高等油層工程 Advanced Reservoir Engineering

高等油層工程高等油層工程Advanced Reservoir EngineeringAdvanced Reservoir Engineering

開課班級開課班級 : : 碩博士班碩博士班 (2007(2007 年秋年秋季季 ))

講授教師：林再興講授教師：林再興

目的目的講授石油及天然氣流體性質，以及生產石講授石油及天然氣流體性質，以及生產石油及天然氣所導致的油層壓力變化原理。油及天然氣所導致的油層壓力變化原理。

討論壓力測試分析討論壓力測試分析 ((井壓測試分析井壓測試分析 ) ) ，以，以及生產資料分析及生產資料分析 (( 生產遞減曲線分析生產遞減曲線分析 ) ) 。。

求得地層參數求得地層參數 // 預測未來產生產率預測未來產生產率 // 計算計算地層的石油或天然的儲量及蘊藏量。地層的石油或天然的儲量及蘊藏量。

Textbooks and referencesTextbooks and references

(A) Dake , L.P., Fundamentals of Reservoir(A) Dake , L.P., Fundamentals of Reservoir Engineering, revised edition, Elsevier ScientificEngineering, revised edition, Elsevier Scientific B.V., Amsterdam, the Netherlands, 2001.B.V., Amsterdam, the Netherlands, 2001.

(B) Ahmed, T., and McKinney, P., Advanced (B) Ahmed, T., and McKinney, P., Advanced Reservoir Engineering, Gulf Publishing Reservoir Engineering, Gulf Publishing Company, Houston, Texas, 2004Company, Houston, Texas, 2004

(B) Craft, B.C., and Hawkins, M.F. , Revised(B) Craft, B.C., and Hawkins, M.F. , Revised by Terry, R.E. , Applied Petroleum Reservoirby Terry, R.E. , Applied Petroleum Reservoir Engineering, Second edition., Prentice Hall ,Engineering, Second edition., Prentice Hall , Englewood Cliffs, New Jersey, 1991.Englewood Cliffs, New Jersey, 1991.

Textbooks and referencesTextbooks and references

(C) Lee, J., Well Testing, SPE Textbook (C) Lee, J., Well Testing, SPE Textbook

series, Society of Petroleum Engineers ofseries, Society of Petroleum Engineers of

AIME, Dallas, Texas, 2002. AIME, Dallas, Texas, 2002.

(D) (D) 林國安等人，石油探採林國安等人，石油探採 (( 第四冊 – 油氣生產第四冊 – 油氣生產 , , Chapter 24 ) Chapter 24 ) ，中國石油股份有限公司訓練教材，中國石油股份有限公司訓練教材叢書，中油訓練所，嘉義市，叢書，中油訓練所，嘉義市， 2004.2004.

(E) Journal papers(E) Journal papers

55

Advanced Reservoir Advanced Reservoir EngineeringEngineering

byby Ahmed, T., and McKinney, PAhmed, T., and McKinney, P

Well testing analysisWell testing analysis Water influxWater influx Unconventional gas reservoirUnconventional gas reservoir Performance of oil reservoirPerformance of oil reservoir Predicting oil reservoirPredicting oil reservoir Introduction to oil fieldeconomicsIntroduction to oil fieldeconomics

大網大網 Introduction to reservoir engineeringIntroduction to reservoir engineering - Gas reservoir- Gas reservoir - PVT analysis for oil- PVT analysis for oil - Material balance applied to oil- Material balance applied to oil

The flow equations of single-phase and two-phase flow of hydrocarbon in The flow equations of single-phase and two-phase flow of hydrocarbon in porous mediaporous media

- Darcy’s law and applications- Darcy’s law and applications - The basic differential equation in a porous medium- The basic differential equation in a porous medium

Solutions to the flow equations of hydrocarbon in porous mediaSolutions to the flow equations of hydrocarbon in porous media - Steady and semi-steady states- Steady and semi-steady states - Unsteady state- Unsteady state

Pressure drawdown and buildup analysis for oil and gas wellsPressure drawdown and buildup analysis for oil and gas wells

Decline curve analysisDecline curve analysis

Case studyCase study

Part 1Part 1 Introduction to Reservoir EngineeringIntroduction to Reservoir Engineering

The primary functions of a reservoir The primary functions of a reservoir

engineer: engineer: the estimation of hydrocarbon in placethe estimation of hydrocarbon in place the calculation of a recovery factor , and the calculation of a recovery factor , and the attachment of a time scale to the recoverythe attachment of a time scale to the recovery

