

Chapter 2
Volumetric gas Reservoir Engineering
1
Gas PVT

Gas is one of a few substances whose state, as
defined by pressure, volume and temperature
(PVT)

One other such substance is saturated steam.
2
The equation of state for gas

Classical ideal gas law








Classical non-ideal gas law
Cubic Equations of State
van der Waals equation of state
Redlich-Kwong equation of state
Soave modification of Redlich-Kwong
Peng-Robinson equation of state
Elliott, Suresh, Donohue equation of state










Non-cubic Equations of State
Dieterici equation of state
Virial Equations of State
Virial Equations of State
The BWRS equation of state
Other Equations of State of Interest
Stiffened equation of state
Ultrarelativistic equation of state
Ideal Bose equation of state
3
The equation of state for an ideal gas
pV  nRT  (1.13)
(Field units used in the industry)
p [=] psia; V[=] ft3; T [=] OR absolute temperature
n [=] lbm moles; n=the number of lbm moles, one lbm mole is
the molecular weight of the gas expressed in pounds.
R = the universal gas constant
[=] 10.732 psia∙ ft3 / (lbm mole∙0R)
Eq (1.13) results from the combined efforts of Boyle, Charles,
Avogadro and Gay Lussac.
4
Note: In this text
1 Darcy = 1.0133×10-8 cm2
or 1 Darcy
10
-8
cm2
or 1 Darcy

10-12 m2 = 1 m 2
1 Darcy
1
m 2
In other book
1 Darcy = ０.986927×10-8 cm2
1 Darcy = ０.986927×10-12 m2
Non-ideal gas law
pV  nzRT  (1.15)

Where z = z-factor =gas deviation factor
=supercompressibility factor
= compressibility factor


Va Actual

z 
Vi
Ideal
volume
volume
of n moles
of n moles
of
of
gas
gas
at
at
T and
T and
P
P
z  f ( P, T , composition)
composition   g  specific gravity(air  1)
7
Determination of z-factor

There are three ways to determine z-factor :

(a)Experimental determination

(b)The z-factor correlation of standing and
katz

(c)Direct calculation of z-factor
8
(a) Experimental determination

n moles of gas

p=1atm; T=reservoir temperature;

pV=nzRT
z=1 for p=1 atm
=>14.7 V0=nRT


=> V=V0


n moles of gas

p>1atm; T=reservoir temperature;
pV=nzRT
pV=z(14.7 V0)
pV




=> V=V
z
p scV0 pV
pV

z
14.7V0
z scT
zT
p scV0
By varying p and measuring V, the isothermal z(p) function can be
readily by obtained.
9
(b)The z-factor correlation of standing and katz






Requirement:
Knowledge of gas composition or gas gravity
Naturally occurring hydrocarbons: primarily
paraffin series CnH2n+2
Non-hydrocarbon impurities: CO2, N2 and H2S
Gas reservoir: lighter members of the paraffin series, C1
and C2 > 90% of the volume.
10
The Standing-Katz Correlation







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
Ppc   ni Pci
i
T pc   ni Tci
i



 Pseudo reduced pressure (Ppr)
Pseudo reduced temperature (Tpr)
T pr 


P
Ppr 
Ppc
 Fig.1.6; p.17
T
 const.(Isothermal )
T pc
 z-factor
11
12
(b’)The z-factor correlation of standing and katz






For the gas composition is not available and the gas gravity
(air=1) is available.
The gas gravity (air=1)
( g )
 fig.1.7 , p18
Pseudo critical pressure (Ppc)
Pseudo critical temperature (Tpc)
13
(b’)The z-factor correlation of standing and katz


 Pseudo reduced pressure (Ppr)
Pseudo reduced temperature (Tpr)

 Fig1.6 p.17



P
Ppr 
Ppc
T pr 
T
 const.(Isothermal )
T pc
z-factor
The above procedure is valided only if impunity (CO2,N2 and
H2S) is less then 5% volume.
14
(c) Direct calculation of z-factor

The Hall-Yarborough equations, developed using the Starling-Carnahan
equation of state, are
z



