Gas Material Balance
• Estimate OGIP
4000
Basic Concepts
(p/z)i
3500
3000
p/z, psia
• Predict recovery
• Identify drive
mechanism
Last measured
data point
2500
2000
extrapolate
1500
1000
(p/z)a
500
G=10 Bscf
Gpa=8.5 Bscf
0
0
1
2
3
4
5
6
7
8
9
10
11
12
Cumulative gas produced,Bscf
GRM-Engler-09
Gas Material Balance
Measured
pressure
Basic Concepts
Cumulative gas
Production @ P
p pi  G p 

 1 
z zi 
G 
4000
(p/z)i
3500
Last measured
data point
Gas property, f(T,P,g)
 Bgi 

RFvol  1 
 Bga 


 pa zi 

 1 
 z a pi 
p/z, psia
3000
2500
2000
extrapolate
1500
1000
(p/z)a
500
G=10 Bscf
Gpa=8.5 Bscf
0
0
1
2
3
4
5
6
7
8
9
10
11
12
Cumulative gas produced,Bscf
GRM-Engler-09
Gas Material Balance
Advanced Topics
• Nonlinear gas material balance
Water drive gas reservoirs
(additional pressure support)
Abnormally pressured gas reservoirs
(rock compressibility)
Low permeability gas reservoirs
(measured pressures don’t achieve ave. press.)
GRM-Engler-09
Gas Material Balance
Advanced Topics
• Comprehensive gas material balance
Geopressured
component


Gas in solution
p
1  c e( p ) ( p i  p ) 
z
5.615
 p  p / z i 
Wp B w  Winj B w  We
G p  G inj  Wp R sw 
  
G 
Bg
 z i

Gas injection




Water drive
component
GRM-Engler-09
Gas Material Balance
Water Drive Reservoirs
Cumulative water influx, rcf

G p Bg  We  5.615B w Wp
G
Bg  Bgi
Cumulative water
production, stb

strength
(p/z)i
(p/z)a
water drive
p/z
How to determine
We?
(p/z)a
Depletion drive
0
GRM-Engler-09
50
100
Gp/G
Gas Material Balance

B gi S gr 


RFwd  1 
 B S 
ga gi 

 p z S gr
 1  a i
 z p S
a i gi

Water Drive Reservoirs
B gi  S gr 1  E 

v
RFwd  1  Ev

B ga  S gi
Ev 






Volumetric sweep
efficiency
Gas saturation
Recovery (water drive) < Recovery (depletion)
45 to 75%
>75%
GRM-Engler-09
Modified Gas Material Balance
Prediction of water influx and reservoir pressure
(Schafer, et al, 1993)
Method accounts for
pressure gradients within
the invaded zone due to
relative permeability
effects resulting from
trapped residual gas
aquifer
Original
Reservoir
boundary
ra
Invaded
zone
reservoir
ro
rt
Water
influx
Current
Reservoir
boundary
G p Bg  G( Bg  Bgi )  Gt ( Bgt  Bg )  We  BwW p
Trapped gas volume
GRM-Engler-09
Modified Gas Material Balance
*
Given Gpn
Input data
Calculate
Calculate
Calculate
ro
pave
Guess
pok
pt
Update
Calculate
Pok+1
Gt
Calculate
Solve for
We 2
We1
Calculate
rt
*
N
We1-We2
< tol
Y
Gpn+1 = Gpn +DGp
GRM-Engler-09
Modified Gas Material Balance
Comparison of reservoir performance from simulation,
Conventional and modified material balance methods
(Hower and Jones, 1991)
GRM-Engler-09
Modified Gas Material Balance
Modified Gas Material Balance - example
4500
We = 831 mstb
Sgr = 24%
4000
3500
measured
p/z,psia
3000
2500
2000
1500
1000
500
0
0
500
1000
1500
Gp, mscf
2000
2500
G=2.16 Bscf
GRM-Engler-09
Gas Material Balance
Water Drive Reservoirs
Linearized Gas Material Balance
We too small
W B
F
G e w
Et
Et
Et = Eg + Ecf = total expansion
Eg = expansion of gas in reservoir
We too large
F/Et,stb
F = total net reservoir voidage
F  G p Bg  Wp B w
We correct
Intercept=G
E g  Bg  Bgi
Ecf = connate water and formation expansion * Bgi
 S wic w  c f 

pi  p 
E cf  Bgi * 
 1  S wi 

We Bw
Et
GRM-Engler-09
Gas Material Balance
Water Drive Reservoirs
Linearized Gas Material Balance - example
5.0
We = 831 mstb
4.8
4.6
y = 1.0602x + 2.1074
4.4
R2 = 0.9843
484mstb
F/Et
4.2
1,427mstb
4.0
3.8
3.6
3.4
3.2
3.0
1.0
G = 2.107 Bscf
1.5
2.0
2.5
3.0
We/Et
GRM-Engler-09
Gas Material Balance
Abnormally pressured gas reservoirs
p  G 
i 1  p 
z 
G 
p
i


z 
 p  p 
1

c




e i
Gas expansion
+
Formation compaction
+
Water expansion
(p/z)i
p/z
Where,
c S  c 




w
wi
f
c 
1 S
e
wi
Gas expansion
Overestimate of G
Gp
GRM-Engler-09
Gas Material Balance
Abnormally pressured gas reservoirs
Method 1 : Assume formation compressibility is known and constant
  c S  c ( p  p ) 

p   w wi f  i
Plotting function
1 
 vs G
z
(1  S )
p

wi


Volumetric (geopressured)
modified p/z, psia
7000
y = -92.336x + 6532.2
R2 = 0.9979
6000
5000
Conventional
overestimates G
by > 25%!
4000
3000
G=70.7
2000
1000
0
0
20
40
60
80
100
Gp, Bcf
GRM-Engler-09
Gas Material Balance
Abnormally pressured gas reservoirs
Method 2 [Roach (1981)] : Simultaneous solution of G and cf
 G

p z 
S c c
p
z

1
p
 i
 1
f
i  wi w

1



 G 
1 S
 p  p   pz
  p  p  pz i 

 i 
wi
 i


 i

Revised material
Balance eq.
Geopressured
G
150
100
c  bx10
f
-1
y, psi x10
-6
y = 13.199x - 17.511
R2 = 0.993
1
1000

