Structural Analysis and Design of Foundations: A Comprehensive Handbook for S...
Numerical Simulation of 2-D Subsurface Reservoir
1. PTE 508 Project Two Report
Je↵rey Daniels
May 8, 2015
1 Reservoir Description
1.1 Reservoir Properties
The oil reservoir under study is approximated by a rectangular prism. The geometry
of this prism is length Lx = 1500 ft, width Ly = 1100 ft and height Z = 50 ft. The
reservoir has a reference pressure-dependent porosity of = 0.22 measured at 1000
psia, an isotropic permeability of k = 11 mD, and a formation compressibility factor of
ct = 2x10 6
psi 1
which is assumed to be constant.
1.2 Injection and Production Wells
Five wells, Well-A and Well-B, Well-C, Well-D and Well-E are drilled in the reservoir
first. Well-C serves as a producer and later as an injector in this reservoir simulation.
All wells have the same wellbore radius rw = 0.25 ft and the same skin factor of S = 0.
1.3 Reservoir Fluid Properties
The fluid flow is two-phase characterized by an initial oil saturation of Soi = 0.75 and
initial water saturation of Swi = 0.25. The fluid properties of oil are viscosity µo =
5 cp, oil compressibility co = 10 5
psi 1
and a reference formation volume factor Bo =
1.25 RB/STB measured at 1000 psia. The oil bubble point pressure is 1000 psia. The
fluid properties of water are viscosity µw = 1 cp, water compressibility cw = 10 6
psi 1
and a reference formation volume factor Bw = 1.02 RB/STB measured at 1000 psia.
2 Creation of Simulation cells
In this study, the reservoir is assumed to be two dimensional and was discretized into
165 simulation cells (i.e characterized by 11x15 grid blocks). Each cell with a well present
had a length of X = 100 ft, a width of Y = 100 ft and a height of Z = 50 ft.
1
2. 3 Partial Di↵erential Equations (PDEs) With Initial and Bound-
ary Conditions
The PDEs to be discretized are as follows:
@
@x
( ox
@p
@x
) +
@
@y
( oy
@p
@y
) 887.53 (x xw) (y yw)
qo,wµB
z
= 158
@
@t
(
So
Bo
)
@
@x
( wx
@p
@x
) +
@
@y
( wy
@p
@y
) 887.53 (x xw) (y yw)
qw,wµB
z
= 158
@
@t
(
Sw
Bw
)
ox =
kxKro(So, Sw)
µoBo
, oy =
kyKro(So, Sw)
µoBo
wx =
kxKrw(So, Sw)
µwBw
, wy =
kyKrw(So, Sw)
µwBw
So + Sw = 1
Boundary conditions: @P
@x
|x=0= 0, @P
@x
|x=1500= 0 @P
@y
|y=0= 0, @P
@y
|y=1100= 0
Initial condition: P(x)|t0= 3000 psia
The PVT relations for formation volume factors Bo, Bw and are defined by the fol-
lowing:
Bo(p) = 1.25 exp( Co(p 1000))
Bw(p) = 1.02 exp( Cw(p 1000))
(p) = 0.22 exp(C (p 1000))
The following plot shows the PVT relations graphically as the reservoir pressure de-
creases from 3000 psia to 1000 psia
1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000
1
1.05
1.1
1.15
1.2
1.25
FormationVolumeFactor,RB/STB
Plot of Bo
, Bw
and φ vs P
1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000
0.22
0.2202
0.2204
0.2206
0.2208
0.221
Porosity,v/v
Bw
Bo φ
2
3. Relative permeability curves, Kro and Krw are defined by two-phase Corey’s model.
Swc = 0.2 and Sor = 0.
Sn =
Sw Swc
1 Sw
=
1 So Swc
1 Swc
Krw(Sw) =
(
S4
n , Sw > Swc
0 , Sw Swc
Kro(Sw) =
(
(1 Sn)2
(1 S2
n) , So > SororSw < 1 Sor
0 , So SororSw 1 Sor
The following plot shows the graphical relation between the relative perameabilities
and water saturation
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Water Saturation
RelativePermeability
Plot of Relative Permeability Curves
krw
kro
4 Implicit Block-Centered Grid System Discretization
4.1 General Discretized Equation
The following equation is used to solve for the grid cell pressures of the discretized
reservoir. It contains the PVT parameters for both water and oil.