0.06125Ppr te
1.2 (1t ) 2
y
(1.20)
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:
 0.06125Ppr te
1.2 (1t )2
y  y2  y3  y4

 (14.76t  9.76t 2  4.58t 3 ) y 2
3
(1  y )
 (90.7t  242.2t 2  42.4t 3 ) y ( 2.182.82t )  0(1.21)
This non-linear equation can be conveniently solved for y using the simple
Newton-Raphson iterative technique.
15
(c) Direct calculation of z-factor


The steps involved in applying thus are:
 making an initial estimate of yk, where k is an iteration counter (which in
this case is unity, e.q. y1=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 Fk.
(3) using the first order Taylor series expansion, a better
estimate of y can be determined as
y

where
k 1
Fk
y 
 (1.22)
k
dF
dy
k
dF k 1  4 y  4 y 2  4 y 3  y 4

 (29.52t  19.52t 2  9.16t 3 ) y
4
dy
(1  y)


 (2.18  2.82t )(90.7t  242.2t 2  42.4t 3 ) y (1.18 2.82t ) (1.23)
(4) iterating, 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) substituting the correct value of y in eq.(1.20)will give the z-factor.
16
The equation of state for real gas

The equation of Van der Waals (for one lb mole of gas
a
( p  2 )(V  b)  RT  (1.14)
V


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
17
Peng-Robinson equation of state
where, ω is the acentric factor of the species
R is the universal gas constant.
18
19
Peng-Robinson equation of state
The Peng-Robinson equation was developed in 1976 in order to satisfy
the following goals:[3]
1. The parameters should be expressible in terms of the critical
properties and the acentric factor.
2. The model should provide reasonable accuracy near the critical point,
particularly for calculations of the Compressibility factor and liquid
density.
3. The mixing rules should not employ more than a single binary
interaction parameter, which should be independent of temperature
pressure and composition.
4. The equation should be applicable to all calculations of all fluid
properties in natural gas processes.
For the most part the Peng-Robinson equation exhibits performance
similar to the Soave equation, although it is generally superior in
predicting the liquid densities of many materials, especially nonpolar
ones. The departure functions of the Peng-Robinson equation are
given on a separate article.
20
Application of the real gas equation of state
pV  nzRT  (1.15)
Equation of state of a real gas

This is a PVT relationship to relate surface to reservoir volumes of
hydrocarbon.
(1) the gas expansion factor E,

E







Vsc volume of n moles of gas at s tan dard conditions

V
volume of n moles of gas at reservoir conditions
Real gas equation for n moles of gas at standard conditions
nz sc RTsc

V

p scVsc  nz sc RTsc
sc
p sc
Real gas equation for n moles of gas at reservoir conditions
nzRT

V

pV  nzRT
p
>
>
E
Vsc

V
nz sc RT sc
nzRT
E  35.35
p sc
p

nz sc RT sc p Tsc p 519.6  


nzRTp sc
zTp sc
zT  14.7
(note : z sc  1)
p
[] surface volume/reservoir volume
zT
[=] SCF/ft3 or STB/bbl
21
Example

Reservoir condition:
P=2000psia; T=1800F=(180+459.6)=639.60R; z=0.865
>
2000
E  35.35
0.865  639.6
 127.8
surface volume/reservoir
or SCF/ft3 or STB/bbl

OGIP  V (1  S wi ) Ei
22
23
24
(2) Real gas density
m  V

m nM

V
V
where n=moles; M=molecular weight)
nM

nzRT
p
MP

zRT

at any p and T

For gas


For air

 gas 
 air
M gas p
 gas 
M gas P
z gas RT
M gas P
z gas RT
M air p

z air RT
M gas
 gas
z gas RT
Z gas
g 

M gas p
M air
 air
Z air
z air RT
( M ) gas
Z
g 
( M ) air
Z
25
(2) Real gas density
g


