Tarek Ahmed. Reservoir engineering handbook
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Step 2. Calculate the areal sweep efficiency at fill-up from Equation 14-63:
at fill-up:
Step 3. Given a mobility ratio M of 0.8 (Example 14-11) and E
A
of
0.324, calculate the conductance ratio at the fill-up from Figure
14-42: .
Step 4. Calculate the initial (base) injection rate from Equation 14-71:
Step 5. Calculate the water injection rate at fill-up i
wf
from Equation 14-74:
Step 6. Calculate the average water injection rate from interference to
fill-up:
Step 7. Calculate the incremented time occurring from interference to
fill-up:
Thus the time to fill-up t
f
is:
Stage 3: Fill-up—Water Breakthrough
The time to fill-up, as represented by point C on Figures 14-44 and
14-15, marks the following four events:
t days
f
=+=80 2 29 5 109 7.. .
∆t
WW
i
days
if ii
w
avg
=
−
(
)
=
−
=
46 550 36 572
338 55
29 5
,,
.
.
i
ii
bbl day
w
avg
wi wf
(
)
=
+
=
+
=
2
418 9 258 2
2
338 55
..
./
i i bbl day
wf base
==
(
)
(
)
=γ 0 96 269 1 258 2.../
i
base
=
(
)
(
)
(
)
(
)
(
)
−
=
0 003541 5 31 5 1 3 000
10
932
1
0 619
269 1
..,
.ln .
. bbl/ day
γ=096.
E
W
A
if
==
143 678
46 550
143 678
0 324
,
,
,
.
E
W
PV S S
WW
A
inj
wBT wi
inj inj
=
(
)
−
(
)
=
−
(
)
=
310 320 0 563 0 10 143 678,. . ,
978 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 978
1. No free gas remaining in the flood pattern
2. Arrival of the oil-bank front to the production well
3. Flood pattern response to the waterflooding
4. Oil flow rate Q
o
equal to the water injection rate i
w
During this stage, the oil production rate is essentially equal to the injec-
tion due to the fact that no free gas exists in the swept flood area. With
continuous water injection, the leading edge of the water bank eventually
reaches the production well, as shown in Figure 14-44, and marks the
time to breakthrough. At breakthrough the water production rises rapidly.
The waterflood performance calculations are given by the following
steps:
Step 1. Calculate cumulative water injected at breakthrough by using
Equation 14-43 or 14-44 :
Step 2. Assume several values of cumulative water injected W
inj
between
W
if
and W
iBT
and calculate the areal sweep efficiency at each
Winj from Equation 14-63:
Step 3. Determine the conductance ratio γ for each assumed value of
W
inj
from Figure 14-42.
Step 4. Calculate the water injection rate at each W
inj
by applying Equa-
tion 14-74:
Step 5. Calculate the oil flow rate Q
o
during this stage from:
Step 6. Calculate cumulating oil production N
P
from the following
expression:
N
WW
B
P
inj if
o
=
−
(
)
14 - 85
Q
i
B
o
w
o
=
(
)
14 - 84
ii
w base
=γ
E
W
PV S S
A
inj
wBT wi
=
(
)
−
(
)
WPVSSE PVQE
iBT wBT wi ABT iBT ABT
=
(
)
−
(
)
=
(
)
(
)
Principles of Waterflooding 979
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 979
Example 14-15
Using the data given in Example 14-14, calculate the flood perfor-
mance from the fill-up to breakthrough. Results of Example 14-14 show:
• Cumulative water injected to fill-up W
if
= 46,550 bbl
• Water injection rate at fill-up i
wf
= 358.2 bbl/day
• Time to fill-up t
f
= 109.7 days
Solution
Step 1. Calculate cumulative water injected at breakthrough from Equa-
tion 14-43:
Step 2. Perform the required computations in the following tabulated form:
(1)
(6)
(9)
W
inj
(2) (3) (4) (5) (7) (8) N
p
=
(Assume) E
A
i
w
(i
w
)
avg
t = tQ
o
= i
w
/B
o
(W
inj
– W
if
)/B
o
46,550 0.324 0.96 258.6 109.7 215.5 0
50,000 0.348 0.95 255.6 257.1 13.27 123.0 213.0 2,844
60,000 0.418 0.94 253.0 254.3 39.32 162.3 210.8 11,177
70,000 0.487 0.94 253.0 253.0 39.53 201.8 210.8 19,511
80,000 0.557 0.93 251.7 251.7 39.73 241.6 208.6 27,844
90,000 0.626 0.92 247.6 249.0 40.16 281.7 206.3 36,177
100,000 0.696 0.92 247.6 247.6 40.39 322.1 206.3 44,511
103,020 0.717 0.91 244.9 246.3 12.26 334.4 204.1 47,027
The above calculations indicate that the time to breakthrough will occur
after 334.4 days from the start of flood with cumulative oil produced of
47,027 STB.