SmPn+1
m Nx + WmPn+1
m 1 + CmPn+1
m + EmPn+1
m+1 + NmPn+1
m+Nx = Rm
where m = 1, 2, 3, ..., 165
Sm = ¯OS(
x
y
)2
Bo + ¯WS(
x
y
)2
Bw
Wm = ¯OW Bo + ¯WW Bw
3
4. Cm = Bo(¯OW +¯OE) (
x
y
)2
Bo(¯OS+¯ON ) Bw(¯WW +¯WE) (
x
y
)2
Bw(¯WS+¯WN )
158
x
t
Ct
Em = ¯OEBo + ¯WEBw
Nm = ¯ON (
x
y
)2
Bo + ¯WN (
x
y
)2
Bw
Rm = 158
x
t
CtPn
m + 887.53
qo,STBBo
x y
|ifwellincellm+887.53
qw,STBBw
x y
|ifwellincellm
To calculate the average mobility ratios
¯o,S =
kKro(So,up)
µoBo(Pm Nx+Pm
2
)
So,up =
(
So,m , Pm Pm Nx
So,m Nx , Pm < Pm Nx
¯o,N =
kKro(So,up)
µoBo(
Pm+Nx +Pm
2
)
So,up =
(
So,m , Pm Pm+Nx
So,m+Nx , Pm < Pm+Nx
¯o,W =
kKro(So,up)
µoBo(Pm 1+Pm
2
)
So,up =
(
So,m , Pm Pm 1
So,m 1 , Pm < Pm 1
¯o,E =
kKro(So,up)
µoBo(Pm+1+Pm
2
)
So,up =
(
So,m , Pm Pm+1
So,m+1 , Pm < Pm+1
Sw = 1 So
¯w,S =
kKrw(Sw,up)
µwBw(
Pm Nx +Pm
2
)
4
5. ¯w,N =
kKrw(Sw,up)
µwBw(
Pm+Nx +Pm
2
)
¯w,W =
kKrw(Sw,up)
µwBw(Pm 1+Pm
2
)
¯w,E =
kKrw(Sw,up)
µwBw(Pm+1+Pm
2
)
4.2 Incorporating no flow boundary conditions
For no flow boundary conditions, the mobility ratios of water and oil are set to 0. At
the Southern Boundary:
¯os = ¯ws = 0
At the Western Boundary:
¯ow = ¯ww = 0
At the Eastern Boundary:
¯oe = ¯we = 0
At the Northern Boundary:
¯on = ¯wn = 0
4.3 Production term discretization
The production rate if a well exists in the grid cell is treated implicitly therefore
qSTB w = Jww(Pn+1
m Pn
m)
qo = Jwo(Pn+1
m Pn
m)
4.4 Finite-Di↵erence Equation for grid cells containing wells
The finite di↵erence equation for the grid cells with wells are as follows:
Well-A at m=34
S34Pn+1
19 + W34Pn+1
18 + C34Pn+1
34 + E34Pn+1
35 + N34Pn+1
49 = R34
S34 = ¯OS(
x
y
)2
Bo + ¯WS(
x
y
)2
Bw
W34 = ¯OW Bo + ¯WSBw
E34 = ¯OEBo + ¯WEBw
5
6. N34 = ¯ON (
x
y
)2
Bo + ¯WN (
x
y
)2
Bw
C34 = Bo(¯OW +¯OE) (
x
y
)2
Bo(¯OS+¯ON ) Bw(¯WW +¯WE) (
x
y
)2
Bw(¯WS+¯WN )
158
x
t
Ct 887.53
Jwo well ABo
x y
887.53
Jww well ABw
x y
R34 = 158
x
t
CtPn
34 887.53
Jwo well APwf well ABo
x y
887.53
Jww well APwf well ABw
x y
Well-B at m=42
S42Pn+1
27 + W42Pn+1
41 + C42Pn+1
42 + E42Pn+1
43 + N42Pn+1
57 = R42
S42 = ¯OS(
x
y
)2
Bo + ¯WS(
x
y
)2
Bw
W42 = ¯OW Bo + ¯WW Bw
E42 = ¯OEBo + ¯WEBw
N42 = ¯ON (
x
y
)2
Bo + ¯WN (
x
y
)2
Bw
C42 = Bo(¯OW +¯OE) ( x
y
)2
Bo(¯OS +¯ON ) Bw(¯WW +¯WE) ( x
y
)2
Bw(¯WS +¯WN )
158
x
t
Ct 887.53
Jwo well BBo
x y
887.53
Jww well BBw
x y
R42 = 158
x
t
CtPn
42 887.53
JwoPwf well BBo
x y
887.53
Jww well BPwfABw
x y
Well-C at m=83
S83Pn+1
68 + W83Pn+1
82 + C83Pn+1
83 + E83Pn+1
84 + N83Pn+1
98 = R83
S83 = ¯OS(
x
y
)2
Bo + ¯WS(
x
y
)2
Bw
W83 = ¯OW Bo + ¯WW Bw
E83 = ¯OEBo + ¯WEBw
N83 = ¯ON (
x
y
)2
Bo + ¯WN (
x
y
)2
Bw
6
7. C83 = Bo(¯OW +¯OE) (
x
y
)2
Bo(¯OS+¯ON ) Bw(¯WW +¯WE) (
x
y
)2
Bw(¯WS+¯WN )
158
x
t
Ct 887.53
Jwo well CBo
x y
887.53
Jww well CBw
x y
R83 = 158
x
t
CtPn
83 887.53
Jwo well CPwf well CBo
x y
887.53
Jww well CPwf well CBw
x y
Well-D at m=124
