Advanced Reservoir Engineering
開課班級: 碩博士班 (2007年秋季)
講授教師： 林再興

目的

講授石油及天然氣流體性質，以及生產石
油及天然氣所導致的油層壓力變化原理。

討論壓力測試分析(井壓測試分析) ，以及
生產資料分析(生產遞減曲線分析) 。

求得地層參數/預測未來產生產率/計算地
層的石油或天然的儲量及蘊藏量。

 (A) Dake , L.P., Fundamentals of Reservoir
Engineering, revised edition, Elsevier Scientific
B.V., Amsterdam, the Netherlands, 2001.
 (B) Ahmed, T., and McKinney, P., Advanced
Reservoir Engineering, Gulf Publishing
Company, Houston, Texas, 2004
 (B) Craft, B.C., and Hawkins, M.F. , Revised by Terry, R.E. , Applied Petroleum Reservoir
Engineering, Second edition., Prentice Hall ,
Englewood Cliffs, New Jersey, 1991.

 (C) Lee, J., Well Testing, SPE Textbook series, Society of Petroleum Engineers of
AIME, Dallas, Texas, 2002.

(D) 林國安等人，石油探採 ( 第四冊 – 油氣生產 ,
Chapter 24 ) ，中國石油股份有限公司訓練教材
叢書，中油訓練所，嘉義市， 2004.

(E) Journal papers

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Well testing analysis
Water influx
Unconventional gas reservoir
Performance of oil reservoir
Predicting oil reservoir
Introduction to oil fieldeconomics
5

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Introduction to reservoir engineering
- Gas reservoir
- PVT analysis for oil
- Material balance applied to oil
The flow equations of single-phase and two-phase flow of hydrocarbon in porous media
- Darcy’s law and applications
- The basic differential equation in a porous medium
Solutions to the flow equations of hydrocarbon in porous media
- Steady and semi-steady states
- Unsteady state
Pressure drawdown and buildup analysis for oil and gas wells
Decline curve analysis
Case study

Part 1
Introduction to Reservoir Engineering
 The primary functions of a reservoir engineer:
 the estimation of hydrocarbon in place
 the calculation of a recovery factor , and
 the attachment of a time scale to the recovery
 Note: pressure/flow rate information → parameters/future flow rate/future pressure

Outlines of Reservoir Engineering
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(1) Introduction
Petrophysical properties ( Rock properties)
Fluid properties (gas, water, crude properties)
Calculations of hydrocarbon volumes
Fluid pressure regimes
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(2) Gas reservoirs
Calculating gas in place by the volumetric method
Calculating gas recovery factor
Material balance calculation (Depletion & Water drive)
Hydrocarbon phase behavior (gas condensate phase behavior)
The gas equivalent of produced condensate and water
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(3) PVT analysis for oil
Definition of the basic PVT parameters
Determination of the basic PVT parameters in the lab. And conversion for field operating conditions.

Outlines of Reservoir Engineering – cont.
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(4) Material balance applied to oil reservoirs
General form of the material balance equation for a hydrocarbon reservoir (Undersaturated and Saturated reservoir)
Reservoir drive mechanisms
Solution gas drive
Gas cap drive
Natural water drive
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(5) Darcy’s law and applications

Outlines of Reservoir Engineering – cont.
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(6) The basic differential equation for radial flow in a porous medium
Derivation of the basic radial flow equation
Conditions of solution
Linearization of radial flow equation
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(7) Well inflow equations for stabilized flow conditions
Semi steady state solution
Steady state solution
Generalized form of inflow equation (for semi steady state)

Outlines of Reservoir Engineering – cont.
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(8) The constant terminal rate solution of the radial diffusivity equation and its application to oil well testing
Constant terminal rate solution
General Transient flow
Semi steady state flow
Superposition theorem; general theory of well testing
The Matthews, Brons, Hazebroek pressure buildup theory
Pressure buildup analysis techniques
Multi-rate drawdown testing
The effects of partial well completion
After-flow analysis