(M
(M
Z
) gas
Z
) air
At standard conditions zair = zgas = 1
 gas
M gas M gas
g 

 (1.28)
 air
M air 28.97


in general
 g  0 .6 ~ 0 .8
(a) If  is
g known, then
or ,
(b) If the gas composition is known, then
M gas
g 
 gas   g   air
28.97
where
 gas   g   air
M gas   g  28.97
(  air ) sc  0.0763 lbm
M gas   ni M i
i
ft 3
26
(3)Isothermal compressibility of a real gas
pV  nzRT
V
nzRT
 nRTzp 1
p
V
z
 nRTz [ p 2 ]  nRTp 1
p
p
(note : z  f ( p))
V
nzRT nRT z
 2 
p
p p
p
V
nzRT 1 1 z
1 1 z

( 
)  V ( 
)
p
p
p z p
p z p
Cg  
1 V
1
1 1 z
  [V ( 
)]
V p
V
p z p
Cg 
1 1 z

p z p
Cg 
1
p
since
1
1 z

p
z p
p.24, fig.1.9
27
28
Exercise 1.1 - 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)
29
30
31
Exercise 1.1 -- solution -1

(1) Molecular weight of the gas
M gas   ni M i  19.15
g 
i
since
g 
M gas
28.97

19.15
 0.661
28.97
 gas
  gas   g   air
 air
  gas  0.661 0.0763(lbm ft 3 )  0.0504(lbm ft 3 )

or from
pV  nzRT
pVM  nMzRT
 mzRT
m pM
 
V zRT

At standard condition
 gas 
Psc M
14.7 19.15

 0.0505(lbm ft 3 )
z sc RTsc 110.73  519.6
32
Exercise 1.1 -- solution -2

(2) gas in the reservoir conditions
pV  nzRT
pVM  nMzRT
 mzRT
m pM
2000 19.15
 

 6.451(lbm ft 3 )
V zRT 0.865 10.73  (459.6  180)

33
Exercise 1.1 -- solution -3
p  gD
dp
dD

dp  gdD
g
 6.451
 (6.451
lbm 1slug
ft
)
32
.
2
s2
ft 3 32.2lbm
slug ft
ft 3 s 2
 6.451
lb f
ft 3
lbf 1 1 ft 2
 6.451 2
ft ft 144in 2

lb f 1
 0.0448 2
in ft
 0.0448 psi
ft
34
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)
35
Fluid Pressure Regimes

Density of sandstone
gm 2.2lbm (0.3048 100cm)3
 2.7 3 
cm 1000 gm
(1 ft)3
 168.202
lbm 1slug

3
ft
32.7lbm
slug
 5.22 3
ft
36
Pressure gradient for sandstone

Pressure gradient for sandstone
p  gD
p
 g
D
 5.22  32.2  168.084
lbf
ft 3
lbf
1 ft 2
lbf
 168.084 2
 1.16 2
( psi / ft )
2
ft  ft 144in
in  ft
37
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
38
Normal hydrostatic pressure



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
) water  D  14.7
dD
dP
) water  0.4335
dD
dP
( ) water  0.4335
dD
[=] psia
psi/ft for pure water
psi/ft for saline water
39
Abnormal hydrostatic pressure
( No continuity of water to the surface)


dP
P  (
) water  D  14.7  C
dD
[=] psia
Normal hydrostatic pressure
c=0


Abnormal (hydrostatic) pressure
c > 0 → Overpressure (Abnormal high pressure)
c < 0 → Underpressure (Abnormal low pressure)
40
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.
41
Are the water bearing sands abnormally pressured ?

If so, what effect does this have on the extent of any
hydrocarbon accumulations?
42
Hydrocarbon pressure regimes

