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PETE 410
NATURAL GAS ENGINEERING
Ibrahim KocabasPetroleum Engineering Department
King Fahd University of Petroleum & Minerals
Reservoir EngineeringReservoir Engineering
Learning Objectives of Lecture 5:Learning Objectives of Lecture 5:
MBE and its use for gas reservoirs Plots for volumetric reservoirs Plots for water drive Over-pressured reservoirsOve p essu ed ese vo s Gas equivalent of produced condensate
MBEMBEThe Material-Balance Equation (MBE)The Material Balance Equation (MBE) provides a simple, but effective, alternative to volumetric methods for estimating 1. not only original gas in place 2. but also gas reserves at any stage of reservoir
depletiondepletion. The MBE is simply a statement of the principle of conservation of mass orprinciple of conservation of mass, or
(original hydrocarbon mass)-(produced hydrocarbon mass) ( g y ) (p y )=(remaining hydrocarbon mass).
MBEMBE
The MBE is derived as a volumetric balance basedThe MBE is derived as a volumetric balance based on the simple assumption that the reservoir volume, which is selected as the control volume,
i t tremains constant.
With thi ti th h i th l fWith this assumption, the changes in the volume of the constituents, namely solids, oil, water and gas phases must be equal to zero at any time. p q y
(See appendix for derivation)pp
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
The general MBE for dry andThe general MBE for dry and wet gas reservoirs (no oil phase g ( pin the reservoir):
gi p gGB G G B
( ) 0(1 )
w wi fe p w gi
wi
c S cW W B GB pS
MBE
The volumetric methods can be used early in a reservoir's life,
MBE can be applied until after some MBE can be applied until after some development and production.
MBE
MBE can estimate only the gas volumes that are in pressure communication with and that may b lti t l d b th d i llbe ultimately recovered by the producing wells.
C l l i i b d Conversely, volumetric estimates are based on the total gas volume in place, part of which may not be recoverable with the existing wellsnot be recoverable with the existing wells because of unidentified reservoir discontinuities or heterogeneities. g
MBEMBE
Therefore comparisons of estimates from both Therefore, comparisons of estimates from both methods can:
1 provide a qualitative measure of the degree of1. provide a qualitative measure of the degree of reservoir heterogeneity and
2. allow a more accurate assessment of gas2. allow a more accurate assessment of gas reserves for a given field-development strategy.
MBEMBE
Another advantage of MBE is that if sufficient Another advantage of MBE is that, if sufficient production and pressure histories are available, application of these methods can provide insight pp p ginto the predominant reservoir drive mechanism,
Whereas the correct use of volumetric methods requires a priori knowledge of the primary source of reservoir energy.
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
rearranging:rearranging:
( )
( )
g gi
w wi f
G B B
c S cGB p W G B W B
( )
(1 )f
gi e p g p wwi
GB p W G B W BS
Estimation of OGIP by MBE :Estimation of OGIP by MBE :For a volumetric reservoir with negligible
( )G B B G B
rock and water expansion MBE reduces to:
( )g gi p gG B B G B
( )p g g giG B G B B
Prediction plot #1
p gG B vs ( )g giB B
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
E ti ti l t Estimation plot
GSlope=G
Gp
B BBg-Bgi
Estimation of OGIP by MBE : :Estimation of OGIP by MBE : :
Continue to manipulate:Continue to manipulate:
( )p g g giG B G B B
(1 )giBG G (1 )p
gG G
B
Formation volume factorFormation volume factor
Tsc
gp zTBT p
gi
ZpTconstB
.
scT piZ
T
ZpTconstBg .
Z
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
Substituting Bg expressions intoSubstituting Bg expressions into
i iB G B(1 ) (1 )gi p gi
pg g
B G BG G
B G B
1 p pi iG GP PP P Pz z G z z z G
i iiz z G z z z G
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
A th di ti Another predictive and diagnostic plot pi i
i i
GP PPz z z G
P vs GPz
vspG
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
Can use for diagnosis as well as prediction and estimation Can use for diagnosis as well as prediction and estimation
P/z Initial p/z
abandonment
GpG
Recovery factor=Gp/G
Estimation of OGIP by MBE :Estimation of OGIP by MBE :
Note that this is an eq ilibri m eq ation It Note that this is an equilibrium equation. It means that only final point on the curve are important and not the path to reach thatimportant and not the path to reach that point
Therefore the same plot is also valid for Therefore, the same plot is also valid for storage of the same gas. That means we can determine how much we need to inject to jraise the pressure to certain value.
