K (II-2) Material Balance Concepts

8/11/2019 K (II-2) Material Balance Concepts

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Copyright 2008, NExT, All rights reserved

Basics of Reservoir Engineering Module II

II.2 Material Balance Concepts


2/91


The Material Balance Concept

What will happen if weremove 10 scf of air

from each tire?

Tractor Tire

Pressure = 45 psia

Bicycle Tire

Pressure = 45 psia


3/91


The Material Balance Concept

Tractor Tire

Pressure = 34 psia

Bicycle Tire

Pressure = 14.6 psia


4/91Copyright 2008, NExT, All rights reserved

What is Material Balance ?

Relationship between reservoir pore volume, reservoir

pressure, and cumulative production/injection

Applications

Estimate volume of hydrocarbons in place Estimate average reservoir pressure

Estimate average fluid saturations in reservoir



Material Balance Applications

Given Find

Volume of fluids produced

Average reservoir pressure

Fluid PVT relationships

Original hydrocarbons In place

Volume of fluids produced

Original hydrocarbons In placeFluid PVT relationships

Average reservoir pressure

Average fluid saturations



Outline

Basic theory and concepts

Material balance analysis

Volumetric oil reservoirs

Volumetric gas reservoirs

Aquifer driven reservoirs



Development of the Equation

The reservoir is filled with fluid (oil, gas, water) at all times;therefore, as fluids are produced:

The change in reservoir pore volume =the change in reservoir oil volume

+ the change in reservoir free gas volume

+ the change in reservoir water volume



How Does the Reservoir Remain Filled With Fluids During

Production?

Fluid expansion

Pore volume compression

Natural water influx

Fluids (water and/or gas) injection



Fluid and Rock Properties

Solution gas/oil ratio (Rs

)

Oil formation volume factor (Bo)

Gas formation volume factor (Bg)

Total formation volume factor (Bt)

Formation compressibility (cf)

Water compressibility (cw)



Solution Gas/Oil Ratio (Rs)

0

100

200

300

400

500

600

700

800

0 500 1000 1500 2000 2500 3000 3500 4000 4500

Pressure, psia

SolutionGas-OilRatio,scf/stb

Bubblepoint Pressure

Undersaturated

Saturated

Oil F ti V l F t (B )



Oil Formation Volume Factor (Bo)

1.000

1.050

1.100

1.150

1.200

1.250

1.300

1.350

1.400

0 500 1000 1500 2000 2500 3000 3500 4000 4500

Pressure, psia

OilForm

ationVolume

Factor,rb/stb

Bubblepoint Pressure

Undersaturated

Saturated

G F ti V l F t (B )



Gas Formation Volume Factor (Bg)

0

5

10

15

20

25

30

0 500 1000 1500 2000 2500 3000 3500 4000 4500

Pressure, psia

GasForm

ationVolumeF

actor,rb/Mcf

C ibilit


13/91


Compressibility

Coefficient of isothermal compressibility (c)

dp

dv

vc

i

1=

p

v

vc

i

= 1

pcvv i =

B i N l t


14/91


Basic Nomenclature

OIP/GIP oil/free gas in place

N/G original OIP/GIP [ Also known as OOIP and OGIP ]Np cumulative oil production

Gp cumulative gas production

Wp cumulative water productionWi cumulative water influx/injection

Gi cumulative gas injection

Note:

All except for OIP/GIP are at standard conditions

B i N l t


15/91


Basic Nomenclature

Bo,w,g formation volume factor of oil, water, and gas, respectively

cf,o,w compressibility of formation, oil, and water, respectively

Rs solution gas-oil ratio

Rp cumulative GOR

Sw,wi average and connate water saturation

m ratio of initial free gas volume to initial oil volume at reservoirconditions

Subscript i when used with fluid properties denotes initial

conditions


16/91


Basic Nomenclature

Free Gas

Oil + Solution Gas

Water

NpGpWp

Sw = Vwi/Vpim = Vgi/Voi

GiWi

Vp

Vg

Vo

Vw

Derivation of MBE


17/91


Derivation of MBE -

Undersaturated Oil Reservoir

Assumptions

P > Pb

No original or final gas cap

No water influx or production

Derivation of MBE


18/91


Derivation of MBE -


Oil volume

(N-Np)Bo

Oil volume

NBoi

Np

Gp

Rock and water

expansion

Original conditions Later conditions

Derivation of MBE


19/91


Derivation of MBE -


From definition of compressibility

Thus, change in reservoir water volume due to pressure

change:

p

V

Vdp

dV

Vc w

wiT

w

w

w

=

=

11

pVcV wiww =

Derivation of MBE


20/91


Derivation of MBE -


As pressure decreases, matrix supporting structure collapses

into pore space

Thus, change in pore volume due to pressure change:

p

V

Vdp

dV

Vc

p

piT

p

p

f

=

=

11

pVcV pifp =

Derivation of MBE


21/91


Derivation of MBE -


Total change in water volume and pore volume:

Note that

Thus

totalpifwiwpw VpVcVcVV =+=+

piwiwi

pww

VSVVSV

==

pVcScV pifwiwtotal +=

Derivation of MBE -


22/91


Derivation of MBE -


Also

Thus

wi

oipi

SNBV= 1

( ) pcScS

NBV fwiw

wi

oitotal +

=

1

Derivation of MBE -


23/91


Derivation of MBE -


The volumetric balance becomes

Solving for N:

( ) ( ) pcScS

NBBNNNB fwiw

wi

oiopoi +

=

1

( ) oiiwi

fwiw

oio

op

Bpp

S

cScBB

BNN

++

=

1

Derivation of MBE -


24/91


Derivation of MBE


To simplify, note:

If Vsc is volume of oil at standard conditions

p

V

Vdp

dV

Vc

T

o=

= 11

=

=

i

oio

oii

scisc

sci ppBB

BppVVVV

VVpV

V111

Derivation of MBE -


25/91


Derivation of MBE


Then

Substituting( )ppBcBB ioiooio

=

( )ppS

cSccB

BNN

i

wi

fwiw

ooi

op

++

=

1

Derivation of MBE -


26/91


Derivation of MBE


Define

Finally

wi

fwiwoio

wi

fwiw

oeS

cScSc

S

cSccc

++=

++=11

( )ppcBBN

Nieoi

op

=

Material Balances - Exercise 1


27/91


Material Balances Exercise 1

Determine the OOIP for the undersaturated reservoir given the data

Np = 1.4*106

STB Bo = 1.46 RB/STB

Boi = 1.39 RB/STB

cw = 3.71*10-6 1/psi cf= 3.52*10

-6 1/psi

Swi = 32%

The reservoir was discovered at an initial pressure of 4,300 psi. Pressure hasdeclined to 2,450 psi

Also calculate N, assuming cf= 0 and compare results

Derivation of MBE -


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Derivation of MBE

Saturated Oil Reservoir

Assumptions

P Pb

No original gas cap

No water influx or production Negligible rock and water expansion

Derivation of MBE -


29/91


Derivation of MBE


Oil volume

(N-Np)Bo

Oil volume

NBoi

Np

Gp

Gas Volume

Original conditions Later conditions

Derivation of MBE -


30/91



Determine final free gas volume by performing a gas balance

Original dissolved gas = NRsi

Final dissolved gas = (N-Np)Rs

Gas produced = Gp

Therefore Final free gas = NRsi - (N-Np)Rs - Gp

Convert to reservoir conditions

Final free gas = (NRsi - (N-Np)Rs Gp ) Bg / 5.614

Derivation of MBE -


31/91



The volumetric balance becomes:

Solving for N

( ) ( )[ ]pspsig

opoi GRNNNRBBNNNB +=614.5

( )

( ) oig

ssio

g

pspop

BB

RRB

BGRNBN

N

+

=

614.5

614.5

Derivation of MBE -


32/91



To simplify, note that

Also, since no gas evolved at Pb

( )

p

p

p

g

ssioT

N

G

R

BRRBB

=

+= 614.5

Tioi BB =

Derivation of MBE -


33/91



Finally

( )

