Tarek Ahmed. Reservoir engineering handbook
Подождите немного. Документ загружается.
Notice that the parameter E is constant, whereas the parameter is
decreasing with continuous water injection. Craig et al. (1955) expressed
the producing water–oil ratio as:
where WOR
s
= surface water–oil ratio, STB/STB
W
iBT
= cumulative water injected at breakthrough, bbl
W
inj
= cumulative water injected at any time after
breakthrough, bbl
f
w2
= water cut at the producing well, bbl/bbl
Note that when the areal sweep efficiency EA reaches 100%, the incremen-
tal oil produced from the newly swept areal is zero, i.e., (N
P
)
newly
= 0,
which reduces the above expression to Equation 14-28:
The recommended methodology for predicting the recovery performance
after breakthrough is summarized in the following steps:
Step 1. Select several values of W
inj
> W
iBT
.
Step 2. Assuming constant injection rate i
w
, calculate the time t required
to inject W
inj
barrels of water.
Step 3. Calculate the ratio W
inj
/W
iBT
for each selected W
inj
.
Step 4. Calculate the areal sweep efficiency E
A
at each selected W
inj
by
applying Equation 14-65 or 14-66:
WOR
f
f
B
B
B
B
f
s
w
w
o
w
o
w
w
=
−
=
−
2
2
2
1
1
1
WOR
fN
fN
B
B
s
wp
wp
o
w
=
−
(
)
−−
(
)
(
)
2
2
1
11
∆
∆
newly
newly
14 - 70
948 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 948
Step 5. Calculate the ratio Q
i
/Q
iBT
that corresponds to each W
inj
/W
iBT
from Table 14-1. The ratio Q
i
/Q
iBT
is a function of E
ABT
and
W
inj
/W
iBT
.
Step 6. Determine the total pore volumes of water injected by multiply-
ing each ratio of Q
i
/Q
iBT
(obtained in step 5) by Q
iBT
, or:
Step 7. From the definition of Q
i
, as expressed by Equation 14-46,
determine the slope (df
w
/dS
w
)
S
w2
for each value of Q
i
by:
Step 8. Read the value of S
w2
, i.e., water saturation at the producing
well, that corresponds to each slope from the plot of
(df
w
/dS
w
)
S
w2
vs. S
w2
(see phase 1, step 7).
Step 9. Calculate the reservoir water cut at the producing well f
w2
for
each S
w2
from Equation 14-24 or 14-37.
or
f
ae
w
w
o
bS
w
2
1
1
2
=
+
µ
µ
f
k
k
w
w
o
ro
rw
2
1
1
=
+
µ
µ
df
dS Q
w
w
S
i
w
=
2
1
Q
Q
Q
Q
i
i
iBT
iBT
=
EE
W
W
E
W
W
A ABT
inj
iBT
ABT
inj
iBT
=+
=+
0 633 0 2749. log . ln
Principles of Waterflooding 949
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 949
Step 10. Determine the average water saturation in the swept area by
applying Equation 14-45:
Step 11. Calculate the displacement efficiency E
D
for each :
Step 12. Calculate cumulative oil production from Equation 14-6:
For 100% vertical sweep efficiency:
Step 13. Calculate cumulative water production from Equation 14-53 or
14-54:
Step 14. Calculate the surface water–oil ratio WOR
s
that corresponds to
each value of f
w2
from Equation 14-70:
W
WNB
B
W
WSSPVE
B
p
inj p o
w
p
inj w wi A
w
=
−
=
−−
(
)
(
)
2
NNEE
pSDA
=
NNEEE
pSDAV
=
E
SS
S
D
wwi
wi
=
−
−
2
1
S
w2
SS
f
df
dS
ww
w
w
w
S
w
22
2
1
2
=+
−
S
w2
950 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 950
Step 15. Calculate the oil and water flow rates from Equations 14-55 and
14-50, respectively:
Steps 1 through 15 could be conveniently performed in the following
worksheet form:
W
inj
E
A
Q
i
S
w2
f
w2
E
D
N
P
W
P
WOR
S
Q
o
Q
w
W
iBT tBT
1.0 E
ABT
1.0 Q
iBT —
S
wBT
f
wBT
E
DBT
—— — ——
•••••• ••••••••••
•••••• ••••••••••
•••••• ••••••••••
Example 14-11
1
An oil reservoir is under consideration for waterflooding. The relative
permeability data and the corresponding water cut are given below:
S
w
0.100 0.300 0.400 0.450 0.500 0.550 0.600 0.650 0.700
k
ro
1.000 0.373 0.210 0.148 0.100 0.061 0.033 0.012 0.000
k
rw
0.000 0.070 0.169 0.226 0.300 0.376 0.476 0.600 0.740
f
w
0.000 0.2729 0.6168 0.7533 0.8571 0.9250 0.9665 0.9901 1.0000
S
w2
df
dS
w
w
Sw
2
Q
Q
i
iBT
W
W
inj
iBT
t
W
i
inj
w
==
Q
i
B B WOR
Q Q WOR
o
w
ow s
wo s
=
+
=
WOR
fN
fN
B
B
s
wP
wP
o
w
=
−
(
)
(
)
−−
(
)
(
)
[]
2
2
1
11
∆
∆
newly
newly
Principles of Waterflooding 951
1
From The Reservoir Engineering Aspects of Waterflooding, Craig, Dallas: Society of Petroleum Engineers, 1971,
p. 116.