How can you determine the reserves? Gpa ?y p
Diagnostic Plots: Water drive reservoirs
Influence of water influx neglecting water
( )G B B W G B W B
Influence of water influx neglecting water and rock effects:
( )g gi e p g p wG B B W G B W B
( )G B W B G B B W ( )p g p w g gi eG B W B G B B W
p g p w e eG B W B W WF
( ) ( )p g p w e e
g gi g gi g g
W WFG GB B B B E E
pF vs t or G Sh ld i h i t lp
gE Should give a horizontal line for zero We
Diagnostic Use of MBEDiagnostic Use of MBE
Diagnostic Use of MBE :Dry GasDiagnostic Use of MBE :Dry Gas
Influence of water influx neglecting water
( )G B W B G B B W
Influence of water influx neglecting water and rock effects:
( )p g p w g gi eG B W B G B B W
( ( ))gi e p wG B W W BG G
assume Wp is negligible and rearrange again
pg
G GB
p
i
egiegip
BGW
BB
BGW
BB
GG
111
giggg BGBBGBG
Diagnostic Use of MBE :Dry GasDiagnostic Use of MBE :Dry Gas1 pGp
G
1
i
e
z Gpz W
G B
The quantity is actually that is the
giG B
gi
e
BGW HCPV
We
fraction of the HCPV that is invaded by the water (<= 1),
Note that if we do not account for We for a given Gp
gi
Note that if we do not account for We, for a given Gp, p/z values will be high due to pressure support.
This shows that p/z vs Gp can also be used as aThis shows that p/z vs Gp can also be used as a diagnostic plot as follows.
Diagnostic Use of MBE 3Diagnostic Use of MBE
Di ti f t f / l t Diagnostic features of p/z plotAquifer of
P/z different strengths
Maximum possible gas recovery
Gp
recovery
Recovery factors with Water influxRecovery factors with Water influx
In the figure of p/z versus Gp/G : as the aquifer In the figure of p/z versus Gp/G : as the aquifer strength increases, the ultimate gas recovery (denoted by circle) decreases. Why?( y ) y
In water-wet rock, water prefers smaller pores. Gas in larger pores is bypassed due to this capillary effect. Sgr, which is typically quite high (40-50%)high (40 50%).
Note: Sgr is a capillary phenomenon, and it is independent of reservoir pressure P. p p
Recovery factors with Water influxRecovery factors with Water influx
Comp ting the n mber of moles left behind Computing the number of moles left behind (use real gas law):
TRnSzP
gr zPn
and since when we have water influx p/z i hi h h b f l f
z z
remains high, the number of moles of gas left behind is high.
Recovery factors with Water influxRecovery factors with Water influx
At b d t f b th l t i d At abandonment for both volumetric and water drive reservoirs, the recovery factor was:
Initial GIP Final GIPGas recovery factor=
grwc
Gas recovery factorInitial GIP
S(1-S )
ga
wc
B(1-S )
giB
B
giB
Recovery factors with Water influxRecovery factors with Water influx
Wh th i t t i fl thWhen there is a strong water influx the pressure remains essentially constant during the whole process i e Bgi=Bgaprocess i.e. Bgi Bga.
(1 S )Swc
wc
(1-S )(1-S )
grSGRF
You compare the recovery factors for any reservoir d tdata
Abnormally pressured reservoirsAbnormally pressured reservoirs
P/z
Abnormal pressureThis behaviour comes about because of the significance of pore g pcompression and fluid expansion
Gp/Gp
MBE plotsMBE plots
I fl f ( )G B B Influence of pore volume reduction and connate water
( )
( )(1 )
g gi
w wi fgi p g
G B B
c S cGB p G BS
connate water expansion
(1 )wiS
1 ( )(1 )
w wi fg gi p g
i
c S cG B B p G BS
(1 )wiS
MBE plotsMBE plots
I fl f Influence of pore volume reduction and connate waterconnate water expansion
1 ( )(1 )
w wi fp g g gi
wi
c S cG B G B B pS
( )wi
1 1 ( )
(1 )p giw wi f
wi g
G Bc S c pG S B
( )wi g
Abnormally pressured reservoirsAbnormally pressured reservoirs
this material balance eq ation : this material balance equation :
( )1 1p giw wc fG Bc S c P
can also be written as:
( )1 11
p giw wc f
wc gG S B
1 1 pt
i
Gp pc pz z G
Where ct is the equivalent compressibility accounting for water expansion and pore g p pcompression.
