chapter3(简)

Fig. 3.1 dry gas
1. gas in-place
From the real-gas law:
气体地下体积： Vgi
zi nRT pi
n Vgi Pi zi RT
气体储量：
Vsc
G
zscnRTsc psc
G piVgi zscTsc ziTpsc
Vgi 43.56 Ah (1 S wi ) (Mft3)
A-acre
1acre(英亩)=43.5Swi )
p i z sc Tsc p sc z i T
(Mft3)
(3.5)
Vgi 43.56 Ah (1 S wi ) (Mft3)
RB
1acre(英 亩)=43.56 *103 ft2
G Vgi 7758Ah(1 Swi )
Mo
R3
The molecular weight of the stock-tank liquid:
Mo
5954 API 8.811
or
Mo
42.43 o 1.008 o
•Gas production from low-pressure separators and stock
tanks often is not measured.
total initial gas in place :
GT
7758Ah(1 Swi )
Bgi
(3.17)
•For a three-stage separation system, the reservoir gas
gravity is:
w
4602 0 R1 1 R2 2
R1
133316 o
F 1 EV
Bgi S gr Bga S gi
1 EV EV
(F=50~70%)
3.2.3 Volumetric Wet-Gas and Gas-Condensate Reservoirs
the primary composition of a wet gas is methane , but has more of the intermediate and heavier-weight hydrocarbon molecules.
Chapter 3 Gas Volumes and MaterialBalance Calculations
3.1 Introduction
This chapter presents methods for estimating original gas in place(地质储量), gas reserves（可采储 量）, and recovery factors（采收率） for a variety of reservoir drive mechanisms。
scf/STB. P108错
•The total gas that will be produced in the gaseous phase is :
G f g GT （Mscf）
• the original oil (condensate) in place is:
N 1000 f gGT Rt
（STB）
(3.25)
对于凝析气：Above equation is valid for P>Pd.
否则Rt不为常数。
Example 3.3 -Calculating Original Gas and Condensate
in Place for a Volumetric Wet-Gas Reservoir.
Bgi
Bgi
(Mscf)
RB/Mft3
Where:
Bgi
1000 psc ziT 5.615 pi z scTsc
5.02 ziT pi
(bbl/Mft3)
psc 14.65 psi, Tsc 60o F 460 520o R
2. gas reserve：
可采储量
Gp G Ga
(3.8)
Pi=5500 psia, A=1000 acres，h=50 ft， T 288 o F
pi
Swi =0.32, 0.21
GT
7758Ah(1 Swi )
Bgi
(Mscf)
fg
Rt
Rt
132800 o
Mo
N 1000 f gGT Rt
G f g GT
Solution:
o
141 .5
3.1.1 Volumetric Dry-Gas Reservoirs
A volumetric reservoir is completely enclosed by lowpermeability or completely impermeable barriers and does not receive pressure support from external sources.
Gp
7758Ah(1 Swi )
Bgi
7758AhS gr
Bga
residual gas
saturation
Gp
7758Ah
(1
S
wi
)
1
Bgi
B
Bg i S ga (1
gr
S
wi
)
F 1
Bg i S gr
the recovery factor,
Bga (1 Swi )
2. gas reserve (Ev≠100%)
131 .5 API
141 .5 131 .5 54.5
0.76.
Mo
5,954 API 8.811
5,954 54.5 8.811
130 .3
lbm/lbm-mol
w
R1 1 4,602 o R3 3
R1
133,316 o
Mo
R3
(59,550)(0.72) (4,602)(0.76) (415)(1.23) 59,550 (133,316)(0.76) 415
Mo
59,965
59,965
132,8000.76
0.99
130.3
G f g GT (0.99)(76.9) 76.1Bcf (106MSCF)
N 1,000 f gGT (1,000 )(0.99)(76.9 10 6 Mscf )
Rt
59,965 scf / STB
1.310 6 STB
RB
(Mscf)
Gp
7758Ah(1 Bgi
Swi )
1
Bgi Bga
F 1 Bgi --recovery factor Bga
(F=80~90%)
RB/Mft3
3.2.2 Dry-Gas Reservoirs With Water Influx
1. gas reserve (water sweep efficiency Ev=100%)
w
R1 1
4602 o
R1 Veq
G pa
Gpa -the additional gas production from the secondary separator and stock tank,
Veq -the vapor equivalent of the primary separator liquid.
0.77
130.3
Use rw to determine: p pc 655psia Tpc 395 R
p pr
pi p pc
5500 655
8.40
Tpr T 288 460 1.89
Tpc
395
zi l.06
Bgi
5.20 ziT pi
(5.02 )(1.06 )( 288 5,500
Gp
7758Ah(1 Swi )
Bgi
EV
7758Ah(1 Swa )
Bga
(1
EV
)
7758Ah
(1
S
wi
)
Bga
Ev—波及的体积百分数。
（MSCF）(3.15)
Gp
7758
Ah (1
Bgi
S
wi
)
1
EV
Bgi Bga
S gr S gi
1 EV EV
recover factor:
3.3 Material-Balance Methods
Material-balance methods provide a simple, but effective, alternative to volumetric methods.
（2）Material-balance concepts.
3.2 Volumetric Methods
Volumetric methods consider the reservoir PV occupied by hydrocarbons at initial conditions and at later conditions after some fluid production and associated pressure reduction.
Mo
R2
R3 3
R3
scf/STB
rw :at standard condition the ratio of gas density to that of air.
•For a two-stage separation system:
w
4602 o R1 1 R3 3
R1
133316 o
460 )
0.72（RB/Mscf）