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 951
Reservoir properties are as follows:
Flood area, acres = 40
Thickness, ft = 5
Average permeability, md = 31.5
Porosity, % = 20
Initial water saturation, % = 10
Connate water saturation, % = 10
Current gas saturation, % = 0
Water viscosity, cp = 0.5
Oil viscosity, cp = 1.0
Reservoir pressure, psi = 1000
Constant B
o
, bbl/STB = 1.20
Flood pattern = 5 spot
Wellbore radius, ft = 1.0
Predict the recovery performance under a constant water injection rate of
269 bbl/day.
Solution
Phase 1. Initial Calculations
Step 1. Calculate pore volume and oil volume at start of flood:
Step 2. Plot f
w
vs. S
w
on a Cartesian scale, as shown in Figure 14-37, and
determine:
S
wf
= S
wBT
= 0.469 Q
iBT
= = 0.463
f
wf
= f
wBT
= 0.798 = 0.563
(df
w
/dS
w
)
Swf
= 2.16
S
wBT
1
216.
PV bbl
N STB
S
(
)
=
(
)
(
)
(
)
=
=−
(
)
=
7758 40 5 0 20 310 320
310 320 1 0 1 1 20 232 740
.
,./.
,
,
952 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 952
Figure 14-37. The f
w
curve for Example 14-11. (Permission to publish by the Soci-
ety of Petroleum Engineers.)
Step 3. Determine k
ro
and k
rw
at S
wi
and from the relative permeabil-
ity data, to give:
Step 4. Calculate the mobility ratio M from Equation 14-61:
Step 5. Calculate the areal sweep efficiency at breakthrough from Equa-
tion 14-64 or Figure 14-35:
E
ABT
= 0 71702.
M ==
04
10
10
05
08
.
.
.
.
.
k
k
ro
rw
@. .
@. .
01 100
0 563 0 40
=
=
S
wBT
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
10 20 30 40 50 60 70
f
w
S
wf
= S
wBT
= 0.469
(df
w
/S
w
)swf = 2.16
f
wf
= f
wBT
= 0.798
S
wBT
= 0.563
S
w
,
WATER SATURATION, % PV
Principles of Waterflooding 953
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 953
Step 6. Select several values of S
w2
between 0.469 and 0.700 and deter-
mine the slope, graphically or numerically, at each selected satu-
ration:
S
w2
f
w2
df
w
/dS
w
0.469 0.798 2.16
0.495 0.848 1.75
0.520 0.888 1.41
0.546 0.920 1.13
0.572 0.946 0.851
0.597 0.965 0.649
0.622 0.980 0.477
0.649 0.990 0.317
0.674 0.996 0.195
0.700 1.000 0.102
Step 7. Plot df
w
/dS
w
vs. S
w2
as shown in Figure 14-38.
Figure 14-38. Derivative curve for Example 14-11. (Permission to publish by the
Society of Petroleum Engineers.)
2.5
2.0
1.5
1.0
0.5
0
40 50 60 70
S
w
,
WATER SATURATION, % PV
df
w
dS
w
954 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 954
Phase 2. Calculation of Recovery Performance to Breakthrough
Step 1. Calculate Q
iBT
using Equation 14-41:
Step 2. Calculate cumulative water injected at breakthrough from Equa-
tion 14-43 or 14-44:
Step 3. Calculate time to breakthrough:
Step 4. Calculate the displacement efficiency at breakthrough E
DAB
from
Equation 14-50:
Step 5. Calculate cumulative oil production at breakthrough by using
Equation 14-51.
Notice that when S
gi
= 0, the cumulative water injected at break-
through W
iBT
will displace an equivalent volume of oil, i.e.:
N
W
B
STB
p
BT
iBT
o
(
)
== =
103 020
12
85 850
,
.
,
NNEE
N STB
p
BT
S DBT ABT
p
BT
(
)
=
(
)
=
(
)
(
)
(
)
=232 740 0 5144 0 717 85 850,. . ,
E
SS
S
E
DBT
wBT wi
wi
DBT
=
−
−
=
−
−
=
1
0 563 0 1
101
0 5144
..
.