Outlines of Reservoir Engineering – cont.
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(9) Gas well testing
- Linearization and solution of the basic differential equation for the radial flow of a real gas
- The Russell, Goodrich, et al. Solution technique
- The Al-Hussainy, Ramey, Crawford solution technique
- Pressure squared and pseudo pressure solution technique
- Non-Darcy flow & determination of the non-darcy coefficient
- The constant terminal rate solution for the flow of a real gas
- General theory of gas well testing
- Multi-rate testing of gas well
- Pressure building testing of gas wells
- Pressure building analysis in solution gas drive reservoirs

Outlines of Reservoir Engineering – cont.
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(10) Natural water influx
- Steady state model
- Unsteady state model
- The van Everdingen and Hurst edge-water drive model
- Bottom – water drive model
- Pseudo steady state model (Fetkovich model)
- Predicting the amount of water influx

Fluid Pressure Regimes
The total pressure at any depth
= weight of the formation rock
+ weight of fluids (oil, gas or water)
[=] 1 psi/ft * depth(ft)

Fluid Pressure Regimes
 Density of sandstone

2 .
7 gm cm 3

2 .
2 lbm
1000 gm
( 0 .
3048

100 cm )
3
( 1 ft ) 3

168 .
202 lbm ft
3

1 slug
32 .
7 lbm

5 .
22 slug ft
3

Pressure gradient for sandstone
 Pressure gradient for sandstone p
  gD p
D
  g

5 .
22

32 .
2

168 .
084 lbf ft
3

168 .
084

1 .
16 ( psi / ft lbf
2  ft )
1 ft
2 ft 144 in
2

1 .
16 in lbf
2  ft

Overburden pressure
 Overburden pressure (OP)
= Fluid pressure (FP) + Grain or matrix pressure (GP)
 OP=FP + GP
 In non-isolated reservoir
PW (wellbore pressure) = FP
 In isolated reservoir
PW (wellbore pressure) = FP + GP’ where GP’<=GP

Normal hydrostatic pressure
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In a perfectly normal case , the water pressure at any depth
Assume :(1) Continuity of water pressure to the surface
(2) Salinity of water does not vary with depth.
P

 dP
( dD
) water

D

14 .
7 [=] psia dP
( dD
) water

0 .
4335
( dP dD
) water

0 .
4335 psi/ft for pure water psi/ft for saline water


Abnormal hydrostatic pressure
( No continuity of water to the surface)
P

 dP
( dD
) water

D

14 .
7

C [=] psia
 Normal hydrostatic pressure c = 0
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Abnormal (hydrostatic) pressure c > 0 → Overpressure (Abnormal high pressure) c < 0 → Underpressure (Abnormal low pressure)

Conditions causing abnormal fluid pressures
 Conditions causing abnormal fluid pressures in enclosed water bearing sands include

Temperature change ΔT = +1℉ → ΔP = +125 psi in a sealed fresh water system
 Geological changes – uplifting; surface erosion
 Osmosis between waters having different salinity, the sealing shale acting as the semi permeable membrane in this ionic exchange; if the water within the seal is more saline than the surrounding water the osmosis will cause the abnormal high pressure and vice versa.

Are the water bearing sands abnormally pressured ?
 If so, what effect does this have on the extent of any hydrocarbon accumulations?

Hydrocarbon pressure regimes
 In hydrocarbon pressure regimes
 dP
( dD
) water

0 .
45 psi/ft

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( dP dD
) oil

0 .
35 dP
( dD
) gas

0 .
08 psi/ft psi/ft

Pressure Kick
5000x0.45+15
5000
5100
5200
5300
5400
5500
D
2265Psi 2369Psi
P
Pg=P
0
=2385Psi
Pg=P w
=2490Psi
GOC
OWC
OIL
GAS
GOC (5200ft)
OWC (5500ft)
Water
5500x0.45+15
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Assumes a normal hydrostatic pressure regime Pω= 0.45 × D + 15
In water zone at 5000 ft Pω(at5000) = 5000 × 0.45 + 15 = 2265 psia at OWC (5500 ft) Pω(at OWC) = 5500 × 0.45 + 15 = 2490 psia