S124Pn+1
109 + W124Pn+1
123 + C124Pn+1
124 + E124Pn+1
125 + N124Pn+1
139 = R124
S124 = ¯OS(
x
y
)2
Bo + ¯WS(
x
y
)2
Bw
W124 = ¯OW Bo + ¯WW Bw
E124 = ¯OEBo + ¯WEBw
N124 = ¯ON (
x
y
)2
Bo + ¯WN (
x
y
)2
Bw
C124 = Bo(¯OW +¯OE) (
x
y
)2
Bo(¯OS+¯ON ) Bw(¯WW +¯WE) (
x
y
)2
Bw(¯WS+¯WN )
158
x
t
Ct 887.53
Jwo well DBo
x y
887.53
Jww well DBw
x y
R124 = 158
x
t
CtPn
124 887.53
JwoPwf well DBo
x y
887.53
Jww well DPwf well DBw
x y
Well-E at m=132
S132Pn+1
117 + W132Pn+1
131 + C132Pn+1
132 + E132Pn+1
133 + N132Pn+1
147 = R132
S132 = ¯OS(
x
y
)2
Bo + ¯WS(
x
y
)2
Bw
W132 = ¯OW Bo + ¯WW Bw
E132 = ¯OEBo + ¯WEBw
N132 = ¯ON (
x
y
)2
Bo + ¯WN (
x
y
)2
Bw
7
8. C132 = Bo(¯OW +¯OE) (
x
y
)2
Bo(¯OS+¯ON ) Bw(¯WW
¯WE) (
x
y
)2
Bw(¯WS+¯WN )
158
x
t
Ct 887.53
Jwo well EBo
x y
887.53
Jww well EBw
x y
R132 = 158
x
t
CtPn
132 887.53
Jwo well EPwf well EBo
x y
887.53
Jww well EPwf well EBw
x y
4.5 Well Productivity Index Calculation using Peaceman’s method
Radial reservoir
re = 0.14 ⇤
p
X2 + Y 2
re = 0.14 ⇤
p
1002 + 1002 = 19.799ft
Well productivity index
Jwo = 7.08 ⇤ 10 3 kkroh
µoBo
⇤
1
ln re
rw
+ S
Jww = 7.08 ⇤ 10 3 kkrwh
µwBw
⇤
1
ln re
rw
+ S
For all wells
S = 0
5 Leakproof Test
In this test performed, the well flow rates are specified to be zero for all the wells and
a simulation is run for 100 days. This is done by setting both Jwoil
and Jwwater to zero.
The time step used in this simulation is 10 days. The objective of this test is to ascer-
tain that the PDEs for the simulation cells have been discretized correctly. The pres-
sure, oil saturation and water saturation distributions are to remain at initial reser-
voir values of 3000 psia, 0.75 and 0.25 everywhere in the reservoir at all the simulation
times as shown in the plots below.
8
12. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Reservoir Length,ft
ReservoirWidth,ft
Water saturation distribution at t=0 days
WaterSaturation,v/v
0.2
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Water saturation distribution at t=50 days
Reservoir Length,ft
ReservoirWidth,ft
WaterSaturation,v/v
0.2
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
12
13. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Water saturation distribution at t=100 days
Reservoir Length,ft
ReservoirWidth,ft
WaterSaturation,v/v
0.2
0.21
0.22
0.23
0.24
0.25
0.26
0.27
0.28
0.29
6 Symmetry Test
In this test performed, the bottomhole flowing pressure for the wells are set to 1000
psia thereby configuring all the wells to be producers. A simulation is then run for 100
days. The time step used in this simulation is 10 days. The reservoir pressure distribu-
tion for this test is expected to be symmetric about the center (x = 750ft, y = 550ft)
of the reservoir as shown in the plots below. The pressure results are also published in
the tables in the additional results section.
