Advanced Reservoir Engineering
Textbooks and references
(A) Dake , L.P., Fundamentals of Reservoir Engineering, revised edition, Elsevier Scientific B.V., Amsterdam, the Netherlands, 2001. (B) Ahmed, T., and McKinney, P., Advanced Reservoir Engineering, Gulf Publishing Company, Houston, Texas, 2004 Rev ised (B) Craft, B.C., and Hawkins, M.F. , Revised by Terry, R.E. , Applied Petroleum Reservoir Engineering, Second edition., Prentice Hall , Englewood Cliffs, New Jersey, 1991.
Advanced Reservoir Engineering by Ahmed, T., and McKinney, P
Well testing analysis
Water influx
Unconventional gas reservoir Performance of oil reservoir Predicting oil reservoir Introduction to oil fieldeconomics 3
Introduction to reservoir engineering - Gas res reserv reservoir ervoir oir - PVT analysis analysi anal ysiss for oil - Material Material balance balancee applied balanc applied to oil The flow equations of singlesingle-phase - phase phase and twotwo-phase - phase phase flow of hydrocarbon in porous media - Darcy¶s Darcy¶s law and applic app applications licati ations ons - The basic differenti differential al equation equation in a porous medium medium Solutions to the flow equations of hydrocarbon in porous media - Steady Steady and semisemii-steady sem -steady states - Unsteady Unste Uns teady ady state statee stat
Pressure drawdown and buildup analysis for oil and gas wells
Decline curve analysis
Case study
Part 1 Introduction to Reservoir Engineering
The primary functions of a reservoir engineer:
the estimation of hydrocarbon in place
the calculation of of a recovery factor , and
the attachment of a time scale to the recovery
Note: pressure/flow rate information parameters/future flow rate/future pressure
Outlines of Reservoir Engineering (1) Introduction
Petrophysical properties ( Rock properties) Fluid properties (gas, water, crude properties) Calculations of hydrocarbon volumes Fluid pressure regimes
(2) Gas reservoirs
Calculating gas in place by the volumetric method Calculating gas recovery factor Material balance calculation (Depletion & Water drive) Hydrocarbon phase behavior ( behavior (gas condensate phase behavior) The gas equivalent of produced condensate and water
(3)) PVT analysis for oil (3
Definition of the basic PVT parameters Determination of the basic PVT parameters in the t he lab. And conversion for field operating conditions.
Outlines of Reservoir Engineering ± cont.
(4) Material balance applied to oil reservoirs
General form of the material balance equation for a hydrocarbon reservoir ( reservoir (Undersaturated and Saturated reservoir) Reservoir drive mechanisms Solution gas drive Gas cap drive Natural water drive
(5) Darcy¶s law and applications
Outlines of Reservoir Engineering ± cont.
(6) The basic differential equation for for radial flow in a porous medium
Derivation of the basic radial flow equation Conditions of solution Linearization of radial flow equation
(7) Well inflow equations e quations for stabilized flow conditions
Semi steady state solution Steady state solution Generalized form of inflow equation (for semi steady state)
Outlines of Reservoir Engineering ± ± cont. cont. (8) The constant terminal rate solution of the radial diffusivity diffusivit y equation and its application to oil well testing
Constant terminal rate solution General Transient flow Semi steady state flow Superposition theorem; general theory of well testing The Matthews, Brons, Hazebroek pressure buildup theory Pressure buildup analysis techniques Multi--rate Multi -rate drawdown testing The effects of partial well completion After--flow After -flow analysis
Outlines of Reservoir Engineering ± ± ccont. ont.
(9) Gas well testing - Linear Linearizatio Linearization ization n and and solution solution of the the basic differential differential equation equation for the radial flow of a real gas - The Russell, Russell Russ ell,, Goodrich, Goodrich, Goodri ch, et et al. Solution Solution Solut ion technique techniqu techn iquee - T The he AlAl-Hussainy, -Hussainy, Ramey, Crawford solution technique - Pre Pressur Pressure ssuree squared squared and pseudo pseudo pressure pressure pres sure solution solutio solu tion n technique techniquee techniqu - NonNon-Darcy -Darcy flow & determination of the non non--darcy -darcy coefficient - The constant constant terminal terminal rate rate solution solution for the flow of a real gas g as - Gen General eral the theory ory of gas gas well well testing testing test ing - M Multiulti-rate -rate testing of gas well - Pre Pressur Pressure ssuree building building buildi ng testing testing test ing of gas gas wells wells - Press Pressure ure building building buildin g analysis analysis analys is in solution solution gas drive drive reservoirs reservoirs reservoi rs
Outlines of Reservoir Engineering ± ± ccont. ont.
