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Chapter 3
Material Balance Applied to Oil Reservoirs
3.1 Introduction
-The Schilthuis material balance equation- Basic tools of reservoir engineering
=> Interpreting and predicting reservoir performance.
-Material balance
1. zero dimensionthis chapter
2. multi-dimension (multi-phase)reservoir simulation
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3.2 General form of the material balance equation for a
hydrocarbon reservoir
Underground withdrawal (RB)
= Expansion of oil and original dissolved gas (RB)(A)
+ Expansion of gascap gas (RB) (B)
+ Reduction in HCPV due to connate water expansion and decrease in
the pore volume (RB) .(C)
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)(1)1()( STBBSVSTBplaceinoilinitialN oi
wc
bbl
bblor
ft
ftor
SCF
SCFconst
oiltheofVolCHinitial
gascaptheofVolCHinitialm
3
3
][.)(...
...
)(STBproductionoilcumulativeNp
)(.
)(.
STBproductionoilcum
SCFproductiongascumratooilgascumulativeRp
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Expansion of oil & originally dissolved gas
)1.3(][])[(
)(.)(.)exp(exp
RBSTB
RBSTBBBNNBNB
patliqpatliqansionoilansionLiquid
oiooio
i
)2.3(][][)(
)]()([exp
RBSCF
RB
STB
SCFSTBBRRNBNRBNR
patgassolutionpatgassolutionansiongasLiberated
gssigsgsi
i
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Expansion of the gascap gas
Expansion of the gascap gas =gascap gas (at p)gascap (at pi)
RBSCF
RBSCFB
BmNBpatgasofAmount
SCF
SCFRB
RB
B
mNBGor
RBSCF
RBSTB
SCF
SCFmNBgasgascapofvolumetotalThe
g
gi
oi
gi
oi
oi
][][1
)(
][1
][1
][][
)3.3()()1(
][
RBB
BmNB
RBmNBB
B
mNBgasgascaptheofExpansion
gi
g
oi
oi
gi
g
oi
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Change in the HCPV due to the connate water expansion &
pore volume reduction
pS
cScNBm
pcScS
HCPVpcScVpVcSVc
SVvolwaterconnatetheVpVcVc
SHCPVvolporetotaltheVdVdVHCPVd
wc
fwcw
oi
fwcw
w
fwcwfffwcfw
wcfwffww
wffw
)1
()1(
)()1(
)()(
.)(
)1/(.)(
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Underground withdrawal
])([)(
)()()(
)()(Pr
gspopgsppop
gspppop
ppp
BRRBNBRRNBNwithdrawaldUndergroun
gasRBBRNRNoilRBBNwithdrawaldUndergroun
gasSCFRNoilSTBNsurfaceatoduction
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The general expression for the material balance as
wpe
wc
fwcw
oi
gi
g
oigssioiogspop
BWWpS
cScNBm
B
B
mNBBRRNBBNBRRBN
)()1
()1(
)1()()(])([
)7.3()(
1)1(1
)()(])([
wpe
wc
fwcw
gi
g
oi
gssioiooigspop
BWW
pS
cScm
B
Bm
B
BRRBBNBBRRBN
pmeasuringdifficultyMain
pVcdV
f luidsreservoirofExpansionoductionformSimple
tpfW
pfBRBNote
e
gso
:
Pr:
),(
)(,,:
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where
)12.3()( , wewfgo BWmEmEENF
STBRBp
ScScBmE
STBRB
B
BBE
STB
RBBRRBBE
RBBWBRRBNF
wc
fwcwoiwf
gi
g
oig
gssioioo
wpgspop
)1
()1(
][)1(
][)()(
][])([
,
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No initial gascap, negligible water influx
With water influx eq(3.12) becomes
Eq.(3.12) having a combination drive-all possible sources of energy.