Pressure Kick
5000x0.45+15
5000
5100
5200
5300
5400
5500
D
2265Psi 2369Psi
P
Pg=P
Pg=P w
0
=2385Psi
=2490Psi
GOC
OWC
OIL
GAS
GOC (5200ft)
OWC (5500ft)
Water
5500x0.45+15
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In oil zone Po = 0.35 x D + C at D = 5500 ft , Po = 2490 psi
→ C = 2490 – 0.35 × 5500 = 565 psia
→ Po = 0.35 × D + 565 at GOC (5200 ft) Po (at GOC) = 0.35 × 5200 + 565 = 2385 psia

Pressure Kick
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In gas zone Pg = 0.08 D + 1969 (psia) at 5000 ft Pg = 0.08 × 5000 + 1969 = 2369 psia

Pressure Kick
2265Psia
P
5000
5100
5200
5300
5400
5500 hydrostatic pressure
P
0
=P w
=2490Psia
D
OIL
GAS
GOC
OWC
Water
2450Psia
5000
5100
5200
5300
5400
5500
Gas pressure gradient
P g
=P w =2490Psia
P
D
GAS
Water
GWC
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In gas zone Pg = 0.08 D + C
At D = 5500 ft, Pg = Pω = 2490 psia
2490 = 0.08 × 5500 + C
C = 2050 psia
→ Pg = 0.08 × D + 2050
At D = 5000 ft
Pg = 2450 psia

GWC error from pressure measurement
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Pressure = 2500 psia Pressure = 2450 psia at D = 5000 ft at D = 5000 ft in gas-water reservoir in gas-water reservoir
GWC = ? GWC = ?
Sol. Sol.
Pg = 0.08 D + C Pg = 0.08 D + C
C = 2500 – 0.08 × 5000 C = 2450 – 0.08 × 5000
= 2100 psia = 2050 psia
→ Pg = 0.08 D + 2100 → Pg = 0.08 D + 2050
Water pressure Pω = 0.45 D + 15 Water pressure Pω = 0.45 D + 15
At GWC Pg = Pω At GWC Pg = Pω
0.08 D + 2100 = 0.45 D + 15 0.08 D + 2050 = 0.45 D + 15
D = 5635 ft (GWC) D = 5500 ft (GWC)

Results from Errors in GWC or GOC or OWC
 GWC or GOC or OWC location affecting volume of hydrocarbon OOIP affecting
OOIP or OGIP affecting development plans

Volumetric Gas Reservoir Engineering
 Gas is one of a few substances whose state, as defined by pressure, volume and temperature
(PVT)
 One other such substance is saturated steam.

The equation of state for an ideal gas pV
 nRT      ( 1 .
13 )
(Field units used in the industry) p [=] psia; V[=] ft 3 ; T [=] O R absolute temperature n [=] lbm moles; n=the number of lb moles, one lb mole is the molecular weight of the gas expressed in pounds.
R = the universal gas constant
[=] 10.732 psia∙ ft3 / (lbmmole∙0R)
Eq (1.13) results form the combined efforts of Boyle, Charles,
Avogadro and Gay Lussac.

The equation of state for real gas
 The equation of Van der Waals (for one lb mole of gas
( p
 a
V
2
)( V
 b )

RT     ( 1 .
14 )

 where a and b are dependent on the nature of the gas.
The principal drawback in attempting to use eq. (1.14) to describe the behavior of real gases encountered in reservoirs is that the maximum pressure for which the equation is applicable is still far below the normal range of reservoir pressures

The equation of state for real gas

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 the Beattie-Bridgeman equation the Benedict-Webb-Rubin equation the non-ideal gas law

Non-ideal gas law pV
 nzRT     ( 1 .
15 )
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Where z = z-factor =gas deviation factor
=supercompressibility factor
 z