(10) Natural water influx - St Stea Steady eady dy state stat st atee model mode mo dell - Un Unst Unsteady stea eady dy state stat st atee model model mod el - Th Thee van Everdingen Ever Ev erdi dinge ngen n and and Hurst Hurst Hur st edgeedge-water edge -water drive model - B Bottom ottom ± ± wat water er dri drive ve model model - Ps Pseu Pseudo eudo do steady stea st eady dy state stat st atee model modell (Fetkovich model) mode - Pre Predic Predicting dictin ting g the the amount amount amou nt of water water influx influx infl ux
Fluid Pressure Regimes The total pressure at any depth = weight of the formation rock + weight of fluids (oil, gas or water) [=] 1 psi/ft * depth(ft)
Fluid Pressure Regimes
Density of sandstone ! 2.7
gm cm
3
! 168.202
v
2.2lbm (0.3048 048 v 100cm) 1000 gm
lbm ft 3
v
1 slug 32.7lbm
slug ! 5.22 3 ft
(1 ft )
3
3
Pressure gradient for sandstone
Pressure gradient for sandstone p p D
! V gD ! V g
! 5.22 v 32.2 ! 168.084 ! 168.084
lbf 2
lbf ft 3
1 ft 2
ft ft 144in
! 1.16( psi / ft )
2
! 1.16
lbf in 2 ft
Overburden pressure
P) Overburden pressure (O (OP) = Fluid pressure (FP) + Grain or matrix pressure (GP)
OP=FP + GP
non-isolated -isolated reservoir In nonPW (wellbore pressure) = FP
In isolated reservoir PW (wellbore pressure) = FP + GP¶ where GP¶<=GP
Normal hydrostatic pressure
In a perfectly normal case , the water pressure at any depth Assume :( :(1) Continuity of water pressure to the surface d epth. (2) Salinity of water does not vary with depth.
P [
(
(
!(
dP dD
dP dD
dP dD
)
) water v D 14.7
ter
wa
) water
! 0 .4335 "
0.4335
[=] psia psi/ft for pure water
psi/ft
for saline water
Abnormal hydrostatic pressure ( No continuity of water to the surface)
P [
!(
dP dD
)
ter
wa
v D 14 .7 C [=] psia
Normal hydrostatic pressure c=0
Abnormal (hydrostatic) pressure c > 0 Overpressure (Abnormal high pressure) c < 0 Underpressure (Abnormal low pressure)
Conditions causing abnormal fluid pressures
Conditions causing abnormal fluid pressures in enclosed enc losed water bearing sands include
Temperature change T = +1 P = +125 Temperature +125 psi in a sealed fresh water system Geological changes ± ± upli uplifti uplifting; fting; ng; surface surface erosion erosion Osmosis between waters having different salinity, the sealing shale acting as the semi permeable membrane in this ionic exchange; if the water within the seal is more saline than the surrounding water the osmosis will cause the abnormal high pressure and vice versa.
Are the water bearing sands abnormally pressured ?
h ave on the extent of any If so, what effect does this have hydrocarbon accumulations?