Note: Note:

pressure/flow rate information → pressure/flow rate information →

parameters/future flow rate/future pressureparameters/future flow rate/future pressure

Outlines of Reservoir EngineeringOutlines of Reservoir Engineering (1) Introduction(1) Introduction

Petrophysical properties ( Rock properties)Petrophysical properties ( Rock properties) Fluid properties (gas, water, crude properties)Fluid properties (gas, water, crude properties) Calculations of hydrocarbon volumesCalculations of hydrocarbon volumes Fluid pressure regimesFluid pressure regimes

(2) Gas reservoirs (2) Gas reservoirs Calculating gas in place by the volumetric methodCalculating gas in place by the volumetric method Calculating gas recovery factor Calculating gas recovery factor Material balance calculation (Depletion & Water drive)Material balance calculation (Depletion & Water drive) Hydrocarbon phase behavior (gas condensate phase behavior)Hydrocarbon phase behavior (gas condensate phase behavior) The gas equivalent of produced condensate and waterThe gas equivalent of produced condensate and water

(3) PVT analysis for oil(3) PVT analysis for oil Definition of the basic PVT parametersDefinition of the basic PVT parameters Determination of the basic PVT parameters in the lab. And conveDetermination of the basic PVT parameters in the lab. And conve

rsion for field operating conditions.rsion for field operating conditions.

Outlines of Reservoir Engineering – cont.Outlines of Reservoir Engineering – cont. (4) Material balance applied to oil(4) Material balance applied to oil reservoirsreservoirs

General form of the material balance equation for a hydrocaGeneral form of the material balance equation for a hydrocarbon reservoir (Undersaturated and Saturated reservoir)rbon reservoir (Undersaturated and Saturated reservoir)

Reservoir drive mechanisms Reservoir drive mechanisms Solution gas drive Solution gas drive Gas cap driveGas cap drive Natural water driveNatural water drive

(5) Darcy’s law and applications(5) Darcy’s law and applications

Outlines of Reservoir Engineering – cont.Outlines of Reservoir Engineering – cont. (6) The basic differential equation for radial flow (6) The basic differential equation for radial flow in a porous mediumin a porous medium

Derivation of the basic radial flow equationDerivation of the basic radial flow equation Conditions of solutionConditions of solution Linearization of radial flow equationLinearization of radial flow equation

(7) Well inflow equations for stabilized flow (7) Well inflow equations for stabilized flow conditions conditions

Semi steady state solutionSemi steady state solution Steady state solutionSteady state solution Generalized form of inflow equation (for semi steady state)Generalized form of inflow equation (for semi steady state)

Outlines of Reservoir Engineering – cont.Outlines of Reservoir Engineering – cont. (8) The constant terminal rate solution of the radial (8) The constant terminal rate solution of the radial diffusivity equation and its application to oil welldiffusivity equation and its application to oil well testingtesting

Constant terminal rate solutionConstant terminal rate solution General Transient flow General Transient flow Semi steady state flowSemi steady state flow Superposition theorem; general theory of well testingSuperposition theorem; general theory of well testing The Matthews, Brons, Hazebroek pressure buildup theoryThe Matthews, Brons, Hazebroek pressure buildup theory Pressure buildup analysis techniquesPressure buildup analysis techniques Multi-rate drawdown testingMulti-rate drawdown testing The effects of partial well completionThe effects of partial well completion After-flow analysisAfter-flow analysis

Outlines of Reservoir Engineering – cont.Outlines of Reservoir Engineering – cont. (9) Gas well testing(9) Gas well testing - Linearization and solution of the basic differential equation- Linearization and solution of the basic differential equation for the radial flow of a real gasfor the radial flow of a real gas - The Russell, Goodrich, et al. Solution technique - The Russell, Goodrich, et al. Solution technique - The Al-Hussainy, Ramey, Crawford solution technique- The Al-Hussainy, Ramey, Crawford solution technique - Pressure squared and pseudo pressure solution technique- Pressure squared and pseudo pressure solution technique - Non-Darcy flow & determination of the non-darcy- Non-Darcy flow & determination of the non-darcy coefficientcoefficient - The constant terminal rate solution for the flow of a real gas- The constant terminal rate solution for the flow of a real gas - General theory of gas well testing- General theory of gas well testing - Multi-rate testing of gas well - Multi-rate testing of gas well - Pressure building testing of gas wells- Pressure building testing of gas wells - Pressure building analysis in solution gas drive reservoirs- Pressure building analysis in solution gas drive reservoirs

