EE Or Or GGp ＝ ＝ GG － （－ （ G/EG/Ei －－ WWe ）） EE where We= the cumulative amount of water influx resultingwhere We= the cumulative amount of water influx resulting from the pressure drop.from the pressure drop. Assumptions:Assumptions: No difference between surface and reservoir volumes ofNo difference between surface and reservoir volumes of water influxwater influx Neglect the effects of connate water expansion and pore Neglect the effects of connate water expansion and pore volume reduction.volume reduction. No water productionNo water production

6464

Water drive reservoirsWater drive reservoirs With water productionWith water production

where where We*EWe*Ei //GG represents the fraction of the initial hydrocarbon represents the fraction of the initial hydrocarbon

pore volume flooded by water and is, pore volume flooded by water and is,

therefore, always less then unity.therefore, always less then unity.

EBWWE

GGG wpe

ip

)41.1(1

1

GEWG

G

z

p

z

p

ie

p

i

i

6565

Water drive reservoirsWater drive reservoirs

sincesince

)41.1(

1

1

G

EW

G

G

z

p

z

p

ie

p

i

i

11

G

EW ie

in water flux reservoirs

G

G

z

p

z

p p

i

i 1

Comparing

G

G

z

p

z

p p

i

i 1 in depletion type reservoir

6666

Water drive reservoirsWater drive reservoirs

In eq.(1.41) the following two parameters to be determinedIn eq.(1.41) the following two parameters to be determined

G and WG and We

History matching or “aquifer fitting” to find WHistory matching or “aquifer fitting” to find We

Aquifer model for an aquifer whose dimensions are of the same Aquifer model for an aquifer whose dimensions are of the same order of magnitude as the reservoir itself.order of magnitude as the reservoir itself.

Where W=the total volume of water and depends primary on theWhere W=the total volume of water and depends primary on the

geometry of the aquifer.geometry of the aquifer.

ΔP=the pressure drop at the original reservoir –aquifer boundaryΔP=the pressure drop at the original reservoir –aquifer boundary

)41.1(

1

1

G

EW

G

G

z

p

z

p

ie

p

i

i

pWcWe

6767

Water drive reservoirsWater drive reservoirs The material balance in such a case would be as shown by plot A The material balance in such a case would be as shown by plot A

in fig1.11, which is not significantly different from the depletion in fig1.11, which is not significantly different from the depletion lineline

For case B & C in fig 1.11For case B & C in fig 1.11 （（ p.30p.30 ） ） =>Chapter 9 =>Chapter 9

6868

Bruns et. al methodBruns et. al method This method is to estimate GIIP in a water drive reservoir This method is to estimate GIIP in a water drive reservoir From Eq. (1.40) such asFrom Eq. (1.40) such as

i

ea

i

e

i

p

i

e

i

p

epi

ei

p

ei

p

ei

p

E

E

EWGGor

E

E

EWG

E

E

Gor

E

E

EW

E

E

GG

EWGE

EG

EWE

EGG

EWE

GEGG

EWE

GGG

1

11

11

1

1

)40.1(

)(

1a

i

p Gor

E

E

G

i

e

E

E

EW

1is plot as function of

6969

Bruns et. al methodBruns et. al method

The result should be a straight line, provided the correct aquifer model has been The result should be a straight line, provided the correct aquifer model has been selected.selected.

The ultimate gas recovery depends both on The ultimate gas recovery depends both on (1) the nature of the aquifer ,and (1) the nature of the aquifer ,and (2) the abandonment pressure.(2) the abandonment pressure.

The principal parameters in gas reservoir engineering:The principal parameters in gas reservoir engineering: (1) the GIIP(1) the GIIP (2) the aquifer model(2) the aquifer model (3) abandonment pressure(3) abandonment pressure (4) the number of producing wells and their mechanical define(4) the number of producing wells and their mechanical define

)(

1a

i

p Gor

EE

G

i

e

E

E

EW

1is plot as function of

7070

Hydrocarbon phase behaviorHydrocarbon phase behavior

7171

Hydrocarbon phase behaviorHydrocarbon phase behavior

7272

Hydrocarbon phase behaviorHydrocarbon phase behavior

Residual saturation (flow ceases)Liquid H.C deposited in the reservoirRetrograde liquid Condensate

C--------- ＞ D-------------- ＞ E

Re-vaporization of the liquid condensate ?NO!Because H.C remaining in the reservoir increaseComposition of gas reservoir changed Phase envelope shift SE direction Thus, inhibiting re-vaporization.

E--------------- ＞ F

producing Wet gas (at scf)Dry gas

injection

displace the wet gas

Δp smallKeep p above dew pt.

until dry gas break through occurs in the producing wells

Condensate reservoir, pt. c,

7373

Equivalent gas volumeEquivalent gas volume The material balance equation of eq(1.35) such asThe material balance equation of eq(1.35) such as

Assume that a volume of gas in the reservoir was Assume that a volume of gas in the reservoir was produced as gas at the surface.produced as gas at the surface.