Hydrocarbon pressure regimes r egimes
In hydrocarbon pressure regimes (
(
dP dD
) water
dP dD
) oil
dP ) g s ( dD a
$
$
0.45
0.35
$ 0 . 08
psi/ft psi/ft psi/ft
Pressure Kick 5 0 0 0 x 0.45 0.45 + 15
2 2 6 5 P si
2 3 6 9 P si
P 5000
G A S
5100 P g = P 0 = 2385 P si
5200
G O C
G O C
5300
(5200 ft)
O IL
5400 O W C
5500 P g = P w = 2490 P si
O W C
(5500 ft)
W ater
D 5 5 0 0 x 0.45 0.45 + 15
Assumes a normal hydrostatic hydrostatic pressure regime regime P= 0.45 0.45 × D + 15 15
In water zone
at 5000 ft
at OWC (55 (5500 00 ft) P P((at OWC) = 55 5500 00 × 0.4 0.45 5 + 15 15 = 2490 psia
P((at P at5 5000) = 5000 × 0.4 0.45 5 + 15 15 = 2265 2265 psia
Pressure Kick 5 0 0 0 x 0.45+ 15
2 2 6 5 P si
2 3 6 9 P si
P 5000
GAS
5100 P g = P 0 = 2 3 8 5 P si
5200
GO C
GOC
5300
(5200 ft)
O IL
5400 OW C
5500 P g = P w = 2 4 9 0 P si
OW C
(5500 ft)
W ater
D 5 5 0 0 x 0.45 + 15
In oil zone Po = 0. 0.35 35 x D + C at D = 55 5500 00 ft , Po = 2490 psi C = 2490 ± ± 0. 0.35 0.35 × 55 5500 00 = 565 psia Po = 0.35 0.35 × D + 565 at GO GOC (5 (5200 200 ft) Po (at GO GOC) = 0.35 0.35 × 5200 + 565 = 2385 2385 psia
Pressure Kick
0.0 8 D + 196 1969 ( psia) psia) In gas zone Pg = 0.08 at 5000 ft Pg = 0.0 0.08 1969 = 236 2369 9 psia 8 × 5000 + 196
Pressure Kick 2450Psia
2265Ps 2265Psia
P
P 5000
5000
G AS
5100 hyd hydrostatic pressure
5200 5300
5100 GO C
5300 5400
O IL
5400
OW C
5500 P0=Pw =2490Ps =2490Psia
In gas zone Pg = 0.0 0.08 8D+C At D = 55 5500 00 ft, Pg = P = 2490 psia 2490 = 0.08 0.08 × 55 5500 00 + C C = 205 2050 psia Pg = 0.08 0.08 × D + 205 2050 At D = 5000 ft Pg = 245 2450 psia
GAS
GW C
5500 Pg=Pw=2490Ps =2490Psia
W ater
D
Gaspressure gradient
5200
D
W ater
GWC error from pressure measurement measurement
Pressure = 25 2500 psia at D = 5000 ft in gasgas-water -water reservoir GWC = ? Sol. Pg = 0.08 0.08 D + C C = 25 2500 ± ± 0.0 0.08 0.08 × 5000 = 2100 psia Pg = 0.08 0.08 D + 2100
Water pressure P = 0.45 0.45 D + 15 15
Water pressure P = 0.45 0.45 D + 15 15
At GWC Pg = P 0.08 0.0 8 D + 2100 = 0.45 0.45 D + 15 15 D = 5635 ft (GWC)
At GWC Pg = P 0.08 0.0 8 D + 205 2050 = 0.45 0.45 D + 15 15 D = 55 5500 00 ft (GWC)
Pressure = 24 245 50 psia at D = 5000 ft in gasgas-water -water reservoir GWC = ? Sol. Pg = 0.0 0.08 8D+C C = 24 245 50 ± ± 0.0 0.08 0.08 × 5000 = 20 205 50 psia Pg = 0.0 0.08 8 D + 205 2050
Results from Errors in GWC or GOC or OWC
GWC or GOC or OWC location affecting volume of hydrocarbon OOIP affecting OOIP or OGIP
affecting development plans
Volumetric Gas Reservoir Engineering
Gas is one of a few substances whose whose state, as defined by pressure, volume and temperature (PVT)
One other such substance is saturated steam.
The equation of state for an ideal gas
pV ! nRT .
. . . .
(1.13)
(Field units used in the industry) p [=] psia; V[=] ft3; T [=] OR absolute temperature n [=] lbm moles; n=the number of lb moles, moles, one lb mole is the molecular weight of the gas expressed in pounds. R = the universal gas constant [=] 10.73 10.732 2 psia· ft3 ft3 / (lbmmole·0R) Eq (1.1 1.13 3) results form the combined efforts of Boyle, Charles, Avogadro and Gay Lussac.
The equation of state for real gas
The equation of Van der Waals (for one lb mole of gas
( p
a
2
V
)(V b) ! RT .
. . .
(1.14)
where a and b are dependent on the nature of the gas. The principal drawback in attempting to use eq. (1.14) to describe the behavior of real gases encountered in reservoirs is that the maximum pressure for which the equation is applicable is still far below the normal range of reservoir pressures
The equation of state for real gas
the BeattieBeattie-Bridgeman -Bridgeman equation
the BenedictBenedict-Webb -Webb Webb--Rubin -Rubin equation
the nonnon-ideal -ideal gas law
Non--ideal Non -ideal gas law
pV ! nzRT .
. . .