.
tWi
t days
BT iBT w
BT
=
==103 020 269 383,/
WPVQE
W bbl
iBT iBT ABT
iBT
=
(
)
=
(
)
(
)
(
)
=310 320 0 463 0 71702 103 020,.. ,
QSS
Q
iBT wBT wi
iBT
=−
(
)
=−
(
)
=0 563 0 1 0 463...
Principles of Waterflooding 955
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 955
Step 6. Calculate the surface water cut WOR
s
exactly at breakthrough
from Equation 14-70:
Step 7. Set up the following table to describe the oil recovery perfor-
mance to breakthrough (remember, S
gi
= 0):
W
inj
WOR
s
Q
w
=Q
o
WOR
s
Wp
bbl days STB STB STB/STB STB/day STB
0000000
20,000 74.34 16,667 224 0 0 0
40,000 148.7 33,333 224 0 0 0
60,000 223.0 50,000 224 0 0 0
80,000 297.4 66,667 224 0 0 0
103,020 383.0 85,850 224 1.49 334 0
Phase 3. Oil Recovery Calculations After Breakthrough
A step-by-step description of the oil recovery calcuations as well as a
convienent worksheet to perform the computations after breakthrough
are given below:
Column 1: Select several values of W
inj
.
Column 2: For a constant injection rate, calculate the time t required to
W
inj
barrels of water.
Column 3: Divide values of W
inj
in column 1 by W
iBT
.
Q
i
B
o
w
o
==
N
W
B
P
inj
o
==t
W
i
inj
w
==
E
SS
ES S
W
W
NE
WOR
fN
wf wi
ABT wBT wi
iBT
inj
P
s
wf P
=
−
−
=
−
−
=
== =
== =
=
−
(
)
(
)
(
)
(
)
(
)
(
)
[]
0 469 0 1
0 717 0 563 0 1
1 1115
0 2749 0 2749
103 020
103 020
0 2749
1 1115 0 2749 0 30555
1
..
.. .
.
..
,
,
.
.. .
λ
λ∆
∆
newly
newly
111
0 798 1 0 30555
1 0 798 1 0 30555
12
1
149
−−
=
−
−−
=
(
)
[]
[]
[]
fN
B
B
STB STB
wf P
o
w
∆
newly
..
..
.
./
956 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:37 Page 956
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16)
W
inj
E
A
S
W2
f
W2
E
D
N
p
W
p
WOR
S
Q
o
Q
w
(Assumed) (days) Eq. Table Fig. Fig. Eq. Eq. Eq. Eq. Eq. Eq. Eq.
14-65 14-1 14-38 14-37 14-45 14-10 14-6 14-53 14-70 14-55 14-56
103,020 383 1.0 0.717 1.000 0.463 2.159 0.470 0.800 0.563 0.514 85,850 0 1.49 99.6 149.4
123,620 460 1.2 0.767 1.193 0.552 1.810 0.492 0.843 0.579 0.532 100,292 3,270 2.03 83.3 169.1
144,230 536 1.4 0.809 1.375 0.636 1.570 0.507 0.870 0.590 0.544 106,986 15,847 2.55 71.73 182.9
164,830 613 1.6 0.846 1.548 0.717 1.394 0.524 0.893 0.601 0.557 113,820 28,246 3.12 62.3 194.4
185,440 689 1.8 0.879 1.715 0.794 1.259 0.534 0.905 0.610 0.567 119,559 41,969 3.63 55.7 202.2
206,040 766 2.0 0.906 1.875 0.869 1.151 0.543 0.920 0.163 0.570 128,417 51,940 4.24 49.4 209.5
257,550 958 2.5 0.969 2.256 1.046 0.956 0.562 0.937 0.628 0.587 136,618 93,608 5.56 39.8 221.3
309,060 1,149 3.0 1.000 2.619 1.214 0.823 0.575 0.944 0.637 0.597 138,946 142,325 22.33 11 255
412,080 1,532 4.0 1.000 3.336 1.545 0.647 0.597 0.963 0.653 0.614 142,902 240,598 31.23 8 259
515,100 1,915 5.0 1.000 4.053 1.877 0.533 0.611 0.973 0.660 0.622 144,764 341,383 43.24 6 262
618,120 2,298 6.0 1.000 4.770 2.208 0.453 0.622 0.980 0.664 0.627 145,928 443,006 58.8 4 264
824,160 3,064 8 1.000 6.204 2.872 0.348 0.637 0.985 0.676 0.640 148,954 645,415 78.8 3 265
1,030,200 3,830 10 1.000 7.638 3.536 0.283 0.650 0.990 0.683 0.648 150,816 849,221 119 2 267
1,545,300 5,745 15 1.000 11.223 5.199 0.192 0.677 0.995 0.697 0.663 154,307 1,360,132 239 1 268
S
w2
df
dS Q
w
wi
==
1
Q
Q
Q
i
iBT
iBT
Q
Q
i
iBT
W
W
inj
iBT
t
W
i
inj
w
==
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 957