material balance analysis

8/13/2019 Material Balance Analysis

1/25

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


2/25

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


11/25

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


12/25

3.5 Solution gas drive

(a) above the B.P. pressure (b) below the B.P. pressure


13/25

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

)()([


14/25

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


15/25

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


17/25

Rs as Function of Pressure


18/25

Bg and E as Function of Pressure


19/25

Producing Gas-oil Ratio (R) as Function of Pressure


20/25

%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


21/25

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


22/25

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..


23/25

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


24/25


25/25

(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

material balance analysis

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