(1.15)
Where z = zz-factor -factor =gas deviation factor
=supercompressibility factor
V Actual volume o f n moles o f g a s z ! ! V i Ideal volume o f n moles o f g a s a
t T and P
a
t T and P
a
z ! f ( P , T , com po sitio n) compo sitio n K g ! specific gr avity (air ! 1)
Determination of zz-factor -factor
There are three ways to determinatio determination n zz-factor -factor :
(a)Experimental determination
( b)The b)The zz-factor -factor correlation of standing and
katz
(c)Direct calculation of zz-factor -factor
(a) Experimental determination
n mole s of gas
p=1atm; T=reservoir temperature; => V=V0
pV=nzRT z=1 for p=1 atm =>14.7 =>14. 7 V0=nRT
n mole of gas p>1atm; T=reservoir temperature; pV=nzRT pV=z((14. pV=z 14.7 7 V0) pV
z
!
14.7V 0
=> V=V
p scV 0 z scT
!
pV zT
z !
pV p scV 0
By varying p and measuring V, the isothermal z( z ( p) p) function can be readily by obtained.
b)The zz -factor -factor correlation of standing and katz ( b)The
Requirement: Knowledge of gas composition or gas gravity Naturally occurring hydrocarbons: primarily paraffin series CnH2n+2
Non--hydrocarbon Non -hydrocarbon impurities: CO C O2, N2 and H2 Gas reservoir: lighter members of the paraffin series, C1 and C2 > 90% of the volume.
The Standing Standing--Katz -Katz Correlation
knowing Gas composition (ni) Critical pressure (Pci) Critical temperature (Tci) of each component P.16 ) ( Table (1.1) and P.16 Pseudo critical pressure (Ppc) Pseudo critical temperature (Tpc) for the mixture P pc
§ n P ! § n T !
T pc
ci
i
ci
i
i
i
Pseudo reduced pressure (Ppr) Pseudo reduced temperature (Tpr)
!
T pr !
P pr
Fig.1.6; p.17 Fig.1.6 p.17
z--factor zfactor
P P pc T T pc
! const .( Isothermal )
b¶)The zz -factor -factor correlation of standing and katz ( b¶)The
For the gas composition is not available and the gas gravity (air=1) is available. The gas gravity (air=1) ( K g )
Pseudo critical pressure (Ppc)
Pseudo critical temperature (Tpc)
fig.1.7 , p18 fig.1.7 p18
( b¶)The b¶)The zz -factor -factor correlation of standing and katz
Pseudo reduced pressure (Ppr) Pseudo reduced temperature (Tpr)
P pr
!
P P pc
T pr !
T T pc
! const .( Isothermal )
Fig1.6 p.17 z-factor z-factor
The above procedure is valided only if impunity (CO2,N2 and H2S) is less then 5% volume.
z -factor -factor (c) Direct calculation of z
The Hall Hall--Yarborough -Yarborough equations, developed using the Starling Starling--Carnahan -Carnahan equation of state, are z
!
0.06125 125 P pr te 1.2 (1t ) y
2
(1.20)
. . . .
where Ppr= the pseudo reduced pressure
t=1/Tpr Tpr=the pseudo pseudo reduced tempera temperature ture
y=the ³reduced´ density which can be obtained as the solution of the equation as followed: 0.06125 125 P pr te
(90.7t
1.2 ( 1 t )
2
y y 2
3
y y
(1 y )
3
4 2
(14.76t 9.76t
242.2t 2 42.4t 3 ) y ( 2.18
2.82 t )
! 0.
4.58t 3 ) y 2
. . .
(1.21)
This nonnon-linear -linear equation can be conveniently solved for y using the simple Newton--Raphson Newton -Raphson iterative technique.
z -factor -factor (c) Direct calculation of z
The steps involved in applying thus are: k iteration counter ( counter (which in this make an initial estimate of y , where k is an iteration case is unity, e.q. y1=0.001 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 non--zero -zero value Fk . (3)) using the first order Taylor series expansio (3 expansion, n, a better estimate of y can be determined as y
k 1
! y
k
F
dF
where
k
. . . .
k
(1.22)
y d dF
k
d y
!
1 4 y 4 y
2
3
4 y y
(1 y )
4
4 2
3
( 29.52t 19.52t 9.16t
) y
2.82t )(90.7t 242.2t 2 42.4t 3 ) y (1.18 2.82 t ) . . . . (1.23) (4) iterate, using eq. (1.21) and eq. (1.22), until satisfactory convergence is obtained(5 obtained(5)) substitution of the correct value of y in ( 2.18
eq.(1.20)will give the zeq.( z-factor. -factor. (5)) substitution of the correct value of y in eq.( (5 eq. (1.20)will give the zz-factor. -factor.