In hydrocarbon pressure regimes

dP
( ) water  0.45
dD
psi/ft

dP
( ) oil  0.35
dD
psi/ft

dP
( ) gas  0.08
dD
psi/ft
43
Pressure Kick – Oil and Water
5000
oil
OWC
water
D=5500ft
5200
5500
5600
Pw=2265
Pw=2355
Po=2315
P(psia)
Po=2385
Pw=Po=2490
Pw=2535
Depth(ft)
Pw  0 . 45 * D  15 [  ] psia in water zone
Pw ( at D  5600 ft )  0 . 45 * 5600  15  2535 psia
Pw ( at D  5500 ft or at OWC )  0 . 45 * 5500  15  2490 psia
Po ( at D  5500 ft or at OWC )  2490
 0 . 35 * D  C o
or C o  2490 - 0.35 * 5500  565
 P o  0.35 * D  565 in oil zone
Po ( at D  5200 ft )  0 . 35 * 5200  5 65  2385 psia
Pw ( at D  5200 ft )  0 . 45 * 5200  15  2355 psia
Po ( at D  5000 ft )  0 . 35 * 5000  565  2315 psia
Pw ( at D  5000 ft )  0 . 45 * 5000  15  2265 psia
44
pressure kick-gas and water
Gas
D=5500ft
GWC
5200
Pw=2355
P(psia)
Pg=2450
Pg=2466
5500
5600
water
Pw  0.45 * D  15
5000
Pw=2265
Pw=Pg=2490
Pw=2535
Depth(ft)
in
water
zone
Pw ( at
D  5600 ft )  2535 psia
Pw ( at
D  5500 ft
GWC )  2490 psia
Pg ( at
D  5500 ft
GWC )  2490
 0.08 * D  C g
or
C g  2490  0.08 * 5500  2050
 Pg  0.08 * D  2050
in
gas
Pg ( at
D  5000 ft )  0.08 * 5200  2050  2466 psi
Pw ( at
D  5200 ft )  2355 psia
Pg ( at
D  5000 ft )  0.08 * 5000  2050  2450 psia
Pw ( at
D  5000 ft )  2265 psia
zone
45
pressure kick-gas, oil and water
5000
Gas
oil
D=5500ft
Pw=2355
Pw=2400
Pw=2445
5200
5300
5400
5500
5600
GOC
D=5300ft
OWC
water
Pg=2396
Pw=2265
P(psia)
Pg=2412
Po =Pg=2420
Po=2455
Pw= Po=2490
Pw=2535
Depth(ft)
p w  0.45 * D  15
in
water
zone
p w ( at
D  5600 ft )  2535 psi
p w ( at
D  5500 ft
OWC )  2490 psia
po ( at
D  5500 ft
OWC )  2490 psia
 0.35 * D  Co
or
 po  0.35 * D  565
in
Co  565
oil
zone
po ( at
D  5400 ft )  0.35 * 5400  565  2455 psia
p w ( at
D  5400 ft )  0.45 * 5400  15  2445 psia
po ( at
D  5300 ft
GOC )  0.35 * 5300  565  2420 psia
p w ( at
D  5300 ft
GOC )  0.45 * 5300  15  2400 psia
po ( at
D  5300 ft
GOC )  p g ( at
D  5300 ft
GOC )  2420 psia
 0.08 * D  C g
or
C g  1996
 p g  0.08 * D  1996
p g ( at
D  5200 ft )  0.08 * 5200  1996  2412 psia
p w ( at
D  5200 ft )  2355 psia
p g ( at
D  5000 ft )  0.08 * 5000  1996  2396 psia
p w ( at
D  5000 ft )  2265 psia
46
47
Pressure Kick
5000x0.45+15
2265Psi
2369Psi
P
5000
5100
5200
5300
5400
5500
GAS
Pg=P0 =2385Psi
GOC (5200ft)
GOC
OIL
OWC
Pg=Pw =2490Psi
OWC (5500ft)
Water
D
5500x0.45+15




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
48
Pressure Kick
5000x0.45+15
2265Psi
2369Psi
P
5000
5100
5200
5300
5400
5500
GAS
Pg=P0 =2385Psi
GOC (5200ft)
GOC
OIL
OWC (5500ft)
OWC
Pg=Pw =2490Psi
Water
D
5500x0.45+15





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
49
Pressure Kick


In gas zone Pg = 0.08 D + 1969 (psia)
at 5000 ft Pg = 0.08 × 5000 + 1969 = 2369 psia
50
Pressure Kick
2450Psia
2265Psia
P
P
5000
5100
5200
5300
5400
5500
GAS
hydrostatic
pressure
GOC
OIL
OWC
P0=Pw =2490Psia






Gas pressure
gradient
GAS
GWC
Pg=Pw=2490Psia
Water
D

5000
5100
5200
5300
5400
5500
Water
D
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
51
GWC error from pressure measurement