Abnormally pressured reservoirsAbnormally pressured reservoirs
Rearranging: Rearranging:
1 1 pi Gp pc p
1 1ti
c pz z G
Plotting vs. gives slope =1/G and
Plotting vs. gives slope 1/G and
intercept = ct
Gas Equivalent of produced condensatefor both dry and wet gases
G h i th i b t d t Gas phase in the reservoir, but condensates observed at surface conditions
A
ReservoirSurface
Gs1 Gs2 Gs3
Sep.1 Sep.2 Tank
B
Reservoir conditioncondition
C Np
Gas Equivalent of produced condensate
G i l t l f N l l t d Gas equivalent volume of Np calculated assuming ideal gas law. Let’s find GE for 1 STB of oil onlySTB of oil only
1 62.4 *5.615 350.5o omass of bbl oil
11o
mass of bbl oilbbl
350.511 o
o o
mass of bbl oiln in bbl oilM M
350.5 133,000sc o sc o
sc o sc o
nRT RTGE Vp M p M
Gas Equivalent of produced condensate
Th th t t l d d Then the total gas produced:
psssp NGEGGGG 321
Similarly for water:
350.5 1 10.73 520 7390 /18 14.7
scw
sc
nRT x xGE x SCF surface barrelP
sc
Appendix Derivation of MBEAppendix Derivation of MBE
The material balance equation is simpler toThe material balance equation is simpler to derive for a dry gas reservoir.
It is instructive to go through the derivation.
The constituents of the gas reservoir are,Gas, water and rock. i.e no oil phase in theGas, water and rock. i.e no oil phase in the
reservoir.
Volume changes of constituentsVolume changes of constituents
Change in gas phase volume +Change in gas phase volume + Change in water phase volume + g pChange in solid phase volume = 0
Change in gas phase volume :Change in gas phase volume :
Initial reservoir gas giGB
SCF gas SCF initial SCF produced
remained = - gas gas
t & tpG G
at p & t
Change in reservoir( )
gas volume gi p gGB G G B
Change in water phase volume :Change in water phase volume : Initial reservoir
,nitial ese voi
Wwatervolume bbl
expansionvolume of Remember compressibility
Eqn’s integration,
freservoirvolume of initial
initialinitial water at p bbl amount due to
amountp drop
(1 ), w
reservoirvolumeW c p
initial water at p bbl
, evolumewater
Wencroached bbl
, p wCumulative
W Bwater produced bbl
Change in water phase volume :Change in water phase volume :
Revolume of Increase in duction in Revolume of Increase in duction inreservoirvolume
water at p Volume of Volume ofinitial water
in control Water due to water due toat p bbl
,
infat p bbl
volume lux production
(1 )
volume ofwater at p
W c p W W B
(1 )w e p wW c p W W Bin controlvolume
Change in water phase volume :Change in water phase volume :
Ch i V l f t i th t lChange in Volume of water in the control volume:
Ch i (1 )w e p w
Change inW W c p W W B
watervolume
Change in rock volume :Change in rock volume :Change in the rock volume affects theChange in the rock volume affects the pore space available for fluids to reside.
But we usually measure and use changeBut we usually measure and use change in pore volume using the compressibility equation:
Change in rock volume :Change in rock volume :
Th h h i l iThen the change in pore volume ischange in
f pg
c V pporevolume
Change in rock volume is negative of change in pore volumechange in pore volume.
change inc V p
f pc V prock volume
Change in rock and water volumecombined:
Combining them results in:
Change inwater
e p w w f pand W W B Wc p c V prock volume
Change in rock and water volumecombined:
R i i th t
S
Recognizing that:
p wiW V S
Remembering that
(1 )gi
pi
GBV
S
(1 )wiS
Change in rock and water volumecombined:
W b iWe obtain
( )w wi fi
Change inwaterc S cand W W B GB p
( )
(1 )e p w giwi
and W W B GB pS
rock volume
MBE :MBE :
Summing up changes in all ofSumming up changes in all of the constituents and equating q gthem to zero gives:
( ) 0
gi p g
w wi f
GB G G B
c S cW W B GB
( ) 0(1 )
w wi fe p w gi
wiW W B GB p
S