Application of the real gas equation of state
p
Equation of state of a real gas
! nz
T . . . .
This is a PVT relationship to relate surface to reservoir volumes volumes of hydrocarbon. (1) the gas expans expansion ion fact factor or E,
E !
sc
!
volume o f n
moles o f
g a s
at
s tan dar d itions d cond iti
volume o f n
moles o f
g a s
at
reservoir cond iti itions
Real gas equation for n moles of gas at standard conditions nz sc T sc p sc sc ! nz sc T sc ! sc p sc
Real gas equation for n moles of gas at reservoir conditions
p
>
! nz E !
V sc V
T
nz sc !
nz
>
E ! 35.35
T sc
p sc
T
p zT
p
!
nz sc
V
T sc p
nz T p sc
!
!
nz
T sc p
zT p sc
T
p !
519.6 v V
zT 14.7
( note : z sc ! 1)
surface surface v volume/reservoir olume/reservoir volume volume
[ !] [=] SCF/ft3 SCF/ft3 or STB/bbl
(1.15)
Example Reservoir condition: P=2000psia; T=18 T=1800F= 00F=((180+4 0+45 9.6 )=63 9.6 z=0.865 59. 6)= 639. 60R; z=0.865 > 2000 127.8 surface volume/reservoir E ! 35.35 ! 127 0.865 v 639 639.6 or SCF/ft3 SCF/ft3 or STB/bbl
P ! V J (1 OG I
S wi ) E i
(2) Real gas density
! VV
m
V
!
m
!
nM
V V where n=moles; M=molecular weight) V
nM nzRT p
!
!
MP
!
V g s
zRT
a
M g s P a
z g s RT a
at any p and T
For gas
V g s a
!
M g s P a
z g s RT a
For air
V
M ir p
ir !
a
a
z ir RT a
M g s p z g s RT
M g s
a
V g s a
V
ir
a
! K g !
a
M g s p z ir RT a
a
!
( M ) g s K g ! z ( M ) ir Z a
a
Z g s a
M ir Z ir a
a
a
(2) Real gas density
K
a
V
!
z
(
Z
)
g a s
)
air
At standard conditions z air = zgas = 1 V g s
g
(
! K g !
ir
a
M g s a
M ir
!
a
M g s a
2 8 .9 7
(1.28)
K g } 0.6 ~ 0.8
in general (a) If
. . . .
K g is known, then
M g s a
where
V g s ! K g V
g a s
28.97 ( V
a
)
ir sc
a
! 0.0763 lbm
ft 3
a
M g s !
( b) b) If the gas composit composition ion is known, then
K g !
V g s ! K g V
! K g 2or 8.97 , a
§ n M i
i
ir
a
i
ir
a
(3)Isothermal compressibility of a real gas
pV ! nz xV x p xV x p
V
T
! n T z[ p 2 ] n T p 1
nz
!
g !
C g
C
T
p 1
xV
V x p
!
g
1 p
}
1 x z
!
nz p
xV
x p
x p
!
1 1 x z [ V ( )] p z x p V
z x p
! n T zp 1
x z
1 ( p
)!
T
V (
1
p
1 x z
z x p
(note : z ! f ( p ))
!
nzRT p 2
nRT x z p
x p
)
1
1 x z z x p 1 p
since
1 p
""
1 x z z x p
p.24, fig.1.9
Exercise 1.1 - P Problem roblem
Exercise1.1 Gas pressure gradient in the reservoir
(1) Calculate the density of the gas, at
standard conditions, whose composition compos ition is listed in the table 11-1. -1.
(2) what is the gas pressure gradient in
the reservoir at 2000psia and 1800F(z=0.865)
Exercise 1.1 -- ssolution olut ol utio ion n -1 -1
(1) Molecular weight of the gas g a s
!
§
ni
i
K g !
! 19.91
i
since
K g !
V g s a
V
V g s ! K g V a
M g s a
28.97
!
19.91 28.97
! 0.687
ir
a
ir
a
V g a s ! 0.687 0.0763(lbm ft 3 ) ! 0.0524(lbm ft 3 )
or from
pV
V !
pV ! nzRT ! n zRT m V
!
! mzRT
p zRT
At standard condition
V g s a
!
P sc z sc RT sc
!