Pressure = 2500 psia
at D = 5000 ft
in gas-water reservoir
GWC = ?
Sol.
Pg = 0.08 D + C
C = 2500 – 0.08 × 5000
= 2100 psia
→ Pg = 0.08 D + 2100

Water pressure Pω = 0.45 D + 15
Water pressure Pω = 0.45 D + 15
At GWC Pg = Pω
0.08 D + 2100 = 0.45 D + 15
D = 5635 ft (GWC)
At GWC Pg = Pω
0.08 D + 2050 = 0.45 D + 15
D = 5500 ft (GWC)











Pressure = 2450 psia
at D = 5000 ft
in gas-water reservoir
GWC = ?
Sol.
Pg = 0.08 D + C
C = 2450 – 0.08 × 5000
= 2050 psia
→ Pg = 0.08 D + 2050
52
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
53
Gas Material Balance: Recovery Factor

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)
54
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 ) Ei
[] SCF
pi
where Ei  35.37
z i Ti



Material Balance （at standard conditions）
Production ＝ GIIP － Unproduced gas
（SC）
（SC） （SC）
G
G p  G  
 Ei

[] SCF / ft 3

 

E (1.33)
Where G/Ei = GIIP in reservoir volume or reservoir volume filled with gas ＝
HCPV
55
Volumetric depletion reservoirs -- 2

Gp
G
 1
E
 (1.34)
Ei
E  35.37
sin ce
p
zT
p
Gp
zT  1 

 1
p
G
35.37 i
z i Ti
35.37
Gp
pi 
p
1 
 
z
zi 
G
where
Gp
G
p
z
pi
zi

SCF
ft 3
note :T  Ti  const .

(1.35)

 the fractional gas re cov ery at any stage during depletion
 Gas re cov ery factor

pi  pi 1 
p





Gp


z
zi
 zi G 
56
In Eq.（1.33）




HCPV 
G
 const .
Ei
?
HCPV≠const.
because:
1. the connate water in reservoir will expand
2. the grain pressure increases as gas
(or fluid) pressure declines
57
OP  FP  GP  (1.3)
d ( FP)   d (GP)
p.3 ~ p.4
 d ( HCPV )  d (G / Ei )
 dVw  dV f (1.36)

where
Vw  initial connate water volume
V f  initial pore volume
negative sign ""  exp ansion of water leads
to a reduction in HCPV
58
1 V f
cf  
V f  GP
GP
1 V f
 cf  
V f (p)
1 V f
 cf 
V f p
GP
pore vol.
Vf
GP
GP
 dV f  c f  V f  dp
59
Vw
1 Vw
1 dVw
cw  

Vw d FP 
Vw dp
 dVw  c w  Vw  dp
FP
FP
Vf
FP
FP=gas pressure
FP
FP
FP
Vw
FP
FP=gas pressure
FP
60
G
d 
 Ei

  d HCPV   c wVw dp  c f V f dp

Since
HCPV
G
V f  PV 

1  S wc  Ei 1  S wc 
Vw  PV  S wc
HCPV
G S wc

S wc 
1  S wc 
Ei 1  S wc 
G
G S wc
G
 d    c w
dp  c f
dp
Ei 1  S wc 
Ei 1  S wc 
 Ei 
G
G G

S wc
1 
  
     
 cf
c w
 p
1  S wc 
 Ei  initial  Ei  t  Ei  initial  1  S wc 

c w S wc  c f p
 G
G
   
  
1  S wc 
 t  Ei  initial  Ei  initial
 c w S wc  c f p 
G G
     
1 



E
E
1

S
wc
 i  t  i  initial 

G
 
 Ei
61
G
G p  G  E  (1.33)
Ei
 cw S wc  c f p 
1 
E
1  S wc  

Gp
 cw S wc  c f  E

 1  1 

G
1  S wc  Ei

G
 Gp  G 
Ei
For
 1
Gp
cw  3 10 6 psi 1; c f  10 10 6 psi 1
cw S wc  c f
1  S wc
and
S wc  0.2
 1  0.013  0.987
E

