500 Gas Material Balance Basic Concepts G=10 Bscfpetro/faculty/Kelly/GRM-chap1-matbal.pdf · Gas...
Transcript of 500 Gas Material Balance Basic Concepts G=10 Bscfpetro/faculty/Kelly/GRM-chap1-matbal.pdf · Gas...
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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0 1 2 3 4 5 6 7 8 9 10 11 12
Cumulative gas produced,Bscf
p/z
, p
sia
(p/z)i
(p/z)a
Gpa=8.5 BscfG=10 Bscf
Last measured
data point
extrapolate
Gas Material Balance Basic Concepts
GRM-Engler-09
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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
, p
sia
(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 gas
Production @ P
ipiz
az
ap1
gaB
giB1volRF
Gas Material Balance Advanced Topics
GRM-Engler-09
• 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.)
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
component Gas injection
Gas in solution
p/z
Gp/G
(p/z)i
(p/z)a
Depletion drive
water drive
strength
(p/z)a
0 100 50
Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
giBgB
pWwB615.5eWgBpGG
Cumulative water influx, rcf Cumulative water
production, stb
How to determine
We?
Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
giS
grS
ip
iz
az
ap
giS
grS
gaB
giB
wdRF
1
1
Gas saturation
Recovery (water drive) < Recovery (depletion)
45 to 75% >75%
vE
vE
giS
grS
gaB
giB
vEwdRF1
1
Volumetric sweep
efficiency
Method accounts for
pressure gradients within
the invaded zone due to
relative permeability
effects resulting from
trapped residual gas
rt
ra
ro reservoir
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-09 Trapped gas volume
Modified Gas Material Balance
Input data
Calculate
ro
Guess
pok
Calculate
We1
Calculate
rt
Calculate
pt
Calculate
pave
Calculate
Gt
Solve for
We 2
We1-We
2 < tol N Y
Gpn+1 = Gp
n +DGp
Update
Pok+1
Given Gpn
*
*
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
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4500
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Gp, mscf
p/z
,psia
G=2.16 Bscf
measured
We = 831 mstb
Sgr = 24%
GRM-Engler-09
F/E
t,stb
Intercept=G
tE
wBeW
We correct
We too small
We too large
Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
tE
wBeWG
tE
F
Linearized Gas Material Balance
wBpWgBpGF
giBgBgE
pipwiS1
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
Gas Material Balance Water Drive Reservoirs
GRM-Engler-09
Linearized Gas Material Balance - example
y = 1.0602x + 2.1074
R2 = 0.9843
3.0
3.2
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/E
t
G = 2.107 Bscf
484mstb
1,427mstb
We = 831 mstb
p/z
Gp
(p/z)i
Gas expansion
Gas expansion
+
Formation compaction
+
Water expansion
Overestimate of G
Gas Material Balance Abnormally pressured gas reservoirs
GRM-Engler-09
pi
pe
c1
G
pG
1
iz
ip
z
p
wiS1
fc
wiS
wc
ec
Where,
GRM-Engler-09
Gas Material Balance Abnormally pressured gas reservoirs
Method 1 : Assume formation compressibility is known and constant
Volumetric (geopressured)
y = -92.336x + 6532.2
R2 = 0.9979
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1000
2000
3000
4000
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6000
7000
0 20 40 60 80 100
Gp, Bcf
mo
difie
d p
/z, p
sia
G=70.7
pGvs
)wi
S1(
)pi
p(f
cwi
Sw
c
1z
p
Plotting function
Conventional
overestimates G
by > 25%!
GRM-Engler-09
Gas Material Balance Abnormally 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, p
si-1
x1
0-6
Bscf8.75199.13
1000
slope
1G
1psi
610x5.12
wc
wiS)
wiS1(
610bx
fc
GRM-Engler-09
Gas Material Balance Abnormally 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
aqM
NNPMM
non-net pay water and
pore volumes
external water volume
found in limited aquifers
GRM-Engler-09
Gas Material Balance Abnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)]
Back-calculate ce from,
pi
p
1
G
pG
1z/p
iz/p1
atedbackcalculec
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10
15
20
25
30
35
40
45
50
0 2000 4000 6000 8000 10000
pressure, psia
ce, p
si- 1
ce(back) assuming OGIP
ce(p) generated from rock &
w ater properties
Compare with rock and fluid derived ce
GRM-Engler-09
Gas Material Balance Abnormally pressured gas reservoirs
Method 3 [Fetkovich (1991)]
Results
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7000
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Gp, Bscf
p/z
, p
sia
historical performance data
model M = 2.25
Cf = 3.2x10-6 psi-1
G = 72 Bscf
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs
G p
p/z
G G p
p/z
G
Expanding
Radius
Rebound
Effect
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs
Gp
p/z
Conventional response
G
Tight gas response
(p/z)i
(p/z)intm
1
m2 = m
1
?
Gp
p/z
Conventional response
G
Tight gas response
(p/z)i
(p/z)intm
1
m2 = m
1
?m2 = m
1
?pG
)z/p(m
D
D
m
1*
scT
scTPhcV
)wS1(Ah43560hcV
Hydrocarbon volume
Estimate area
slope
GRM-Engler-09
Gas Material Balance Low-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 A m2
G1 G2
GRM-Engler-09
Gas Material Balance Low-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 B m2
G1= G2
(p/z)int
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs
• Different slope and
intercept
• Estimate using other
cases
Case B < actual < Case A
(p/z)
(p/z)i
Gp
CASE C m2
G1 G2
(p/z)int
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs
P/Z vs. Cumulative Production
No. 114
y = -0.8367x + 552.88
R2 = 0.9580
200
400
600
800
1000
1200
1400
1600
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Cumulative Production, mmscf
P/Z
, p
sia
Example 1.8
??? 11
gi, cp 0.0134
h, ft 40
cti, psi-1
x 10-4
5.77
gg 0.67
Tr , deg F 106
Sw, % 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
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs
0
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1000
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2000
2500
3000
3500
4000
4500
0 100 200 300 400 500 600 700
Cumulative production, mmscf
flo
w r
ate
, m
scf/
mo
nth
GIP = 700 mmscf
Gp = 526 mmscf
RF = 75%
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs
1
10
100
1000
0 5 10 15 20 25
time, years
pro
du
cti
on
ra
te,
msc
f/m
o
0
200
400
600
800
1000
1200
SIB
HP
, p
si
simulated
measured
Single well simulation model
• Pwf =
250 150 psia
•Model area =
86 acres
GRM-Engler-09
Gas Material Balance Low-permeability gas reservoirs Problem 4
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600
800
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1200
1400
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cumulative production, mmscf
p/z
, p
sia
, % 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?