14.7 v 19.91 1 v 10.73 v 519.6
! 0.0524(lbm ft 3 )
Exercise 1.1 -- ssolution olut ol utio ion n -2 -2
(2) gas in the reservoir conditions
pV ! nzRT V !
m V
!
p z
T
!
pVM ! nMzRT 2000 v 19.91
0.865 v 10.73 v ( 459.6 180)
! m zRT ! 6.707(lbm ft 3 )
Exercise 1.1 -- ssolution olut ol utio ion n -3
p d p
D d
! V gD !
d p
! (6.707
V g
! 6 . 707
! 6 . 707 ! 6.707
d D ! Vg d
slug
ft
ft 3 32.2lbm
)32.2
ft 2 s
ft
ft 3 s lb
lbm 1 slug
2
f 3
2 lbf 1 1 ft
ft 2 ft 144in 2
0465 ! 0.0465
lb f 1 in 2 ft
! 0 . 0465 0465
psi ft
Gas Material Balance: Recovery Factor
Material balance
Production = OGIP (GIIP) - Un Unpr Unproduced prod oduc uced ed gas gas
(SC)
(SC)
(SC)
Case 1no water influx (volumetric depletion reservoirs) Case 2water influx (water drive reservoirs)
Volumetric depletion reservoirs -- 1
No water influx into into the reservoir reservoir from the adjoining aquifer. Gas initially in place (GII IIP) P) or I or Initial gas in placeIGIP
G
Original gas in place OGIP [=] Standard Condition Volume
G ! V J (1 s c ) E i
CF [!] S CF
w
here
w
! 35.37
i
pi
z i T i
3 CF / ft [!] S CF
Material Balance at standard conditions
Production
SC G p
GII IIP P
Unproduced gas
SC SC
¨ G ¸ ! G ©© ¹¹ v ª E i º
E . . . .
(1 .33 )
Where G/Ei = GII GIIP P in reservoir volume or reservoir volume filled with gas HCPV
Volumetric depletion reservoirs -- 2 G p
G
G
E
.
E i
.
E ! 35 . 37
sin ce G p
!1
35.37
! 1 35.37
.
zT
p
here
w
Gp G
?! A
! 1 z
z i T i
z i
p i
. . .
! the fr actional g a s re cov ery
z
!
p i z i
ft
3
note :T ! T i
! const .
(1 . 35 ) t
a
! Ga s re cov ery f act or
p
SC F
p
zT pi
p i
(1 . 34 )
p
¨ G p ¸ ©© 1 ¹¹ . ! z z i ª G º p
.
¨ p i 1 ¸ ©© ¹¹ G p ª z i G º
ny st a ge during depletion
a
In Eq.1.33
C P V V H C
!
G
E i
! const .
?
V const. H H C C P V
1. the connate water in reservoir will expand
2. the grain pressure increases as gas
because:
(or fluid) pressure declines
! F P GP . . d ( F P ) ! d ( GP )
OP
.
p . 3
) ! d (G / E i ) d ( H C C PV ! dV dV f . w
here
w
V
w
V f
! initi ! initi
(1 . 3 )
.
l (connate
a
. . .
wa
p . 4
(1.36)
ter volume )
l pore volume
a
neg ative sign ""
exp ansion t o
a
o f
ter leads
wa
C PV reduction in H C
c f !
xV f
1
V f x G P
c f ! c f !
1
xV f
V f
(x p )
1 V f
GP
GP
pore vol.
V f
xV f x p
V f ! c f V f d p d V
GP
GP
V w
cw !
d V V w
1
xV w
V w d FP
!
1
d V V w
V w
p d
! c w V w d p
FP
FP
V f
FP
FP=gas pressure FP FP
FP
V w
FP
FP=gas pressure FP
¨ G ¸ ©© ¹¹ ! d H C C P V V ! c ª E i º
d
p c f V f d p V w d
w
Since V f
! P V V !
C P V H C V
1 S wc
V w
! P V V v S wc !
!
G
E i 1 S wc
C P V V H C
1 S
G
S wc !
S wc
E i 1 S wc
wc
¨ G ¸ S c G G © ¹ d d © ¹!c p c f p E i 1 S c E i 1 S c ª E i º ¨ G ¸ ¨ G ¸ ¨ G ¸ « » S c 1 ¹ © ¹ ©© ¹¹ ! c c © ( p ¬ ¼ f © ¹ © ¹ 1 S c ½ ª E i º initi l ª E i º t ª E i º initi l - 1 S c c S c c f ( p ¨ G ¸ ¨ G ¸ ¨ G ¸ ¹¹ ©© ¹¹ ! ©© ¹¹ © © ª E i º t ª E i º initi l ª E i º initi l 1 S c ¨ G ¸ ¨ G ¸ « c S c c f ( p » ©© ¹¹ ! ©© ¹¹ 1 ¬ ¼ E E S 1 c ª i º t ª i º initi l ½ w
d
w
w
w
w
w
a
w
a
w
a
a
w
a
w
w
w
w
w
G p
G
! G E . E i
. . .