Outlines of Reservoir Engineering – cont.Outlines of Reservoir Engineering – cont. (10) Natural water influx (10) Natural water influx - Steady state model- Steady state model - Unsteady state model- Unsteady state model - The van Everdingen and Hurst edge-water drive - The van Everdingen and Hurst edge-water drive modelmodel - Bottom – water drive model- Bottom – water drive model - Pseudo steady state model (Fetkovich model)- Pseudo steady state model (Fetkovich model) - Predicting the amount of water influx- Predicting the amount of water influx

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)

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

Pressure gradient for sandstonePressure gradient for sandstone

Pressure gradient for sandstonePressure gradient for sandstone

gD

p

gDp

3084.1682.3222.5

ft

lbf

)/(16.1

16.1144

1084.168

22

2

2

ftpsi

ftin

lbf

in

ft

ftft

lbf

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

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

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)(

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 se℉Temperature change ΔT = +1 → ΔP = +125 psi in a se℉aled fresh water systemaled fresh water system

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

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

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?

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

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


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


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


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

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)

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

Volumetric Gas Reservoir EngineeringVolumetric Gas Reservoir Engineering

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.

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 [=] lbm moles; n=the number of lb moles, one lb mole isn [=] lbm moles; n=the number of lb moles, one lb 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∙ ft3 / (lbmmole∙0R)[=] 10.732 psia∙ ft3 / (lbmmole∙0R)

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

)13.1(nRTpV

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. (1.1The principal drawback in attempting to use eq. (1.1

4) to describe the behavior of real gases encountered i4) to describe the behavior of real gases encountered in reservoirs is that the maximum pressure for which tn reservoirs is that the maximum pressure for which the equation is applicable is still far below the normal he equation is applicable is still far below the normal range of reservoir pressures range of reservoir pressures

)14.1())((2

RTbVV

ap

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

the Beattie-Bridgeman equationthe Beattie-Bridgeman equation the Benedict-Webb-Rubin equationthe Benedict-Webb-Rubin equation the non-ideal gas lawthe non-ideal gas law

Non-ideal gas lawNon-ideal gas law

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

)15.1(nzRTpV

PandTatgasofmolesnofvolumeIdeal

PandTatgasofmolesnofvolumeActual

V

Vz

i

a

),,( ncompositioTPfz )1( airgravityspecificncompositio g

Determination of z-factorDetermination of z-factor

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

(a) Experimental determination(a) Experimental determination n mole s of gas n mole s of gas

p=1atm; T=reservoir temperature; => V=V0p=1atm; T=reservoir temperature; => V=V0

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

n mole of gas n mole 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

(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 CnH2n+2 paraffin series CnH2n+2 Non-hydrocarbon impurities: CO2, N2 and H2Non-hydrocarbon impurities: CO2, N2 and H2 Gas reservoir: lighter members of the paraffin series, C1Gas reservoir: lighter members of the paraffin series, C1 and C2 > 90% of the volume.and C2 > 90% of the volume.

The Standing-Katz CorrelationThe Standing-Katz Correlation

knowing Gas composition (ni)knowing Gas composition (ni) Critical pressure (Pci)Critical pressure (Pci) Critical temperature (Tci) of each component Critical temperature (Tci) of each component ( Table (1.1) and P.16 ) ( Table (1.1) and P.16 ) Pseudo critical pressure (Ppc) Pseudo critical pressure (Ppc) Pseudo critical temperature (Tpc) for the mixturePseudo critical temperature (Tpc) for the mixture

Pseudo reduced pressure (Ppr) Pseudo reduced pressure (Ppr) Pseudo reduced temperature (Tpr) Pseudo reduced temperature (Tpr)

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

iciipc

iciipc

TnT

PnP

pcpr P

PP

).(IsothermalconstT

TT

pcpr

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

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

Pseudo critical pressure (Ppc)Pseudo critical pressure (Ppc) Pseudo critical temperature (Tpc)Pseudo critical temperature (Tpc)

g

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

Pseudo reduced pressure (Ppr)Pseudo reduced pressure (Ppr)

Pseudo reduced temperature (Tpr)Pseudo reduced temperature (Tpr)

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

z-factorz-factor The above procedure is valided only if impunity (CO2,N2 and HThe above procedure is valided only if impunity (CO2,N2 and H

2S) is less then 5% volume.2S) is less then 5% volume.

pcpr P

PP

).(IsothermalconstT

TT

pcpr

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

uation of state, areuation of state, are

where Ppr= the pseudo reduced pressurewhere Ppr= the pseudo reduced pressure t=1/Tpr Tpr=the pseudo reduced temperature t=1/Tpr Tpr=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 NeThis non-linear equation can be conveniently solved for y using the simple Newton-Raphson iterative technique.wton-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

