grm-chap1-matbal
TRANSCRIPT
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Gas Material Balance Basic Concepts
Estimate OGIP
Predict recovery
Identify drive
mechanism
GRM-Engler-09
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500
1000
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3000
3500
4000
0 1 2 3 4 5 6 7 8 9 10 11 12
Cumulative gas produced,Bscf
p/z,psia
(p/z) i
(p/z)a
Gpa=8.5 BscfG=10 Bscf
Last measured
data point
extrapolate
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Gas Material Balance Basic Concepts
GRM-Engler-09
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500
1000
1500
2000
2500
3000
3500
4000
0 1 2 3 4 5 6 7 8 9 10 11 12
Cumulative gas produced,Bscf
p/z,psia
(p/z)i
(p/z)a
Gpa=8.5 BscfG=10 Bscf
Last measured
data point
extrapolate
G
pG1
izip
z
p
Gas property, f(T,P,g)
Measured
pressure Cumulative gasProduction @ P
ipiz
azap1
gaB
giB
1volRF
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Gas Material Balance Advanced Topics
GRM-Engler-09
Nonlinear gas material balanceWater drive gas reservoirs
(additional pressure support)
Abnormally pressured gas reservoirs
(rock compressibility)
Low permeability gas reservoirs(measured pressures dont achieve ave. press.)
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Gas Material Balance Advanced Topics
GRM-Engler-09
Comprehensive gas material balance
eWwBinjWwBpWgB
615.5swRpWinjGpG
G
iz/p
iz
p
)pip()p(ec1z
p
Geopressured
component
Water drive
componentGas injection
Gas in solution
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p/z
Gp/G
(p/z)i
(p/z)a
Depletion drive
water drive
strength
(p/z)a
0 10050
Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
giBgB
pWwB615.5eWgBpGG
Cumulative water influx, rcfCumulative water
production, stb
How to determineWe?
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Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
giS
grS
ip
iz
az
ap
giS
grS
gaB
giBwdRF
1
1
Gas saturation
Recovery (water drive) < Recovery (depletion)
45 to 75% >75%
vE
vE
giS
grS
gaB
giB
vEwdRF1
1
Volumetric sweep
efficiency
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Method accounts for
pressure gradients within
the invaded zone due to
relative permeability
effects resulting from
trapped residual gas
rt
ra
roreservoir
aquifer
Invaded
zone
Original
Reservoir
boundary
Current
Reservoir
boundary
Water
influx
Modified Gas Material Balance
Prediction of water influx and reservoir pressure
(Schafer, et al, 1993)
pWwBeWgBgtBtGgiBgBGgBpG )()(
GRM-Engler-09Trapped gas volume
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Modified Gas Material Balance
Input data
Calculate
ro
Guess
pok
CalculateWe
1
Calculate
rt
Calculatept
Calculate
pave
Calculate
Gt
Solve forWe
2
We1-We
2< tolN Y
Gpn+1= Gp
n+DGp
Update
Pok+1
Given Gpn
*
*
GRM-Engler-09
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Modified Gas Material Balance
Comparison of reservoir performance from simulation,
Conventional and modified material balance methods(Hower and Jones, 1991)
GRM-Engler-09
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Modified Gas Material Balance
Modified Gas Material Balance - example
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 500 1000 1500 2000 2500
Gp, mscf
p/z,psia
G=2.16 Bscf
measured
We = 831 mstb
Sgr = 24%
GRM-Engler-09
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F/Et,stb
Intercept=G
tE
wBe
W
Wecorrect
Wetoo small
Wetoo large
Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
tE
wBeWGtE
F
Linearized Gas Material Balance
wBpWgBpGF
giBgBgE
pip
wiS1
fcwcwiS*giBcfE
F = total net reservoir voidage
Et= Eg+ Ecf= total expansion
Eg= expansion of gas in reservoir
Ecf = connate water and formation expansion * Bgi
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Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
Linearized Gas Material Balance - example
y = 1.0602x + 2.1074
R2= 0.9843
3.03.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
1.0 1.5 2.0 2.5 3.0
We/Et
F/Et
G = 2.107 Bscf
484mstb
1,427mstb
We = 831 mstb
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p/z
Gp
(p/z)i
Gas expansion
Gas expansion
+
Formation compaction
+
Water expansion
Overestimate of G
Gas Material BalanceAbnormally pressured gas reservoirs
GRM-Engler-09
p
i
p
e
c1
G
pG
1
iz
ip
z
p
wiS1
fc
wiS
wc
e
c
Where,
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GRM-Engler-09
Gas Material BalanceAbnormally pressured gas reservoirs
Method 1 : Assume formation compressibility is known and constant
Volumetric (geopressured)
y = -92.336x + 6532.2
R2= 0.9979
0
1000
2000
3000
4000
5000
6000
7000
0 20 40 60 80 100
Gp, Bcf
modifie
dp/z,psi
G=70.7
pGvs
)wi
S1(
)pi
p(f
cwi
Sw
c
1z
p
Plotting function
Conventional
overestimates Gby > 25%!