(1.33)
« c S c c f ( p » G p ! G ¬1 E ¼ 1 S c ½ E i G p « c S c c f » E !1 ¬1 ¼ E 1 G S c ½ i G
w
w
w
w
w
w
F or
1
G p G
cw ! 3v106 psi1; c f !10v106 psi1 cwS wc c f 1 S wc
and
S wc ! 0.2
!1 0.013 013! 0.987
E !1 0.987 E i
com puting
1.3%
with
d ifference ifference
G p G
E
!1 E i
p/z plot From Eq. (1. 1.35 35)) such as
¨ G p ¸ p ! ©©1 ¹¹ . z z i ª G º pi
p
!
z
pi
z i
pi z i G
p/z
. . .
(1.35)
G p
Abandon pressure pab 0
p In
v s .
z
G p
Gp
G
pl ot p/z
Y=a+mx y ! x
m
a
! ! !
p z G
p
p i
z i G p i
0
G p / G= R F
1.0
A straight line in p/z v.s Gp plot means that the reservoir is a depletion type
Water drive reservoirs
exp ansion of If the reduction in reservoir pressure leads to an expansion adjacent aquifer water, and consequent influx into the reservoir, the material balance equation must then be modified as: Production = GII GIIP P Unproduced gas SC SC SC Gp G HCPV--We -WeE HCPV Or Gp G G/Ei WeE where We= the cumulative amount of water influx resulting from the pressure drop. Assumptions: No difference between surface and reservoir volumes of water influx Neglect the effects of connate water expansion and pore volume reduction. No water production
Water drive reservoirs
With water production «¨ G W e W p B G p ! G ¬©© -ª E i ¨ G p ¸ ©©1 ¹¹ G º p z i ª ! .
w
¸» ¹¹¼ E º½
pi
z
1
W e E i
. . .
(1.41)
G
where W e*Ei e*Ei /G represents the fraction of the initial hydrocarbon pore volume flooded by water and is, therefore, always less then unity. un ity.
Water drive reservoirs ¨ G p ¸ ©©1 ¹¹ G º p z i ª . ! W E z ¨ ¸ ©1 e i ¹ G º ª pi
since
. . .
¨ W e E i ¸ ©1 ¹ 1 G º ª
¨ G p ¸ " ©©1 ¹¹ z z i ª G º p
(1.41)
p i
in water flux reservoirs
Comparing
¨ G p ¸ ©©1 ¹¹ ! z z i ª G º
p
p i
in depletion type reservoir
Water drive reservoirs ¨ G p ¸ ©©1 ¹¹ G º p z i ª . ! W E z ¨ ¸ ©1 e i ¹ G º ª pi
. . .
(1.41)
eq.(1.41) the following two parameters to be determined In eq.( G; We
History matching or ³aquifer fitting´ to find We Aquifer modelfor modelfor an aquifer whose dimensions are of the same order of magnitude as the reservoir itself.
W e
! cW ( p
Where W=the total volume of water and depends primary on the geometry of the aquifer. P=the pressure drop at the original reservoir reservoir ±aquifer reservoir ± aquifer boundary
Water drive reservoirs
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 line For case B & C in fig 1.11 p. p.3 30 =>Chapter 9
Bruns et. al method
This method is to estimate GII G IIP P in a water drive reservoir
From Eq. (1.40) such as G p
¨G
! G ©© ª E i
G p ! G
G E
E i
¨
G p ! G©©1
ª
¨
G©©1
ª
G!
or
or
W e
. . .
