chapter 2 volumetric gas reservoir engineering
DESCRIPTION
Chapter 2 Volumetric gas Reservoir Engineering. Gas PVT. Gas is one of a few substances whose state, as defined by pressure, volume and temperature (PVT) One other such substance is saturated steam. The equation of state for gas. Classical ideal gas law Classical non-ideal gas law - PowerPoint PPT PresentationTRANSCRIPT
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
PandTatgasofmolesnofvolumeIdealPandTatgasofmolesnofvolumeActual
VVz
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 VpVz
0
0
VppVz
zTpV
TzVp
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
).(IsothermalconstTTT
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
).(IsothermalconstTTT
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 tpr
)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
yyyydy
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 RTbVVap
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 hydrocarbon.This is a PVT relationship to relate surface to reservoir volumes of 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
conditionsreservoiratgasofmolesnofvolumeconditionsdardsatgasofmolesnofvolume
VV
E sc tan
scscscsc RTnzVp sc
scscsc p
RTnzV
nzRTpV pnzRTV
)1:(7.14
6.519
scsc
sc
sc
scscsc
scsc
sc znotezTzTp
pTnzRTp
pRTnz
pnzRT
pRTnz
VV
E
][35.35 zTpE
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
nMVm
zRTMP
pnzRT
nM
RTzPM
gas
gasgas
RTzPM
gas
gasgas
RTzpM
air
airair
air
air
gas
gas
air
gas
gas
gas
gair
gas
ZM
ZM
RTzpM
RTzpM
air
gasg
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
gasg
ZM
ZM
)(
)(
)28.1(97.28
gas
air
gasg
air
gas MMM
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
pznRTppnRTz
pV
12 ][
pz
pnRT
pnzRT
pV
2
)11()11(pz
zpV
pz
zppnzRT
pV
)]11([11pz
zpV
VpV
VCg
pz
zpCg
11
pCg
1
since pz
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 15.19 661.097.2815.19
97.28 gas
g
M
airggasair
gasg
)(0504.0)(0763.0661.0 33 ftlbmftlbmgas
nzRTpV nMzRTpVM mzRT
zRTpM
Vm
)(0505.06.51973.101
15.197.14 3ftlbmRTzMP
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 3ftlbmzRTpM
Vm
3333
Exercise 1.1 -- solutionExercise 1.1 -- solution -3 -3
gDp gdDdp
gdDdp 23 2.32)
2.321451.6( s
ftlbm
slugft
lbm
23451.6sft
ftslug
3451.6ftlb f
2
2
2 14411451.6
inft
ftftlbf
ftinlb f 10448.0 2 ft
psi0448.0
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(
10002.27.2
ftcm
gmlbm
cmgm
lbmslug
ftlbm
7.321202.168 3
322.5ft
slug
3636
Pressure gradient for sandstonePressure gradient for sandstone
Pressure gradient for sandstonePressure gradient for sandstone
gDp
gDp
3084.1682.3222.5ftlbf
)/(16.1144
1084.168 22
2
2 ftpsiftin
lbfin
ftftft
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)( DdDdPP water
4335.0)( waterdDdP
4335.0)( waterdDdP
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)
CDdDdPP 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)( waterdDdP
35.0)( oildDdP
08.0)( gasdDdP
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
psiaftDatPpsiaftDatP
psiftDatP
zonegasinDP
Cor
CD
GWCftDatPpsiaGWCftDatP
psiaftDatPzonewaterinDP
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
psiaftDatppsiaftDatp
zoneoilinDpCor
CDpsiaOWCftDatppsiaOWCftDatp
psiftDatpzonewaterinDp
w
o
o
o
o
o
w
w
w
2445155400*45.0)5400(24555655400*35.0)5400(
565*35.0565
*35.02490)5500(2490)5500(
2535)5600(15*45.0
psiaftDatp
psiaftDatppsiaftDatp
psiaftDatp
Dp
Cor
CD
psiaGOCftDatpGOCftDatppsiaGOCftDatppsiaGOCftDatp
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)
GasGOCoil
OWC
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(EEGGG
ip
5555
Volumetric depletion reservoirs -- 2Volumetric depletion reservoirs -- 2
)34.1(1 i
p
EE
GG
337.35sinft
SCFzTpEce
.:137.35
37.351 constTTnote
zpzp
TzpzTp
GG
i
i
i
ii
i
p
)35.1(1
GG
zp
zp p
i
i
factoreryreGas
depletionduringstageanyateryregasfractionaltheGG
where p
cov
cov
pi
i
i
i GGz
pzp
zp
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
?.constEGHCPV
i
5757
wherewhere
4.~3.)()()3.1(
ppGPdFPdGPFPOP
)36.1()/()(
fw
i
dVdVEGdHCPVd
HCPVinreductionatoleadswaterofansionsignnegative
volumeporeinitialVvolumewaterconnateinitialV
f
w
exp""
5858
dpVcdV
pV
Vc
pV
Vc
GPV
Vc
fff
f
ff
f
ff
f
ff
1
)(1
1
pore vol.
GP
GPGP
GP
fV
5959
dpVcdV
dpdV
VFPdV
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
SpcSc
EG
EG
SpcSc
EG
EG
EG
pS
cS
Sc
EG
EG
EG
dpSE
GcdpS
SEGc
EGd
SS
EGS
SHCPVSPVV
SEG
SHCPVPVV
Since
dpVcdpVcHCPVdEGd
11
1
11
1
11
11
11
6161
differenceEE
GG
withcomputingEE
GG
ScSc
SandpsicpsicFor
EE
ScSc
GG
ES
pcScEGGG
EEGGG
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
pzp
zp
GG
zp
zp
)35.1(1
In plotGpsv
zp .
Y=a+mx
i
i
i
i
p
zp
a
Gzp
m
Gxzpy
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.
EBWWEGGG wpe
ip
)41.1(1
1
GEWGG
zp
zp
ie
p
i
i
6565
Water drive reservoirsWater drive reservoirs
sincesince
)41.1(1
1
GEWG
Gzp
zp
ie
p
i
i
11
GEW ie
in water flux reservoirs
GG
zp
zp p
i
i 1
Comparing
GG
zp
zp 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
GEWG
Gzp
zp
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
EEEW
GGor
EEEW
G
EE
Gor
EEEW
EE
GG
EWGEEG
EWEEGG
EWEGEGG
EWEGGG
1
11
11
1
1
)40.1(
)(1
a
i
p Gor
EE
G
i
e
EEEW
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
)(1
a
i
p Gor
EE
G
i
e
EEEW
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.
GG
zp
zp 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
VmolelbmlbmM
ftVft
lbm
MV
n owo 00
3030 4.62
/
4.62
volumegasEquivalent
bblsNMN
V
MN
pRT
MN
pnRTV
bblsNwhereMN
n
bblft
MbblsVn
pp
sc
p
sc
scp
scsc
pp
05
00
0
300
1033.1
7.1452073.10
5.3505.350
][5.350
161458.54.62
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 Gp’ by withdraw gas in stages from the cell, and measure gas 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.
GG
zp
pz
GG
zp
zp
GG
zp
pZ
p
i
i
p
i
i
p
i
iphase
'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