V a
V i

Actual
Ideal volume volume of of n moles n moles of of gas gas at at
T
T and and P
P z
 f ( P , T , compositio n ) compositio n
  g
 specific gravity ( air

1 )

 There are three ways to determination z-factor :
 (a)Experimental determination
 (b)The z-factor correlation of standing and katz
 (c)Direct calculation of z-factor

(a) Experimental determination
 n mole s of gas
 p=1atm; T=reservoir temperature; => V=V0
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 pV=nzRT z=1 for p=1 atm
=>14.7 V
0
=nRT
 n mole of gas
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 p>1atm; T=reservoir temperature; => V=V pV=nzRT pV=z(14.7 V
0
) z
 pV p sc
V
0
14 .
7 V
0 z sc
T
 pV zT
 z
 p sc
V
0
By varying p and measuring V, the isothermal z(p) function can be readily by obtained.
pV

(b)The z-factor correlation of standing and katz
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Requirement:
Knowledge of gas composition or gas gravity
Naturally occurring hydrocarbons: primarily paraffin series CnH2n+2
Non-hydrocarbon impurities: CO2, N2 and H2
Gas reservoir: lighter members of the paraffin series, C1 and C2 > 90% of the volume.

The Standing-Katz Correlation
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 knowing Gas composition (ni)

Critical pressure (Pci)
Critical temperature (Tci) of each component

( Table (1.1) and P.16 )


Pseudo critical pressure (Ppc)
Pseudo critical temperature (Tpc) for the mixture
P pc
T pc

  i
 i n i
P ci n i
T ci

Pseudo reduced pressure (Ppr)
Pseudo reduced temperature (Tpr)
P
 pr
T pr

T
T pc
P
P pc
 const .( Isothermal )


Fig.1.6; p.17
 z-factor

(b’)The z-factor correlation of standing and katz
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For the gas composition is not available and the gas gravity
(air=1) is available.
The gas gravity (air=1)
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 fig.1.7 , p18
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Pseudo critical pressure (Ppc)
Pseudo critical temperature (Tpc)

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
(b’)The z-factor correlation of standing and katz
Pseudo reduced pressure (Ppr)
P
 pr
P
P pc
Pseudo reduced temperature (Tpr)
T pr

T
T pc
 const .( Isothermal )


Fig1.6 p.17
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 z-factor
The above procedure is valided only if impunity (CO2,N2 and
H2S) is less then 5% volume.

(c) Direct calculation of z-factor
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The Hall-Yarborough equations, developed using the Starling-Carnahan equation of state, are z

0 .
06125 P pr te

1 .
2 ( 1
 t )
2
    ( 1 .
20 ) y where Ppr= the pseudo reduced pressure t=1/Tpr Tpr=the pseudo reduced temperature y=the “reduced” density which can be obtained as the solution of the equation as followed:

0 .
06125 P pr te

1 .
2 ( 1
 t )
2

( 90 .
7 t

242 .
2 t
2
 y
 y
2 
( 1
 y
3 y )
3
 y
4

( 14 .
76 t

9 .
76 t
2 
4 .
58 t
3
) y
2

42 .
4 t
3
) y
( 2 .
18

2 .
82 t ) 
0     ( 1 .
21 )
This non-linear equation can be conveniently solved for y using the simple
Newton-Raphson iterative technique.

(c) Direct calculation of z-factor
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The steps involved in applying thus are:
 make an initial estimate of y k , where k is an iteration counter (which in this case is unity, e.q. y 1 =0.001
 substitute this value in Eq. (1.21);unless the correct value of y has been initially selected, Eq. (1.21) will have some small, non-zero value F k .
(3) using the first order Taylor series expansion, a better estimate of y can be determined as


 where dF k dy
F k y k

1  y k  dF k

1

4 y

4 y 2
( 1

 y ) 4
4 y 3 dy

    ( 1 .
22 ) y 4

( 29 .
52 t

19 .
52 t 2 
9 .
16 t 3 ) y

( 2 .
18

2 .
82 t )( 90 .
7 t

242 .
2 t
2 
42 .
4 t
3
) y
( 1 .
18

2 .
82 t )     ( 1 .
23 )
(4) iterate, using eq. (1.21) and eq. (1.22), until satisfactory convergence is obtained(5) substitution of the correct value of y in eq.(1.20)will give the z-factor.
(5) substitution of the correct value of y in eq.(1.20)will give the z-factor.