TiT

g

sipTp

BB

BRRBN

N

+

=

614.5

General MBE


34/91


))()(1(1)()(

)()(

RiRfww

wi

oi

gi

gig

oigssioio

giweipgsppop

PPccSmS

B

B

BB

mBBRRBB

BGBWWWBRNGBN

N

+++

++

++

=

Material Balance Analysis


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y

Data requirements

Assembling the data set

Data QC

Black oil material balance

Water influx

Material Balance Analysis


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Data requirements

Estimates of average reservoir pressure vs time

Reservoir fluid PVT relationships

Reservoir cumulative production and injection volumes

Average Reservoir Pressure


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Average Reservoir Pressure

0

500

1000

1500

2000

2500

3000

3500

4000

0 50 100 150 200 250

Hours Since Shut-In

Bottom-HolePressure,psia

Fluid PVT Relationships


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Fluid PVT Relationships

Methods for obtaining relationships

Reservoir fluid study (laboratory analysis)

Correlations

Cumulative Production/Injection Data


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Cumulative Production/Injection Data

Source

Operator records of monthly production and injection

Potential errors

Date of first record date of first production Inaccurate reporting of noncommercial phases

Wrong set of wells

Data Preparation


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Data Preparation

Convert all pressure data to common datum

Plot pressure vs. time for all wells

Calculate cumulative production/injection from reservoir

Assemble fluid PVT data

Converting Pressures to Common Datum


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Converting Pressures to Common Datum

)( measureddatumf

measureddatum

hh

pp

+

=

Plot of Pressure vs. Time


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Plot of Pressure vs. Time

0

500

1000

1500

2000

2500

3000

3500

Jan-90 Jan-91 Jan-92 Jan-93 Jan-94

Date

MeasuredPress

ure,psia

Well #1

Well #2

Well #3

Well #4

Black Oil Material Balance


43/91


Black Oil Material Balance

Straight line analysis techniques (Havlena-Odeh)

Assumptions Analysis techniques

Common pitfalls

Basic Assumptions


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Basic Assumptions

Volumetric reservoir model

Closed system (no fluid influx across boundary ofreservoir)

Measured pressures represent average reservoir pressure

Black oil fluid PVT relationships are accurate

Reservoir Models


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Reservoir

Volumetric Reservoir

Reservoir Models


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Reservoir Aquifer

Aquifer Driven Reservoir

Reservoir Models


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Compartmental Reservoir

Reservoir 3 Reservoir 1 Reservoir 2

Straight-Line Analysis Techniques


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g y q

The material balance equation as a straight line

Introduced by Havlena and Odeh

Typical straight-line techniques OOIP vs. cumulative oil production

F vs. Etotal

F/EO vs. Eg/Eo

MBE as a Straight Line


49/91


g

)( fwgo EmEENF ++=

ortotalNEF=

MBE as a Straight Line


50/91


wpgsppop BWBRNGBNF ++=

gssioioo BRRBBE )()( +=

= 1

gi

g

oig

B

BBE

( ) pS

cScBmE

wi

fwiw

oifw

++=

1

1

Typical Straight-Line TechniquesF/E vs Cumulative Oil Production


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F/Etotal vs. Cumulative Oil Production

OOIP vs. Cum Oil - Example Reservoir

3,000

3,500

4,000

4,500

5,000

5,500

6,000

0 200 400 600 800 1000 1200 1400 1600 1800

Cum Oil Production, Mstb

OOIP,

Mstb

OOIP = 4,833.9 Mstb (32.3%)

OGIP = 6,187.3 MMscf (49.2%)

Typical Straight-Line TechniquesF vs E


52/91


F vs. Etotal

F vs. Etotal - Example Reservoir

0

1000

2000

3000

4000

5000

6000

7000

8000

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Etotal, rb/stb

F,Mrb

OOIP = 5,034.8 Mstb (31.0%)

OGIP = 6,444.5 MMscf (47.3%)