If, due to surface separation, small amounts of liquid If, due to surface separation, small amounts of liquid hydrocarbon are produced, the cumulative liquid hydrocarbon are produced, the cumulative liquid volume must be converted into an equivalent gas volume must be converted into an equivalent gas volume and added to the cumulative gas production to volume and added to the cumulative gas production to give the correct value of Gp for use in the material give the correct value of Gp for use in the material balance equation.balance equation.

G

G

z

p

z

p p

i

i 1

7474

Equivalent gas volumeEquivalent gas volume If n lbIf n lbmm –mole of liquid have been produced, of molecular –mole of liquid have been produced, of molecular

weight M, then the total mass of liquid is weight M, then the total mass of liquid is

where γwhere γ00 = oil gravity (water =1) = oil gravity (water =1) ρρww ＝ ＝ density of water density of water （（ =62.43 lb=62.43 lbmm/ft/ft33 ））

volumeliquidnM wo

M

V

molelbmlbmM

ftVft

lbm

M

Vn owo 00

3030

4.62

/

4.62

volumegasEquivalent

bblsNM

NV

M

N

p

RT

M

N

p

nRTV

bblsNwhereM

Nn

bbl

ft

M

bblsVn

pp

sc

p

sc

scp

scsc

pp

05

00

0

300

1033.1

7.14

52073.105.3505.350

][5.350

1

61458.54.62

7575

Condensate ReservoirCondensate Reservoir

The dry gas material balance equations can also be The dry gas material balance equations can also be applied to gas condensate reservoir, if the single applied to gas condensate reservoir, if the single phase z-factor is replaced by the ,so-called ,two phase phase z-factor is replaced by the ,so-called ,two phase z-factor. This must be experimentally determined in z-factor. This must be experimentally determined in the laboratory by performing a constant volume the laboratory by performing a constant volume depletion experiment.depletion experiment.

Volume of gas Volume of gas ＝＝ G scf , as charge to a PVT cellG scf , as charge to a PVT cell PP ＝＝ PPii ＝＝ initial pressure initial pressure （（ above dew pointabove dew point ）） TT ＝＝ TTrr ＝＝ reservoir temperaturereservoir temperature

7676

Condensate ReservoirCondensate Reservoir

p decrease p decrease by withdraw gas in stages from the cell, and measure gas by withdraw gas in stages from the cell, and measure gas Gp’Gp’

Until the pressure has dropped below the dew pointUntil the pressure has dropped below the dew point

The latter experiment, for determining the single phase z-factor, implicitly The latter experiment, for determining the single phase z-factor, implicitly assumes that a volume of reservoir fluids, below dew point pressure, is assumes that a volume of reservoir fluids, below dew point pressure, is produced in its entirety to the surface.produced in its entirety to the surface.

G

G

z

p

pz

G

G

z

p

z

p

G

G

z

p

pZ

p

i

i

p

i

i

p

i

i

phase

'1

)35.1('

1

)46.1('

12

7777

Condensate ReservoirCondensate Reservoir

In the constant volume depletion experiment, however, allowance is made In the constant volume depletion experiment, however, allowance is made for the fact that some of the fluid remains behind in the reservoir as liquid for the fact that some of the fluid remains behind in the reservoir as liquid condensate, this volume being also recorded as a function of pressure condensate, this volume being also recorded as a function of pressure during the experiment. As a result, if a gas condensate sample is analyzed during the experiment. As a result, if a gas condensate sample is analyzed using both experimental techniques, the two phase z-factor determined using both experimental techniques, the two phase z-factor determined during the constant volume depletion will be lower than the single phase z-during the constant volume depletion will be lower than the single phase z-factor.factor.

This is because the retrograde liquid condensate is not included in the This is because the retrograde liquid condensate is not included in the cumulative gas production Gp’ in equationcumulative gas production Gp’ in equation （（ 1.461.46 ）） , which is therefore , which is therefore lower than it would be assuming that all fluids are produced to the surface, lower than it would be assuming that all fluids are produced to the surface, as in the single phase experiment.as in the single phase experiment.

7878

7979

油層工程油層工程 蘊藏量評估蘊藏量評估

體積法 體積法 物質平衡法物質平衡法 衰減曲線衰減曲線 油層模擬油層模擬

壓力分析壓力分析 (( 隨深度變化，或壓力梯度隨深度變化，或壓力梯度 ), ), 例如例如 , , 求氣水界面。求氣水界面。 物質平衡法物質平衡法

井壓測試分析井壓測試分析 (( 暫態暫態 )) （求（求 kk 、、 ss 、、 rere 、、 xfxf 、氣水界面、地層異質性）、氣水界面、地層異質性） Pressure buildupPressure buildup Pressure drawdownPressure drawdown

水驅計算水驅計算 (water drive)(water drive)

8080