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Reservoir Length,ft
ReservoirWidth,ft
Pressure distribution at t=0 days
Pressure,psia
2999
2999.2
2999.4
2999.6
2999.8
3000
3000.2
3000.4
3000.6
3000.8
3001
13
14. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure distribution at t=50 days
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1650
1700
1750
1800
1850
1900
1950
2000
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure distribution at t=100 days
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1300
1320
1340
1360
1380
1400
1420
1440
1460
7 Oil Initially In Place Determination and Material Balance
Check
The wells are configured to be producers by setting their bottomhole flowing pressures
Pwf = 1000psia. A simulation is then run for 100 days. The time step used in this
simulation is 10 days. The Oil-Initially-In-Place (OIIP) was calculated to be approx-
imately 1986558 STB. The cumulative production of the reservoir along with the oil
in place is calculated for each time step as well as the error between Oil-Initially-In-
Place and the sum of cumulative oil production and Oil-In-Place at the end of each
14
15. time step. It is expected that there should be oil mass conservation hence the error is
expected to be within 0.1%.
The oil initially in place was determined by the following equation with initial reservoir
and PVT parameters at Pi = 3000psia:
OIIP =
1
5.615
⌃
NxNy
m=1 [
x y z (Pi)So,i
Bo(Pi)
]m
Nx = 15
Ny = 11
The Oil In Place at the end of each time step was calculated using the following equa-
tion:
OIPn+1
=
1
5.615
⌃
NxNy
m=1 [
x y z n+1
Sn+1
o
Bn+1
o
]m
The cumulative oil production is calculated as follows
V n+1
o,prod = V n
o,prod + qn+1
o t = ⌃T
n=1qn+1
o t
The error from oil mass conservation is determined by the following equation:
Error =
|OIIP (V n+1
o,prod + OIPn+1
)|
OIIP
⇤ 100
Material Balance Check
Time (Days) Cumulative Pro-
duction (STB)
Oil In Place
(STB)
Error (%)
10 7557 1978998 0.000154
20 13785 1972767 0.000282
30 19097 1967453 0.000390
40 23644 1962904 0.000483
50 27541 1959005 0.000563
60 30886 1955664 0.000631
70 33746 1952798 0.000689
80 36202 1950341 0.000739
90 38308 1948234 0.000783
100 40115 1946426 0.000819
15
16. 8 Primary Recovery Study
In this study recovery of hydrocarbons from the reservoir is simulated without water
injection. The wells are configured to be producers with their flowing pressures set to
1000 psia. The simulation is run until the oil production rate from Well-A falls below 1
STB/D. This occurs after 313 days of producing the reservoir. A time step of 1 day is
used in the simulation.
From the simulation results obtained, all the wells have an initial oil production rate of
197 STB/D and an initial water production rate of 0.021 STB/D however, after about
11 days the the oil and water production rate of all the well C slightly decreases while
the other wells continue to have equivalent oil and water production rates. This trend
continues until t = 171days, after which all the wells have equivalent oil and water
production rates again.
0 50 100 150 200 250 300 350
0
20
40
60
80
100
120
140
160
180
200
Time, days
Oilproductionrate,STB/D
Oil production rate vs time
Oil producion rate for Well A
Oil producion rate for Well B
Oil producion rate for Well C
Oil producion rate for Well D
Oil producion rate for Well E
16
17. 0 50 100 150 200 250 300 350
0
0.005
0.01
0.015
0.02
0.025
Time, days
Waterproductionrate,STB/D
Water production rate vs time
Water producion rate for Well A
Water producion rate for Well B
Water producion rate for Well C
Water producion rate for Well D
Water producion rate for Well E
The cumulative oil production rate of the reservoir after 313 days of production is 50684
STB as show in the plot below. The recovery factor is 2.5% which is significantly low.
Such a scenario recommends the need for pressure maintenance.
0 50 100 150 200 250 300 350
0
1
2
3
4
5
6
x 10
4
Cumulativeproduction,STB/D
Time, days
Cumulative Production vs time
The reservoir pressure at the wells decreases from 3000 psia to 1260 psia while the pres-
sure at the boundaries drops to 1420 psia after 100 days of production. At the end of
production, the pressure drops to 1008 psia at the wells while it drops to 1013 psia at
the boundaries at the end of production. The pressure decline in the first t=100 days
is the fastest decline as compared to other 100 day intervals. However, the decreasing
pressure drop in the reservoir as time progresses is expected and explains the decreas-
ing oil and water production rates as time progresses.