Application of the real gas equation of state
Equation of state of a real gas pV
 nzRT     ( 1 .
15 )
 This is a PVT relationship to relate surface to reservoir volumes of hydrocarbon.
(1) the gas expansion factor E,
E

V sc
V
 volume volume of of n n moles moles of of gas gas at at s tan dard reservoir conditions conditions
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Real gas equation for n moles of gas at standard conditions p sc
V sc
 nz sc
RT sc

V sc
 nz sc p
RT sc sc
Real gas equation for n moles of gas at reservoir conditions pV
 nzRT 
V
 nzRT p nz sc
RT sc 
>
E

V sc
V
 nzRT p p sc
 nz sc
RT sc nzRTp sc p

T sc p zTp sc

519 .
6 zT
 

14 .
7
( note : z sc
 >
E

35 .
35 p zT
[

] surface volume/reservoir volume
[=] SCF/ft3 or STB/bbl

1 )

Example
 Reservoir condition:
P=2000psia; T=1800F=(180+459.6)=639.60R; z=0.865

>
E

35 .
35
2000
0 .
865

639 .
6

127 .
8 surface volume/reservoir or SCF/ft3 or STB/bbl

OGIP

V

( 1

S wi
) E i


(2) Real gas density m
 
V   m
 nM
V V where n=moles; M=molecular weight)
  nM nzRT p

MP zRT
 gas

M gas
P z gas
RT
 at any p and T



For gas
For air
 gas

M gas
P z gas
RT
 air

M air p z air
RT
 gas
 air
  g

M gas p z gas
RT
M gas p z air
RT

M gas
Z gas
M air
Z air
 g

(
M z
) gas
(
M
Z
) air

(2) Real gas density
 g

(
M
(
M z
) gas
Z
) air


At standard conditions z air
 gas   
M gas 
M gas
 g
M 28 .
97
    ( 1 .
28 ) air air
= z gas
= 1 in general
 g

0 .
6 ~ 0 .
8

 M gas
  g

28 .
97
(a) If is known, then or ,
 gas
  g
  air
(b) If the gas composition is known, then
 g

M gas
28 .
97
 gas
  g
  air
M gas
  i n i
M i where
(
 air
) sc

0 .
0763 lbm ft
3

(3)Isothermal compressibility of a real gas pV
 nzRT V
 nzRT p

V
 p
 nRTz [
 p

2
]
 nRTp

1
 z
 p
 nRTzp

1
( note : z
 f ( p ))

V
 p
  nzRT p
2
 nRT p
 z
 p

V
 p
  nzRT p
(
1 p

1 z
 z
 p
)
 
V (
1 p

1 z
 z
 p
)
C g
 
1
V

V
 p
 
1
V
[

V (
1 p

1 z
 z
 p
)]
C g

1 p

1 z
 z
 p
C g

1 p since
1 p

1 z
 z
 p p.24, fig.1.9

- Problem
 Exercise1.1 Gas pressure gradient in the reservoir
 (1) Calculate the density of the gas, at standard conditions, whose composition is listed in the table 1-1.
 (2) what is the gas pressure gradient in the reservoir at 2000psia and
1800F(z=0.865)




Exercise 1.1 -- solution -1
(1) Molecular weight of the gas
M gas since
  gas
  i n i
M i

19 .
91
 g


 gas
 air
  gas
  g
  air
0 .
687

0 .
0763 ( lbm
 g

M gas
28 .
97

19 .
91
28 .
97

0 .
687 ft
3
)