 75.8 Bscf
slope 13.199
50
6
 12.5x10
0
0
2
4
6
8
x, Bscf/psi x 10
10
12
(1  S
6
psi
wi
1
) S
c
wi w
14
-3
GRM-Engler-09
Gas Material Balance
Abnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)] : Addition of gas solubility and total water
c
Defines: c
e( p )

S
tw (p) wi
c
 M[c
f ( p)
(1  S
Where,
Ctw, cumulative total water compressibility
wi
tw (p)
c
f ( p)
]
)
water expansion due to pressure depletion
the release of solution gas in the water
M, associated water-volume ratio given by:
non-net pay water and
pore volumes
M M

NNP
external water volume
found in limited aquifers

M
aq
 1  h / h  h 
nnp 
n g
aq




  h / h  h r
r  n g 
 r  2 
 aq 

 1


 rr 



GRM-Engler-09
Gas Material Balance
Abnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)]
Back-calculate ce from,

 G 


p
/
z

1
p 
c 
i

 1
1

 
G   p  p 
 e  backcalculated  p / z  



  i

Compare with rock and fluid derived ce
ce, psi-1
50
45
c e(p) generated from rock &
w ater properties
40
35
30
25
20
15
10
c e(back) assuming OGIP
5
0
0
2000
4000
6000
8000
10000
pressure, psia
GRM-Engler-09
Gas Material Balance
Abnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)]
Results
7000
historical performance data
model
6000
M = 2.25
p/z, psia
5000
Cf = 3.2x10-6 psi-1
4000
G = 72 Bscf
3000
2000
1000
0
0
20
40
60
80
100
Gp, Bscf
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
Expanding
Radius
p/z
Rebound
Effect
Gp
G
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
slope
m
D(p / z)
DG p
(p/z)i
Hydrocarbon volume
Estimate area
Vhc  43560Ah (1  Sw )
1
(p/z)int
p/z
TPsc 1
Vhc 
*
Tsc m
m ?
2 =
m
Co
nv
e
m
nti
1
Tig
on
ht g
al
as r
res
esp
po
ons
ns
e
e
Gp
G
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
Gas-in-place
p  1
G   i  *
 zi  m
(p/z)i m2
CASE A
Hydrocarbon volume
TP
1
Vhc  sc *
Tsc m
(p/z)
Estimate area
Vhc  43560 Ah  (1  S w )
Gp
G1 G2
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
Find Gas-in-place and m2
1  pi  1
p
G  *
   *
 z int m1  z i  m 2
(p/z)i m2
CASE B
Hydrocarbon volume
TP
1
Vhc  sc *
Tsc m 2
(p/z)int
(p/z)
Gp
G1= G2
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
• Different slope and
intercept
(p/z)i m2
• Estimate using other
cases
CASE C
(p/z)int
(p/z)
Case B < actual < Case A
Gp
G1 G2
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
Example 1.8
11
0.0134
40
5.77
0.67
106
44
0.229
1131
1600
1400
Case B
1200
P/Z, psia
??
gi, cp
h, ft
cti, psi-1 x 10-4
gg
Tr , deg F
Sw, %
rw, ft.
Pi, psi
P/Z vs. Cumulative Production
No. 114
1000
800
600
400
y = -0.8367x + 552.88
R2 = 0.958
200
0
0
100
200
300
400
500
600
700
Cumulative Production, mmscf
1 552.9
 p
G  *

 660 mmscf
 z int m1 .8367
m2 = 2.12 psia/mmscf
Vhc = 7.544 mmrcf
A = 70 acres
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
4500
GIP = 700 mmscf
RF = 75%
flow rate, mscf/month
Gp = 526 mmscf
4000
3500
3000
2500
2000
1500
1000
500
0
0
100
200
300
400
500
600
700
Cumulative production, mmscf
GRM-Engler-09
Gas Material Balance
Low-permeability gas reservoirs
Single well simulation model
1000
measured
100
800
600
10
SIBHP, psi
•Model area =
86 acres
1200
simulated
production rate, mscf/mo
• Pwf =
250  150 psia
1000
400
200
1
0
0
5
10
15
20
25
time, years
GRM-Engler-09
Gas Material Balance
Problem 4
Low-permeability gas reservoirs
Estimate the gas-in-place and drainage area for this well. If
cumulative production was 752 mmscf, what has been the
recovery factor?
, %
gi, cp
11
1400
0.0131
1200
cti, psi-1 x 10-4
6.22
1000
gg
0.67
Tr , deg F
103
Sw, %
44
rw, ft.
0.229
Pi, psi
1045
p/z, psia
67
h, ft
800
600
400
200
0
0
200
400
600
800
1000
cumulative production, mmscf
GRM-Engler-09