Current Pressure = 0 psiCurrent

Typical Straight Line TechniquesF/E vs E /E


53/91


F/Eo vs. Eg/Eo

F/Eo vs. Eg/Eo - Example Reservoir

0

10000

20000

30000

40000

50000

60000

70000

80000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01

Eg/Eo

F/Eo,Mstb

Measured Data

Common PitfallsUsing Only One Analysis Technique


54/91


Using Only One Analysis Technique

F vs. Etotal - Example Reservoir

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 0.05 0.1 0.15 0.2 0.25 0.3

Etotal, rb/stb

F,Mrb

OOIP = 5,034.8 Mstb (31.0%)

OGIP = 6,444.5 MMscf (47.3%)


OOIP vs. Cum Oil - Example Reservoir

10,000

15,000

20,000

25,000

30,000

35,000

40,000

45,000

50,000

55,000

60,000

0 200 400 600 800 1000 1200 1400 1600 1800


OOIP,Mstb

F/Eo vs. Eg/Eo - Example Reservoir

0

10000

20000

30000

40000

50000

60000

70000

80000

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01

Eg/Eo

F/Eo,Mstb

Measured Data

Common PitfallsIncorrect Reservoir Model


55/91


Incorrect Reservoir Model

OOIP vs. Cum Oil - 9700 ft Sand

0

50,000

100,000

150,000

200,000

250,000

300,000

350,000

400,000

0 10000 20000 30000 40000 50000 60000 70000


OOIP,M

stb

OOIP = 101,567.4 Mstb (58.4%)

OGIP = 238,413.0 MMscf (57.3%)

OOIP vs. Cum O il - 9700 ft Sand

0

50,000

100,000

150,000

200,000

250,000

300,000

350,000

400,000

0 10000 20000 30000 40000 50000 60000 70000

Cum O il Production, Mstb

OOIP,M

stb

OOIP = 250,566.0 Mstb (23.7%)

OGIP = 588,163.3 MMscf (23.2%)

Volumetric Reservoir Aquifer Driven Reservoir

Common PitfallsBest-Fit Lines Used Inappropriately


56/91


Best Fit Lines Used Inappropriately

Pressure History Match from Black Oil Model- 9700 ft Sand

2000.0

2500.0

3000.0

3500.0

4000.0

4500.0

5000.0

0 10000 20000 30000 40000 50000 60000 70000Cumulative Oil Production, Mstb

ReservoirPressure,ps

ia

Input Pressure

Calculated Pressure

Pressures not built up

F vs. Etotal - 9700 ft Sand

0

20000

40000

60000

80000

100000

120000

0 0.2 0.4 0.6 0.8 1 1.2

Etotal, rb/stb

F,

Mrb

OOIP = 99,968.4 Mstb (59.4%)

OGIP = 237,519.4 MMscf (57.5%)


Pressure = 2,215.7 psi

Incorrect

Correct

Improper Selection Of Wells


57/91


500

1000

1500

2000

2500

3000

3500

Measured

Pressure,psia

Well #1Well #2Well #3Well #4

Well #2 not in same reservoir

0Jan-90 Jan-91 Jan-92 Jan-93 Jan-94

Date

Physically Impossible Results


58/91


Cumulative production > original in-place volumes

Negative saturations



59/91


It is planned to initiate a water injection scheme in the reservoirwhose PVT properties are defined. The intention is to

maintain pressure at the level of 2,700 psia (pb = 3,330 psia).If the current producing gas-oil ratio of the field is 3,000scf/STB, what will be the initial water injection rate required

to produce 10,000 STB/d of oil?



60/91


An undersaturated reservoir producing above the bubble point had an initial pressure of 5,000

psia, at which pressure the oil volume factor was 1.510 RB/STB. When the pressure had

dropped to 4,600 psia, owing to the production of 100,000 STB of oil, the oil formation

volume factor was 1.520 RB/STB. The connate water saturation was 25%, water

compressibility 3.2x10-6 psi-1, and, based on an average porosity of 16%, the rock

compressibility was 4.0x10-6 psi-1. The average compressibility of the oil between 5,000

and 4,600 psia relative to the volume at 5,000 psia was 17x10-6 psi-1.