(1.40)
W e E
E ¸
¹¹ W e E E i º
E ¸
¹¹ ! G p E i º G p
¨ E ¸ ©©1 ¹¹ ª E i º G p
¨ E ¸ ©©1 ¹¹ ª E i º Ga
¸ ¹¹ E . º
!G
W e E W e E
¨ E ¸ ©©1 ¹¹ ª E i º
!G
W e E
G p
¨ E ¸ ©©1 ¹¹ ª E i º
¨ E ¸ ©©1 ¹¹ ª E i º
W e E
¨ E ¸ ©©1 ¹¹ ª E i º
(or
Ga )
is plot as function of ¨
W e E
E ¸ ©©1 ¹¹ ª E i º
Bruns et. al method G p
¨ E ¸ ©©1 ¹¹ ª E i º
(or
Ga )
is plot as function of
W e E
¨ E ¸ ©©1 ¹¹ ª E i º
The result should be a straight line, line, provided the correct aquifer model has been selected.
The ultimate gas recovery depends both on
(1) the nature of the aquifer ,and
(2) the abandonment pressure.
The principal parameters in gas reservoir engineering:
(1) the GII GIIP P
(2) the aquifer model
(3)) abandonment pressure (3
(4) the number of producing wells and their mechanical define
Hydrocarbon phase behavior
Hydrocarbon phase behavior
Hydrocarbon phase behavior C---------D--------------E
Residual saturation (flow ceases) Liquid H.C deposited in the reservoir Retrograde liquid Condensate E-------------- -F
Re-vaporization of the liquid condensate ? NO! Because H.C remaining in the reservoir increase Composition of gas reservoir changed Phase envelope shift SE direction Thus, inhibiting re-vaporizati re-vaporization. on.
Condensate reservoir, pt. c,
producing
Wet gas (at scf) Dry gas injection displace the wet gas
until dry gas break through occurs in the producing wells
Keep p above dew pt.
p small
Equivalent gas volume
The material balance equation of eq (1.35) such as p z
!
¨ ©© 1 z i ª
p i
G G
p
¸ ¹¹ º
Assume that a volume of gas in the reservoir was produced as gas at the surface.
If, due to surface separation, small amounts of liquid hydrocarbon hydrocarbo n are produced, the cumulative liquid volume must must be converted into an equivalent gas volume and added to the cumulative gas production to give the correct value of Gp for use in the material balance equation.
Equivalent gas volume
± mole of liquid have been produced, produced , of molecular If n lbm ±mole weight M, then the total mass of liquid is
nM ! K o V liquid volume w
where 0 = oil gravity (water =1) w density of water =62.4 2.43 3 lbm/ft3
¨ lbm ¸ K 0 62 .4 ©© 3 ¹¹ V 0 ft 3 62 . 4K 0V 0 K V V ª ft º n! ! ! M M lbm / lbm mole M
o
n!
bbls 5.61458 1458
62.4K 0V 0
!
n RT p sc
ft 3
1 bbl
M
350.5 n ! 350 V sc
o
w
K 0 N p
where
M
! 350 350.5
V sc ! 1.33 v 10 5
K 0 N p
RT sc
M
p sc
K 0 N p
Equivalent g a s volume
N p [ ! ]bbls ! 350 350 .5
K 0 N p 10.73 v 520 M 14 .7
N p ?! Abbls
Condensate Reservoir
The dry gas material balance equations equations can also be applied to gas condensate reservoir, if the single phase zz -factor -factor is replaced by the ,so,so-called -called ,two phase z--factor. zfactor. This must be experimentally experimentally determined in the laboratory by performing a constant volume depletion experiment. Volume of gas
P
T
Pi
G scf , as charge to a PVT cell
initial pressure above dew point
Tr reservoir temperature
Condensate Reservoir
p decrease by withdraw gas from the cell, and measure gas Gp¶ Until the pressure has dropped to the dew point Z 2 pha se !
p
¨ ©© 1 z i ª
p i
p z
¸ ¹¹ º
G p ' G
!
.
.
¨ ©© 1 z i ª
p i
z !
.
.
(1 . 46 )
¸ ¹¹ . º
G p ' G
.
.
.
.
.
.
(1 . 35 )
p
¨ ©© 1 z i ª
p i
¸ ¹¹ º
G p ' G
The latter experiment, for determining the single phase zz -factor, factor, implicitly assumes that a volume of reservoir fluids, below dew point pressure, is produced in its entirety to the surface.
Condensate Reservoir
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 condensate, this this volume being also recorded as a function of pressure during the experiment. As a result, if a gas condensate sample is analyzed using both experimental experimental techniques, the two phase zz-factor -factor determined during the constant volume depletion will be lower than the single phase zzfactor.
This is because the retrograde liquid condensate is not included in the cumulative cumulat ive gas production Gp¶ in equation1.4 1.46 6, which is therefore lower than it would be assuming that all fluids are produced to the surface, as in the single phase experiment.