0 .
0524 ( lbm ft
3
) or from pVM

  m
V
 pV
 nMzRT nzRT
 mzRT pM zRT
At standard condition
 gas

P sc z sc
RT sc

14 .
7

19 .
91
1

10 .
73

519 .
6

0 .
0524 ( lbm ft
3
)


Exercise 1.1 -- solution -2
 (2) gas in the reservoir conditions pV
 nzRT pVM
 nMzRT
 mzRT
  m
V
 pM zRT

2000

19 .
91
0 .
865

10 .
73

( 459 .
6

180 )

6 .
707 ( lbm ft
3
)

 p
 dp dD
 gD
Exercise 1.1 -- solution -3 dp
  gdD

 g

( 6 .
707 lbm ft
3
1 slug
32 .
2 lbm
) 32 .
2 ft s 2

6 .
707 slug ft
3 ft s
2

6 .
707 lb f ft
3

6 .
707 lbf ft
2
1 1 ft
2 ft 144 in
2

0 .
0465 lb f in
2
1 ft

0 .
0465 psi ft

 Material balance

Production = OGIP (GIIP) - Unproduced gas
 (SC) (SC) (SC)
 Case 1 ： no water influx (volumetric
 depletion reservoirs)
 Case 2 ： water influx (water drive reservoirs)

Volumetric depletion reservoirs -- 1

No water influx into the reservoir from the adjoining aquifer.


Gas initially in place (GIIP) or Initial gas in place （ IGIP ）
＝ G ＝ Original gas in place （ OGIP ）
 [=] Standard Condition Volume




G

V

( 1
 s wc
) E i
[

] SCF where E i

35 .
37 p i z i
T i
Material Balance （ at standard conditions ）
[

]
Production ＝ GIIP － Unproduced gas
（ SC ）
G p
（ SC ） （ SC ）

G



G
E i



E     ( 1 .
33 )
SCF / ft
3
 Where G/E i
HCPV
= GIIP in reservoir volume or reservoir volume filled with gas ＝

Volumetric depletion reservoirs -- 2

G p
G

1

E
E i
    ( 1 .
34 ) sin ce E

35 .
37 p SCF


G p
G p z p
35 .
37

1

 zT p i z i p i 35 .
37


1
 z i
T i
G
G p
 where
G p
G
 the fractional zT p
1
 z p i note : T z i


    ( 1 .
35 ) gas re cov ery at any ft
3

T i stage
 const .
during

Gas re cov ery factor depletion

 p
 z p i z i


 p i z i

1
G


G p

In Eq.
（ 1.33
）
HCPV

G
E i
 const .
?
 HCPV
≠ const.
because:
 1. the connate water in reservoir will expand
 2. the grain pressure increases as gas
 (or fluid) pressure declines

OP
 d ( FP )
FP


GP    
 d ( GP )
( 1 .
3 ) p .
3 ~ p .
4
 d ( HCPV )

 d ( G /
 dV w
E i

) dV f
    ( 1 .
36 )
 where
V w
V f
 initial
 initial negative
( connate water volume sign pore
"

" volume
 exp ansion
) of to a water reduction leads in HCPV

c f
 
1
V f


V f
 


 c f
 
1
V f
(

 p c f dV f

1
V f
 c f

V f
 p

V f
V
 f
) dp
GP
GP
V f
GP pore vol.
GP

V w c w

 dV

1
V w
 w d

V
 c w w
 

V w
 
1
V w
 dp dV w dp
FP
FP
V f
FP
FP=gas pressure
FP
FP
FP
V w
FP
FP=gas pressure
FP

d


G
E i


 d

HCPV

 c w
V w dp
 c f
V f dp
Since
V f

PV


HCPV
1

S wc


E i

1
G

S wc

V w

PV

S wc


HCPV
1

S wc
 S wc

G
E i
1
S

 wc
S wc

 d


G
E i


 c w
G
E i
1
S

 wc
S wc
 dp
 c f
E i
1
G


S wc
 dp



G
E i



G
E i



G
E i

 t

 t

 initial









G
E i
G
E i
E i




G

 t initial initial



 1


G
E i


G

E
 c i w

 initial

 c w

1
S
 wc
S wc

 c f

1

1
S wc


 initial
 c w
S

1 wc


S c wc f

S
1
 c
 wc

S wc f


 p



 p


 p

G p

G p

G p
G
For

G

G
E i
E     ( 1 .
33 )