0& wf cc
)13.3()12.3.( oNEFEq
)14.3(o
e
o E
WN
E
F
)12.3()( , wewfgo BWmEmEENF
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3.4 Reservoir Drive Mechanisms
- Solution gas drive
- Gascap drive
-Natural water
drive
- Compaction drive
In terms of
-reducing the M.B to a compact form to
quantify reservoir performance
-determining the main producing
characteristics,
for example, GOR; water cut
-determining the pressure decline in thereservoir
- estimating the primary recovery factor
Reservoir drive mechanism
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3.5 Solution gas drive
(a) above the B.P. pressure (b) below the B.P. pressure
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Above the B.P. pressure
- no initial gascap, m=0
- no water flux, We=0 ; no water production, Wp=0
- Rs=Rsi=Rp
from eq.(3.7)
)7.3()(
1)1(1
)()(])([
wpe
wc
fwcw
gi
g
oi
gssioio
oigspop
BWW
pS
cScm
B
Bm
B
BRRBBNBBRRBN
0;0;0;0)(;0)(: pessisp WWmRRRRNote
ilitycompressibweightedsaturationeffectivetheS
cScSccwhere
SSpcNBBNor
pB
BB
pB
BBp
S
cScScNBBN
dpdB
BdpdV
Vcp
S
cSccNBBN
pS
csc
B
BBNBBN
wc
fwwooe
wcoeoiop
oi
oio
oi
ooi
wc
fwwoo
oiop
o
o
o
o
o
wc
fwwooiop
wc
fwcw
oi
oiooiop
,1
1)18.3(
)()()17.3()
1(
11)]1
([
]1
)()([
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Exercise3.1 Solution gas drive, undersaturated oil reservoir
Determine R.F.
Solution:
FromTable2.4(p.65)
2.0106.8103
)65.(4.2
1616
wfw
bi
Spsicpsic
ptablePVT
pppif
STBRBBpsip
STB
RBBpsip
obb
oii
12511,3330
2417.1,4000
16103.11)33304000(2417.1
2417.12511.1
11
psi
pB
BBc
dp
dB
Bdp
dV
Vc
oi
oiobo
o
o
o
o
o
Eq(3.18)
%5.1015.0
)33304000(108.222511.1
2417.1
..
6
pcB
B
N
NFR
pcNBBN
e
ob
oi
Pb
p
eoiop
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Table 2.4 Field PVT
P(psia) Bo (Rb/STB) Rs(SCF/STB) Bg( Rb/SCF)
4000 (pi) 1.2417 510
3500 1.2480 510
3300 (pb) 1.2511 510 0.00087
3000 1.2222 450 0.00096
2700 1.2022 401 0.001072400 1.1822 352 0.00119
2100 1.1633 304 0.00137
1800 1.1450 257 0.00161
1500 1.1287 214 0.00196
1200 1.1115 167 0.00249900 1.0940 122 0.00339
600 1.0763 78 0.00519
300 1.0583 35 0.01066
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Bo as Function of Pressure
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Rs as Function of Pressure
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Bg and E as Function of Pressure
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Producing Gas-oil Ratio (R) as Function of Pressure
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%7.16167.0
4000
33304000%
108.22
)106.82.01038.0103.11(2.01
1
1
:
6
666
p
S
cScScc
Note
wc
fwwoo
e
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Below B.P. pressure (Saturation oil)
Pgas liberated from saturated oil
16
1
6
16
1616
106.8
103
103.11
10300103003300
111
psic
psic
psic
psicpsiPp
c
f
w
o
g
b
g
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Exercise3.2 Solution gas drive; below bubble point pressure
Reservoir-described in exercise 3.1
Pabandon = 900psia
(1) R.F = f(Rp)? Conclusion?
(2) Sg(free gas) = F(Pabandon)?
Solution:
(1) From eq(3.7)
)7.3()(
1)1(1
)()(
])([
wpe
wc
fwcw
gi
g
oi
gssioio
oigspop
BWW
pS
cSc
mB
B
mB
BRRBB
NBBRRBN
developedisSifnegligibleisp
S
cScNB
WW
capgasinitialnomPBbelowgassolutionfor
g
wc
fwcw
oi
pe
)
1
(
0;0
0..
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Eq(3.7) becomes
)20.3(])()[(])([ gssioiogspop BRRBBNBRRBN
201
344
00339.0)122(0940.1
00339.0)122510()2417.10940.1(..
)(
)()(
..
900
900
ppp
p
p
gspo
gssioiop
RRN
NFR
BRRB
BRRBB
N
N
FR
Conclusion:Rp
RF1
49.0%49500
)55.(3.3.
900
N
N
STBSCFR
pFigFrom
p
p
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(2) the overall gas balance
)21.3()1()]()([
)()1(
)1(1
oi
wcgsppssi
g
gspgppgsi
wc
goi
b
wcwc
oi
NB
SBRRNRRNS
BRNNBRNBNRS
SNB
ppforS
HCPVvolumepore
S
NB
liberated
gas in the
reservoir
total
amount
of gas
gas
produced
at surface
gas still
dissolved
in the oil=
4428.08.000339.02417.1
)]122500(49.0)122510[(
)1(
)]()[()1()]()([
wcg
oi
sp
p
ssi
oi
wcgspssi
g SB
B
RRN
NRR
NB
SBRRNpRRNS