(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:

make an initial estimate of ymake an initial estimate of ykk, where k is an iteration counter (which in this , where k is an iteration counter (which in this case is unity, e.q. ycase is unity, e.q. y11=0.001=0.001

substitute this value in Eq. (1.21);unless the correct value of y has been initisubstitute 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 Fally 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) iterate, using eq. (1.21) and eq. (1.22), until satisfactory (4) iterate, 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) substitution of the correct value of y in eq.(1.20)will give the z-factor. (5) substitution of 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

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/ft3 or STB/bbl[=] SCF/ft3 or 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

ExampleExample

Reservoir condition: Reservoir condition: P=2000psia; T=1800F=(180+459.6)=639.60R; z=0.865P=2000psia; T=1800F=(180+459.6)=639.60R; z=0.865 >> surface volume/reservoirsurface volume/reservoir or SCF/ft3 or STB/bblor SCF/ft3 or STB/bbl

8.1276.639865.0

200035.35

E

iwi ESVOGIP )1(

(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

)(

)(

(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

(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

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)

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 91.19 687.097.28

91.19

97.28 gas

g

M

airggasair

gasg

)(0524.0)(0763.0687.0 33 ftlbmftlbmgas

nzRTpV nMzRTpVM mzRT

zRT

pM

V

m

)(0524.06.51973.101

91.197.14 3ftlbmRTz

P

scsc

scgas

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

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

nzRTpV nMzRTpVM mzRT

)(707.6)1806.459(73.10865.0

91.192000 3ftlbmzRT

pM

V

m

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

gDp gdDdp

gdD

dp 232.32)

2.32

1707.6(

sft

lbm

slug

ft

lbm

23707.6

s

ft

ft

slug

3707.6

ft

lb f

2

2

2 144

11707.6

in

ft

ftft

lbf

ftin

lb f 10465.0

2

ftpsi0465.0

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)

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

No water influx into the reservoir from the adjoining aquifer. No 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

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

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

wherewhere

4.~3.)()(

)3.1(

ppGPdFPd

GPFPOP

)36.1(

)/()(

fw

i

dVdV

EGdHCPVd

HCPVinreductionato

leadswaterofansionsignnegative

volumeporeinitialV

volumewaterconnateinitialV

f

w

exp""

)(

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

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

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

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

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 plotGpsvz

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

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

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

Production = GIIP Production = GIIP －－ Unproduced gasUnproduced gas （（ SCSC ）（）（ SCSC ）（）（ SCSC ）） Gp Gp ＝＝ G G －（－（ HCPV-WeHCPV-We ）） EE Or Or GpGp ＝＝ GG －（－（ G/EiG/Ei －－ WeWe ）） 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

Water drive reservoirsWater drive reservoirs With water productionWith water production

where where We*Ei We*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

G

EW

G

G

z

p

z

p

ie

p

i

i

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

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; WeG; We History matching or “aquifer fitting” to find WeHistory matching or “aquifer fitting” to find We Aquifer modelfor an aquifer whose dimensions are of the same orAquifer modelfor an aquifer whose dimensions are of the same or

der of magnitude as the reservoir itself.der 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

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 lin fig1.11, which is not significantly different from the depletion lineine

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

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

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

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,

Equivalent gas volumeEquivalent gas volume

The material balance equation of eq(1.35) such as The material balance equation of eq(1.35) such as

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

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

G

G

z

p

z

p p

i

i 1

Equivalent gas volumeEquivalent gas volume

If n lbIf n lbmm –mole of liquid have been produced, of molecular weig –mole of liquid have been produced, of molecular weight M, then the total mass of liquid is ht 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

Condensate ReservoirCondensate Reservoir

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

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


p decrease p decrease by withdraw gas from the cell, and measure gas Gp’ by withdraw gas from the cell, and measure gas Gp’ Until the pressure has dropped to the dew pointUntil the pressure has dropped to the dew point

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


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 durincondensate, this volume being also recorded as a function of pressure during the experiment. As a result, if a gas condensate sample is analyzed using g the experiment. As a result, if a gas condensate sample is analyzed using both experimental techniques, the two phase z-factor determined during thboth experimental techniques, the two phase z-factor determined during the constant volume depletion will be lower than the single phase z-factor.e constant volume depletion will be lower than the single phase z-factor.

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

7878

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)

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高等油層工程 Advanced Reservoir Engineering