G

G
E i


 1

 c w
S
1 wc

 c

S
 wc f

 p



E

1



 1
 c w
S wc
 c f
1

S wc c w

3

10

6 psi

1
;



E
E i c f

10

10

6 psi

1

1
 c w
S wc
 c f
1

S wc

1

0 .
013

0 .
987

G p
G

1

0 .
987
E
E i
 computing
1 .
3 % difference with and S wc

0 .
2
G p
G

1

E
E i

p/z plot
 From Eq. (1.35) such as

In p
 z p i z i

 1

G p
G

     ( 1 .
35 )
 p
 z p i z i
 p i z i
G
G p p/z
Abandon pressure p ab
0 p z v .
s Gp plot p/z
Gp G
Y=a+mx y x
 p z

G p m
  p i z i
G a
 p i z i
0 G p
/G = RF 1.0
A straight line in p/z v.s Gp plot means that the reservoir is a depletion type

Water drive reservoirs











If the reduction in reservoir pressure leads to an expansion of adjacent aquifer water, and consequent influx into the reservoir, the material balance equation must then be modified as:
Production = GIIP － Unproduced gas
（ SC ） （ SC ） （ SC ）
Gp ＝ G － （ HCPV-We ） E
Or
Gp ＝ G － （ G/Ei － We ） E where We= the cumulative amount of water influx resulting from the pressure drop.
Assumptions:
No difference between surface and reservoir volumes of water influx
Neglect the effects of connate water expansion and pore volume reduction.
No water production

Water drive reservoirs
With water production
G p

G






G
E i


W e

W p

B w
 




E
 p
 z p i z i

 1

G p
G


    ( 1 .
41 )
1

W e
E i
G
 where We*Ei / G represents the fraction of the initial hydrocarbon pore volume flooded by water and is, therefore, always less then unity.

Water drive reservoirs p
 z p z i


1 i


1

W e
G
G p
G
E i




    ( 1 .
41 )
 since


1
W e
E i
G



1
 p
 z p i z i

 1

G p
G

 in water flux reservoirs
Comparing p
 z p i z i

 1

G p
G

 in depletion type reservoir

Water drive reservoirs p
 z p z i


1 i


1

W e
G
G p
G
E i




    ( 1 .
41 )



In eq.(1.41) the following two parameters to be determined
G; We
History matching or “aquifer fitting” to find We
Aquifer modelfor an aquifer whose dimensions are of the same order of magnitude as the reservoir itself.
W e
 c W
 p
 Where W=the total volume of water and depends primary on the geometry of the aquifer.
ΔP=the pressure drop at the original reservoir –aquifer boundary

Water drive reservoirs


The material balance in such a case would be as shown by plot A in fig1.11, which is not significantly different from the depletion line
For case B & C in fig 1.11
（ p.30
） =>Chapter 9



This method is to estimate GIIP in a water drive reservoir
From Eq. (1.40) such as
G p

G



G
E i

W e

 E     ( 1 .
40 )

G p

G p

G

GE
E i

G

 1


W e
E
E
E i



W e
E

G

 1

E
E i



G p

W e
E

G

G p

 1

E
E i




 1
W e

E
E
E i

 or
G p

 1

E
E i



G

W e
E

 1

E
E i

 or G a

G


 1
W e

E
E
E i


G p



1

E
E i



( or G a
) is plot as function of
W e
E



1

E
E i




G p



1

E
E i



( or G a
) is plot as function of
W e
E



1

E
E i






 The result should be a straight line, provided the correct aquifer model has been selected.
The ultimate gas recovery depends both on
(1) the nature of the aquifer ,and
(2) the abandonment pressure.