 1  0.987
G
Ei
computing
 1.3% difference
with
Gp
E
 1
G
Ei
62
p/z plot
From Eq. (1.35) such as

p pi  G p 
  (1.35)
 1 
z zi 
G 
pi
p pi
 

Gp
z zi zi G
In
p
v.s Gp
z
p/z
Abandon
pressure pab
0
plot
Gp
G
p/z
Y=a+mx
p
z
x  Gp
y

pi
0
Gp/G=RF 1.0
z i G A straight line in p/z v.s Gp plot means that the reservoir is
a depletion type
pi
m
a
zi
63
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
64
Water drive reservoirs
With water production
 G

G p  G    We  W p  Bw  E

 Ei
 Gp 
1 

G 
p
  
 (1.41)
W
E
z
1 e i
G
pi
zi

where We*Ei /G represents the fraction of the initial hydrocarbon
pore volume flooded by water and is,
therefore, always less then unity.
65
Water drive reservoirs
 Gp 
1 

G 
p


 (1.41)
z
 We E i 
1 

G


pi
zi

since
 We Ei 
1 
 1
G


p pi  G p
  1 
z zi 
G



in water flux reservoirs
Comparing
p pi  G p
 1 
z zi 
G



in depletion type reservoir
66
Water drive reservoirs
 Gp 
1 

G 
p


 (1.41)
z
 We E i 
1 

G


pi
zi



In eq.(1.41) the following two parameters to be determined
G and We
History matching or “aquifer fitting” to find We
Aquifer model for an aquifer whose dimensions are of the same
order of magnitude as the reservoir itself.
We  cW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
67
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
68
Bruns et. al method


This method is to estimate GIIP in a water drive reservoir
From Eq. (1.40) such as
G

G p  G    We  E  (1.40)
 Ei

GE
 Gp  G 
 We E
Ei

E
 G p  G1    We E
 Ei 

E
 G1    G p  We E
 Ei 
Gp
We E
G


E 
E
1   1  
 Ei   Ei 
Gp
We E
or
G


E
E
1  
1 
 Ei 
 Ei
or
Ga  G 
We E

E
1  
 Ei 
Gp




E
1  
 Ei 
(or G a )
is plot as function of 
We E
E
1  
 Ei 
69
Bruns et. al method
Gp

E
1  
 Ei 









(or Ga )
is plot as function of
We E

E
1  
 Ei 
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
70
Hydrocarbon phase behavior
71
Hydrocarbon phase behavior
72
Hydrocarbon phase behavior
C---------＞D--------------＞E
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,
pt. c,
producing
Wet gas
(at scf)
Dry gas
injection
displace the wet gas
until dry gas break
through occurs in the
producing wells
Keep p above
dew pt.
Δp small
73
Equivalent gas volume


The material balance equation of eq(1.35) such as
Gp
pi 
p
1 

z
zi 
G




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.
74
Equivalent gas volume

If n lbm –mole of liquid have been produced, of molecular
weight M, then the total mass of liquid is
nM   o  w  liquid volume


where γ0 = oil gravity (water =1)
ρw ＝ density of water （=62.43 lbm/ft3）
n
 o  w  Vo
M

 lbm 
  V0 ft 3
3 
62.4 0V0
 ft 

M lbm / lbm  mole 
M
 0 62.4 
 
62.4 0V0 bbls  5.61458 ft 3
n
M
1 bbl
 0N p
 n  350.5
where N p []bbls
M
 0 N p RTsc
 0 N p 10.73  520
nRT
Vsc 
 350.5
 350.5
psc
M psc
M 14.7
 0N p
M
Equivalent gas volume
 Vsc  1.33 105
N p bbls
75
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＝Pi＝initial pressure （above dew point）
T＝Tr＝reservoir temperature


76
Condensate Reservoir


p decrease  by withdraw gas in stages from the cell, and measure gas Gp’
Until the pressure has dropped below the dew point
Z 2  phase 
p
pi
zi
G '

1  p 
G 

 (1.46)
Gp ' 
pi 
p

  (1.35)

1

z
zi 
G 
p
z
G '
pi 
1  p 
zi 
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.
77
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.
78
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)
79
80