17
18. 0 50 100 150 200 250 300 350
0
0.5
1
1.5
2
2.5
3
Recoveryfactor,%
Time, days
Recovery factor vs time
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Reservoir Length,ft
ReservoirWidth,ft
Pressure distribution at t=0 days
Pressure,psia
2999
2999.2
2999.4
2999.6
2999.8
3000
3000.2
3000.4
3000.6
3000.8
3001
18
20. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure distribution at t=300 days
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1010
1011
1012
1013
1014
1015
1016
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure distribution at end of production
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1008
1008.5
1009
1009.5
1010
1010.5
1011
1011.5
1012
1012.5
1013
The simulation results suggest that the reservoir cannot produce for long under its pri-
mary recovery mechanism alone.
20
21. 9 Secondary Recovery Study
In this study recovery of hydrocarbons from the reservoir is simulated with water injec-
tion for pressure maintenance. Initially, the wells are configured to be producers with
their flowing pressures set to 1000 psia. However after the oil production rate from
Well-A falls below 1 STB/D, Well-C is converted to an injector with a flowing pres-
sure of 2950 psia. The simulation is run until 10000 days. A time step of 1 day is used
in the simulation.
Water injection is started at t = 313days. It is observed from the oil saturation distri-
bution below that there is still a significant amount of hydrocarbons in the reservoir.
However, the due to a low pressure drop as shown in the pressure distribution plot, the
reservoir cannot produce at a prolific rate.
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Oil saturation distribution at start of injection
Reservoir Length,ft
ReservoirWidth,ft
OilSaturation,v/v
0.7485
0.7485
0.7485
0.7485
21
22. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure distribution at start of injection
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1008
1008.5
1009
1009.5
1010
1010.5
1011
1011.5
1012
1012.5
1013
At the end of the simulation (t = 10000days), the oil saturation distribution plot shows
that a significant amount of the reservoir oil has been produced with most of the re-
maining oil in place located at the corners or the reservoir. The pressure distribution
shows a large pressure drop which is able to support the reservoir to produce at a pro-
lific rate. The average pressures at the producing well locations are 1700 psia while the
average pressure at the injection well location is 2400 psia.
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Oil saturation distribution at t=10000 days
Reservoir Length,ft
ReservoirWidth,ft
OilSaturation,v/v
0.1
0.2
0.3
0.4
0.5
0.6
0.7
22
23. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure distribution at t=10000 days
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1700
1800
1900
2000
2100
2200
2300
2400
The cumulative oil production at the end of 10000days is 969687 STB. The recovery
factor is 48.81% which is significantly higher compared to when the reservoir was only
under primary recovery.
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
0
1
2
3
4
5
6
7
8
9
10
x 10
5
Cumulativeproduction,STB/D
Time, days
Cumulative Production vs time
23
24. 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
0
5
10
15
20
25
30
35
40
45
50
Recoveryfactor,%
Time, days
Recovery factor vs time
Also the time for which watercut increased above 20% at Well-A was determined to
be at t=3094 days.The oil saturation distribution shows that a significant amount of
oil remained in the reservoir so continued production with an increasing water cut is
justified. At the end of the simulation the water cut was approximately 90%.
200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Oil saturation distribution at Water Breakthrough Time
Reservoir Length,ft
ReservoirWidth,ft
OilSaturation,v/v
0.1
0.2
0.3
0.4
0.5
0.6
0.7
24
25. 200 400 600 800 1000 1200 1400
100
200
300
400
500
600
700
800
900
1000
Pressure Distribution at Water Breakthrough Time
Reservoir Length,ft
ReservoirWidth,ft
Pressure,psia
1800
1900
2000
2100
2200
2300
2400
2500
The simulation results suggest that the reservoir produces for much longer when there
is pressure maintenance to support production.
10 Conclusion
Water Injection aids in realizing signifcantly higher recovery factors and slower oil pro-
duction rates as time progresses. The simulation studies show that under primary re-
covery alone, the recovery factor was 2.5% as compared to the 60% when there was wa-
ter injection
However, with water injection as serving as pressure maintenance, there is a risk of
higher water cut as time progresses. In the primary recovery study, the water cut was
almost negligible while during the secondary recovery study, the water cut kept increas-
ing after water injection was initiated.
25