Material Balances - Exercise 3 (continued)


61/91


Geologic evidence and the absence of water production indicated a volumetric reservoir.Assuming this was so, what was the calculated initial oil in place?

It was desired to inventory the initial stock tank barrels in place at a second productioninterval. When the pressure had dropped to 4,200 psia, formation volume factor 1.531RB/STB, 205,000 STB had been produced. If the average oil compressibility was17.65x10-6 psi-1, what was the initial oil in place?

When all cores and logs had been analyzed, the volumetric estimate of the initial oil in place

was 7.5 million STB. If this figure is correct, how much water entered the reservoir whenthe pressure declined to 4,600 psia?



62/91


The cumulative oil production, Np, and cumulative gas-oil ratio,

Rp, as functions of the average reservoir pressure over the

first 10 years of production for a gas cap reservoir follow.

Use the Havlena-Odeh approach to solve for the initial oil

and gas (both free and solution) in place.



63/91


Using the following data, determine the original oil in place by

the Havlena-Odeh method. Assume that there is no water

influx and no initial gas cap. The bubble-point pressure is

1,800 psia.

Gas Reservoir Material Balance


64/91


Straight-line analysis techniques

Development Assumptions

Analysis techniques Common pitfalls

Gas Reservoir MBE


65/91


The change in reservoir pore volume =the change in reservoir gas volume

+ the change in reservoir water volume

Gas Reservoir MBE


66/91


pcS

GBV f

wi

gi

p

=1

Change in Reservoir

Pore Volume

Change in Reservoir

Gas Volume gpgig BGGGBV )( =

pcSS

GBV ww

wi

gi

w

=1

Change in Reservoir Water Volume

wgp VVV +=Gas Reservoir MBE

Gas Reservoir MBE


67/91


+= p

S

ccS

B

GBGG

wi

fww

g

gi

p 11

wgp VVV +=Gas Reservoir MBE

Gas Reservoir MBE


68/91


sc

scg

PT

TzP

B =Since:

+

= pS

ccS

zp

zpGGG

wi

fww

i

p11)/(

)/(

Gas Reservoir MBE


69/91


G

G

z

p

z

pp

S

ccS

z

p p

iiwi

fww

=

+

11

Gas Reservoir MBE


70/91


More common forms of the gas reservoir material

balance equation are:

( ) ip

i

ez

p

G

G

z

ppcz

p

=1

i

p

i

z

p

G

G

z

p

z

p

=

Common Gas Reservoir Models


71/91


Volumetric dry gas reservoir

Volumetric wet gas reservoir

Volumetric highly compressible wet gas reservoir

Water influx gas reservoirs

Volumetric Dry Gas Reservoirs


72/91


Gas

GasGas

Initial Conditions Later

i

p

i z

p

G

G

z

p

z

p

+

=

Volumetric Dry Gas Reservoirs


73/91


Assumptions of volumetric dry gas reservoir model

Hydrocarbon pore volume does not change Only dry gas in reservoir

Only dry gas produced

No water influx

Volumetric Wet Gas Reservoirs


74/91


Gas + Condensate

Gas Gas

T = 2Initial Conditions

ieq

eqp

i zp

GG

zp

zp

+

=

,

Geopressured Wet Gas Reservoirs


75/91


Gas + Condensate

GasGas

T = 2Initial Conditions

( )ieq

eqp

i

ezp

GG

zppc

zp

+

= ,1

Geopressured Wet Gas Reservoirs


76/91


Assumptions of geopressured wet gas reservoir model

Constant formation and water compressibility

Only dry gas in reservoir

Only dry gas and condensate are produced

No water influx - or water influx from a small aquifer

Straight-Line Analysis Techniques


77/91


OGIP vs. cumulative gas production

p/z vs. cumulative gas production

p/z vs. cumulative equivalent gas production

p/z(1-cep) vs. cumulative equivalent gas production

Roach plot

OGIP vs. Cumulative Gas Production


78/91


0

20000

40000

60000

80000

100000

120000

0 5,000 10,000 15,000 20,000 25,000 30,000 35,000 40,000 45,000

Equivalent Gas Production, MMscf

OGIP,MM

scf

Geopressured Wet Gas Reservoir Model

Volumetric Dry Gas Reservoir Model

Volumetric Wet Gas Reservoir Model

p/z vs. Cumulative Gas Production


79/91


0.0

1000.0

2000.0

3000.0

4000.0

5000.0

6000.0

7000.0

0 10000000 20000000 30000000 40000000 50000000 60000000 70000000 80000000 90000000

Cumulative Gas Prod, Mscf

P/Z,p

sia

OGIP = 87,674,457 Mscf

Cumulative Recovery = 4.7% OGIP

Ultimate Recovery = 77.6% OGIP

Current Pressure = 8,305.7psi

p/z vs. Cumulative Equivalent Gas Production


80/91


0.0

1000.0

2000.0

3000.0

4000.0

5000.0

6000.0

7000.0

0 10,000 20,000 30,000 40,000 50,000 60,000 70,000 80,000 90,000Cumulative Equivalent Gas, MMscf

P/Z,psia

OGIPeq = 88,507,934 MscfCumulative Recovery = 6.7% OGIPeq

Ultimate Recovery = 79.3% OGIPeq

Current Pressure = 7,881.1 psi

p/z(1-ceP) vs. Cumulative Equivalent GasProduction


81/91


0.0

1000.0

2000.0

3000.0

4000.0

5000.0

6000.0

7000.0

0 10,000 20,000 30,000 40,000 50,000 60,000 70,000 80,000

Cumulative Equivalent Gas, MMscf

(P/Z)(1-Ce

p),psia

OGIPeq = 76,419,899 Mscf

Cumulative Recovery = 7.7% OGIPeq

Ultimate Recovery = 91.8% OGIPeq

Current Pressure = 8,077.7 psi

Common Pitfalls in Gas Reservoir Material Balance


82/91


Wrong reservoir model

Best-fit lines used inappropriately

Improper selection of wells

Physically impossible results

Aquifer Driven Reservoirs


83/91


Aquifer

Reservoir

Aquifer Driven Reservoir Models


84/91


Small aquifer reservoir model

Limited aquifer reservoir model

Infinite aquifer reservoir model

Small Aquifer Model


85/91


Assumes water flows into reservoir instantaneously

Applies only to very small aquifers

(Vp, aq

< 3Vp,res

)

Small Aquifer Model


86/91


Replace Sw in volumetric model with the following relationship:

aqpresp

aqpwresp

wVV

VSV

S,,

,,

+

+

=

Limited and Infinite Aquifer Models


87/91


Aquifer water can expand faster than it can flow into the reservoir

Solutions to the diffusivity equation provide water influx as afunction of reservoir pressure and time

Properties of the aquifer are seldom known

Provides nonunique estimate of original hydrocarbons in place

Limited and Infinite Aquifer Solutions


88/91


Van Everdingen and Hurst method

Carter and Tracy methodFetkovich method

d

d

d

d

dd

d

t

p

r

p

rr

p

=

+

12

2

Aquifer Geometries


89/91


Linear Aquifer ModelRadial Aquifer Model

ro

Reservoir

Aquifer

re

Aquifer

Reservoir

w

L

Outer Boundary Conditions


90/91


Limited aquifer

No flow (closed aquifer)

Constant pressure (aquifer recharge at outcrop)

Infinite aquifer

References


91/91


1. Craft and Hawkins, Applied Petroleum Reservoir

Engineering, 2nd Ed, Prentice Hall 1991.

2. Ahmed, T., Reservoir Engineering Handbook, 2nd Ed, Gulf

Publishing, 2001.

3. Lee and Wattenbarger, Gas Reservoir Engineering, SPETextbook Series No. 5, 1996.

4. Dake, L.P., Fundamentals of Reservoir Engineering,

Elsevier, 1979.

K (II-2) Material Balance Concepts

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