The principal parameters in gas reservoir engineering:
(1) the GIIP
(2) the aquifer model
(3) abandonment pressure
(4) the number of producing wells and their mechanical define

Hydrocarbon phase behavior

Hydrocarbon phase behavior

C--------＞D--------------＞E
Hydrocarbon phase behavior
Residual saturation (flow ceases)

Liquid H.C deposited in the reservoir

Retrograde liquid Condensate
E--------------＞F
Re-vaporization of the liquid condensate ?
NO!
Because H.C remaining in the reservoir increase

Composition of gas reservoir changed

Phase envelope shift SE direction
Thus, inhibiting re-vaporization.
Condensate reservoir, producing Wet gas
(at scf) pt. c,
Dry gas injection displace the wet gas until dry gas break through occurs in the producing wells
Keep p above dew pt.
Δp small

Equivalent gas volume


The material balance equation of eq(1.35) such as p z
 p z i i

 1

G
G p


Assume that a volume of gas in the reservoir was produced as gas at the surface.
 If, due to surface separation, small amounts of liquid hydrocarbon are produced, the cumulative liquid volume must be converted into an equivalent gas volume and added to the cumulative gas production to give the correct value of Gp for use in the material balance equation.

Equivalent gas volume
 If n lb m
–mole of liquid have been produced, of molecular weight M, then the total mass of liquid is nM
  o
 w

 liquid volume


 where γ
0
ρ w
= oil gravity (water =1)
＝ density of water （ =62.43 lb m
/ft 3 ） n

 o
 w

V o
M


0
62 .
4
M

 lbm ft
3
 lbm / lbm



V
0
 mole


62 .
4

0
V
0
M n

62 .
4

0
V
0
 bbls

5 .
61458
M 1 bbl

V sc n


V sc

350 .
5 nRT p sc


0
N p
M
350 .
5

0
N
M where p
RT sc p sc

1 .
33

10 5

0
N
M p ft 3
N p
[

] bbls

350 .
5

0
N p

10 .
73

520
M

14 .
7
N p
  bbls
Equivalent gas volume

Condensate Reservoir
 The dry gas material balance equations can also be applied to gas condensate reservoir, if the single phase z-factor is replaced by the ,so-called ,two phase z-factor. This must be experimentally determined in the laboratory by performing a constant volume depletion experiment.



Volume of gas ＝ G scf , as charge to a PVT cell
P ＝ P i
＝ initial pressure （ above dew point ）
T ＝ T r
＝ reservoir temperature

Condensate Reservoir

 p decrease
 by withdraw gas from the cell, and measure gas Gp’
Until the pressure has dropped to the dew point
Z
2
 phase
 p i z i p

 1

G p
G
'


    ( 1 .
46 ) p z
 p i z i

 1

G p
G
'

        ( 1 .
35 )
 z
 p i z i p

 1

G p
G
'


The latter experiment, for determining the single phase z-factor, implicitly assumes that a volume of reservoir fluids, below dew point pressure, is produced in its entirety to the surface.

Condensate Reservoir
 In the constant volume depletion experiment, however, allowance is made for the fact that some of the fluid remains behind in the reservoir as liquid condensate, this volume being also recorded as a function of pressure during the experiment. As a result, if a gas condensate sample is analyzed using both experimental techniques, the two phase z-factor determined during the constant volume depletion will be lower than the single phase zfactor.
 This is because the retrograde liquid condensate is not included in the cumulative gas production Gp’ in equation （ 1.46
） , which is therefore lower than it would be assuming that all fluids are produced to the surface, as in the single phase experiment.

油層工程






蘊藏量評估




體積法
物質平衡法
衰減曲線
油層模擬
壓力分析(隨深度變化，或壓力梯度),
例如, 求氣水界面。
物質平衡法
井壓測試分析(暫態)
（求 k 、 s 、re、 xf 、氣水界面、地層異質性）
Pressure buildup
Pressure drawdown
水驅計算(water drive)
78