material balance equation
DESCRIPTION
A Gift to All to For Getting commond over MBETRANSCRIPT
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 dimension – this chapter 2. multi-dimension (multi-phase) – reservoir simulation
§ 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)
)(1
)1()( STBB
SVSTBplaceinoilinitialNoi
wc
bbl
bblor
ft
ftor
SCF
SCFconst
oiltheofVolCHinitial
gascaptheofVolCHinitialm
3
3
][.)(...
...
)(STBproductionoilcumulativeN p
)(.
)(.
STBproductionoilcum
SCFproductiongascumratooilgascumulativeRp
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
Expansion of the gascap gas
Expansion of the gascap gas =gascap gas (at p) –gascap (at pi)
RBSCF
RBSCFB
BmNBpatgasofAmount
SCF
SCFRB
RBB
mNBGor
RBSCF
RBSTB
SCF
SCFmNBgasgascapofvolumetotalThe
ggi
oi
gioi
oi
][][1
)(
][1
][1
][][
)3.3()()1(
][
RBB
BmNB
RBmNBB
BmNBgasgascaptheofExpansion
gi
goi
oigi
goi
Change in the HCPV due to the connate water expansion &
pore volume reduction
pS
cScNBm
pcScS
HCPVpcScVpVcSVc
SVvolwaterconnatetheVpVcVc
SHCPVvolporetotaltheVdVdVHCPVd
wc
fwcwoi
fwcww
fwcwfffwcfw
wcfwffww
wffw
)1
()1(
)()1(
)()(
.)(
)1/(.)(
Underground withdrawal
])([)(
)()()(
)()(Pr
gspopgsppop
gspppop
ppp
BRRBNBRRNBNwithdrawaldUndergroun
gasRBBRNRNoilRBBNwithdrawaldUndergroun
gasSCFRNoilSTBNsurfaceatoduction
The general expression for the material balance as
wpe
wc
fwcwoi
gi
goigssioiogspop
BWWpS
cScNBm
B
BmNBBRRNBBNBRRBN
)()1
()1(
)1()()(])([
)7.3()(
1)1(1
)()(])([
wpe
wc
fwcw
gi
g
oi
gssioiooigspop
BWW
pS
cScm
B
Bm
B
BRRBBNBBRRBN
pmeasuringdifficultyMain
pVcdV
fluidsreservoirofExpansionoductionformSimple
tpfW
pfBRBNote
e
gso
:
Pr:
),(
)(,,:
where
)12.3()( , wewfgo BWmEmEENF
STBRBp
S
cScBmE
STBRB
B
BBE
STBRBBRRBBE
RBBWBRRBNF
wc
fwcwoiwf
gi
goig
gssioioo
wpgspop
)1
()1(
][)1(
][)()(
][])([
,
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
§ 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 the reservoir- estimating the primary recovery factor
Reservoir drive mechanism
§ 3.5 Solution gas drive
(a) above the B.P. pressure (b) below the B.P. pressure
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
gssioiooigspop
BWW
pS
cScm
B
Bm
B
BRRBBNBBRRBN
0;0;0;0)(;0)(: pessisp WWmRRRRNote
ilitycompressibweightedsaturationeffectivetheS
cScSccwhere
SSpcNBBNor
pB
BB
pB
BBp
S
cScScNBBN
dp
dB
Bdp
dV
Vcp
S
cSccNBBN
pS
csc
B
BBNBBN
wc
fwwooe
wcoeoiop
oi
oio
oi
ooi
wc
fwwoooiop
o
o
o
oo
wc
fwwooiop
wc
fwcw
oi
oiooiop
,1
1)18.3(
)()()17.3()
1(
11)]
1([
]1
)()([
Exercise3.1 Solution gas drive, undersaturated oil reservoir
Determine R.F.
Solution:
FromTable2.4(p.65)
2.0106.8103
)65.(4.21616
wfw
bi
Spsicpsic
ptablePVT
pppif
STBRBBpsip
STBRBBpsip
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
oo
Eq(3.18)
%5.1015.0
)33304000(108.222511.1
2417.1
..
6
pcB
B
N
NFR
pcNBBN
eob
oi
Pb
p
eoiop
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 0.00107
2701 1.1822 352 0.00119
2702 1.1633 304 0.00137
1800 1.1450 257 0.00161
1500 1.1287 214 0.00196
1200 1.1115 167 0.00249
900 1.0940 122 0.00339
901 1.0763 78 0.00519
300 1.0583 35 0.01066
Bo as Function of Pressure
Rs as Function of Pressure
Bg and E as Function of Pressure
Producing Gas-oil Ratio (R) as Function of Pressure
%7.16167.04000
33304000%
108.22
)106.82.01038.0103.11(2.01
1
1
:
6
666
p
S
cScScc
Note
wc
fwwooe
Below B.P. pressure (Saturation oil)
P<Pb =>gas liberated from saturated oil
16
16
16
1616
106.8
103
103.11
10300103003300
111
psic
psic
psic
psicpsiPp
c
f
w
o
gb
g
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
gssioiooigspop
BWW
pS
cScm
B
Bm
B
BRRBBNBBRRBN
developedisSifnegligibleispS
cScNB
WW
capgasinitialnom
PBbelowgassolutionfor
gwc
fwcwoi
pe
)1
(
0;0
0
..
Eq(3.7) becomes
)20.3(])()[(])([ gssioiogspop BRRBBNBRRBN
201
344
00339.0)122(0940.1
00339.0)122510()2417.10940.1(..
)(
)()(..
900900
ppp
p
p
gspo
gssioiop
RRN
NFR
BRRB
BRRBB
N
NFR
Conclusion:Rp
RF
1
49.0%49500
)55.(3.3.
900
N
NSTB
SCFR
pFigFrom
pp
(2) the overall gas balance
)21.3()1()]()([
)()1(
)1(1
oi
wcgsppssig
gspgppgsiwc
goi
bwcwc
oi
NB
SBRRNRRNS
BRNNBRNBNRS
SNB
ppforS
HCPVvolumepore
S
NB
liberatedgas in the reservoir
totalamountof gas
gasproducedat surface
gas stilldissolvedin the oil
= − −
4428.08.000339.02417.1
)]122500(49.0)122510[(
)1()]()[()1()]()([
wcgoi
spp
ssi
oi
wcgspssig SB
B
RRN
NRR
NB
SBRRNpRRNS