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GRM-Engler-09
Gas Material BalanceAbnormally pressured gas reservoirs
Method 2 [Roach (1981)] : Simultaneous solution of G and cf
Revised material
Balance eq.wi
S1
fc
wc
wiS
ipz
zi
p
pi
p
pG
G
11
ipz
zi
p
pi
p
1
Geopressured
y = 13.199x - 17.511
R2= 0.993
0
50
100
150
0 2 4 6 8 10 12 14
x, Bscf/psi x 10-3
y,psi
-
1x10-6
Bscf8.75199.13
1000
slope
1G
1psi
610x5.12
wc
wiS)
wiS1(
610bx
fc
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GRM-Engler-09
Gas Material BalanceAbnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)] : Addition of gas solubility and total water
Defines:)
wiS1(
])p(f
c)p(tw
c[M)p(f
cwi
S)p(tw
c
)p(ec
Where,
Ctw, cumulative total water compressibility
M, associated water-volume ratio given by:
water expansion due to pressure depletion
the release of solution gas in the water
1
2
rr
aqr
rh
aqh
gh/
nh
gh/
nh1
r
nnp
aqMNNPMM
non-net pay water and
pore volumes
external water volume
found in limited aquifers
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GRM-Engler-09
Gas Material BalanceAbnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)]
Back-calculate cefrom,
pi
p
1
G
pG
1z/p
iz/p1atedbackcalcule
c
0
5
10
15
20
25
30
35
40
45
50
0 2000 4000 6000 8000 10000
pressure, psia
ce,psi-1
ce(back)assuming OGIP
ce(p)generated from rock &
w ater properties
Compare with rock and fluid derived ce
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GRM-Engler-09
Gas Material BalanceAbnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)]
Results
0
1000
2000
3000
4000
5000
6000
7000
0 20 40 60 80 100
Gp, Bscf
p/z,psi
historical performance data
model M = 2.25
Cf = 3.2x10-6psi-1
G = 72 Bscf
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
Gp
p/z
GGp
p/z
G
Expanding
Radius
Rebound
Effect
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
Gp
p/z
Conventionalresponse
G
Tightgasresponse
(p/z)i
(p/z)int m1
m2=m
1
?
Gp
p/z
Conventionalresponse
G
Tightgasresponse
(p/z)i
(p/z)int m1
m2=m
1
?m2=m
1
?pG
)z/p(m
D
D
m
1*
scT
scTPhcV
)wS1(Ah43560hcV
Hydrocarbon volume
Estimate area
slope
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
m
1*
iz
ipG
m
1*
scT
scTPhcV
)1(43560 wSAhhcV
Gas-in-place
Hydrocarbon volume
Estimate area
(p/z)
(p/z)i
Gp
CASE Am2
G1G2
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
2m
1*
iz
ip
1m
1*
intz
pG
Find Gas-in-place and m2
2m
1*
scT
scTPhcV
Hydrocarbon volume
(p/z)
(p/z)i
Gp
CASE Bm2
G1= G2
(p/z)int
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
Different slope and
intercept
Estimate using othercases
Case B < actual < Case A
(p/z)
(p/z)i
Gp
CASE Cm2
G1G2
(p/z)int
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
P/Z vs. Cumulative Production
No. 114
y = -0.8367x + 552.88
R2= 0.958
0
200
400
600
800
1000
1200
1400
1600
0 100 200 300 400 500 600 700
Cumulative Production, mmscf
P/Z,psia
Example 1.8??? 11gi, cp 0.0134h, ft 40
cti, psi-1
x 10-4 5.77
gg 0.67
Tr , deg F 106Sw, % 44
rw, ft. 0.229
Pi, psi 1131
mmscfmz
pG 660
8367.
9.552
1
1*
int
m2 = 2.12 psia/mmscf
Vhc= 7.544 mmrcf
A = 70 acres
Case B
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 100 200 300 400 500 600 700
Cumulative production, mmscf
flow
rate,ms
cf/month
GIP = 700 mmscf
Gp= 526 mmscf
RF = 75%
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs
1
10
100
1000
0 5 10 15 20 25
time, years
productionrate,m
scf/mo
0
200
400
600
800
1000
1200
SIBHP,psi
simulated
measured
Single well simulation model
Pwf =
250 150 psia
Model area =
86 acres
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GRM-Engler-09
Gas Material BalanceLow-permeability gas reservoirs Problem 4
0
200
400
600
800
1000
1200
1400
0 200 400 600 800 1000
cumulative production, mmscf
p/z,psia
, % 11
gi, cp 0.0131
h, ft 67
cti, psi-1 x 10-4 6.22
gg 0.67
Tr , deg F 103
Sw, % 44
rw, ft. 0.229
Pi, psi 1045
Estimate the gas-in-place and drainage area for this well. If
cumulative production was 752 mmscf, what has been the
recovery factor?