Stage 4: Water Breakthrough—End of the Project
After breakthrough, the water–oil ratio increases rapidly with a notice-
able decline in the oil flow rate as shown in Figure 14-45 by point D. The
swept area will continue to increase as additional water is injected. The
incrementally swept area will contribute additional oil production, while
the previously swept area will continue to produce both oil and water.
∆∆
∆∆
t
W
i
inj
w
avg
==
(())
W bbl
iBT
=−
(
)
(
)
=310 320 0 563 0 1 0 717 103 020,. .. ,
980 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 980
As represented by Equation 14-70, the WOR is calculated on the basis
of the amounts of oil and water flowing from the swept region and the oil
displaced from the newly swept portion of the pattern. It is assumed the
oil from the newly swept area is displaced by the water saturation just
behind the stabilized zone, i.e., S
wf
.
The calculations during the fourth stage of the waterflooding process
are given below:
Step 1. Assume several values for the ratio W
inj
/W
iBT
that correspond to
the values given in Table 14-1, i.e., 1, 1.2, 1.4, etc.
Step 2. Calculate the cumulative water injected for each assumed ratio of
(W
inj
/W
iBT
) from:
Step 3. Calculate the areal sweep efficiency at each assumed (W
inj
/W
iBT
)
from Equation 14-65:
Step 4. Calculate the ratio (Q
i
/Q
iBT
) that corresponds to each value of
(W
inj
/W
iBT
) from Table 14-1 or Equation 14-69.
Step 5. Determine the total pore volumes of water injected by multiply-
ing each ratio of Q
i
/Q
iBT
by Q
iBT
, or:
Step 6. From the definition of Q
i
, as expressed by Equation 14-46, deter-
mine the slope (df
w
/dS
w
)
Sw2
for each value of Q
i
by:
Step 7. 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 Example 14-11).
df
dS Q
w
w
Sw
i
=
2
1
Q
Q
Q
Q
i
i
iBT
iBT
=
EE
W
W
A ABT
inj
iBT
=+
0 633. log
W
W
W
W
inj
inj
iBT
iBT
=
Principles of Waterflooding 981
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 981
Step 8. Calculate the reservoir water cut at the producing well f
w2
for
each S
w2
from Equation 14-24 or 14-37:
or
Step 9. Determine the average water saturation in the swept area by
applying Equation 14-45:
Step 10. Calculate the surface water–oil ratio WOR
s
that corresponds to
each value of f
w2
by applying Equation 14-70:
Step 11. Craig, Geffen, and Morse (1955) point out when calculating
cumulative oil production during this stage that one must account
for the oil lost to the unswept area of the flood pattern. To account
for the lost oil, the authors proposed the following expression:
where E
D
is the displacement efficiency and is given by Equa-
tion 14-9 as:
E
SS S
SS
D
wwigi
wi gi
=
−−
−−1
NNEE
PV E S
B
PSDA
Agi
o
=−
(
)
−
(
)
(
)
1
14 - 86
WOR
fN
fN
B
B
s
wP
newly
wP
newly
o
w
=
−
(
)
[]
−−
(
)
[]
2
2
1
11
∆
∆
SS
f
df
dS
ww
w
w
w
S
w
22
2
1
2
=+
−
S
w2
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
982 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 982
Step 12. Calculate cumulative water from the expression:
Water produced = Water injected – Oil produced
– Fill-up volume
or
Step 13. Calculate k
rw
at and determine the mobility ratio M after
breakthrough from Equation 14-62:
Step 14. Calculate the conductance ratio γ from Figure 14-42.
Step 15. Determine the water injection rate from Equation 14-74
.
Step 16. Calculate the oil and water production rates from Equations 14-55
and 14-56, respectively:
Example 14-16
Complete the waterflooding performance calculation for Example 14-11 by
predicting the performance of a producing WOR of 50 STB/STB, given:
W
iBT
= 103,020 bbl
(N
p
)
BT
= 47,027 bbl
t
BT
= 334.4 days
E
ABT
= 0.717
S
wf
= 0.469
= 0.563
S
wi
= 0.10
S
wBT
Q
i
B B WOR
Q Q WOR
o
w
ow s
wo s
=
+
=
ii
w bas
e
=γ
M
kS
kS
rw w
ro wi
o
w
=
@
@
2
µ
µ
S
w2
W
wNBPVS
B
P
inj P o gi
w
=
−−
(
)
Principles of Waterflooding 983
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 983
Solution
The required calculations are conveniently performed in the following
worksheet:
(1) (2) (3) (4) (5) (6) (7) (8) (9)
W
inj
=E
A
Q
i
/Q
iBT
Q
i
=S
w2
f
w2
W
inj
/W
iBT
(1) Eq. Table (4) df
w
/dS
w
Fig. Fig. Eq.
(assume) W
iBT
14-65 14-1 Q
iBT
= 1/Q
i
14-38 14-37 14-45
1.0 103,020 0.717 1.000 0.463 2.159 0.470 0.800 0.563
1.2 123,620 0.767 1.193 0.552 1.810 0.492 0.843 0.579
1.4 144,230 0.809 1.375 0.636 1.570 0.507 0.870 0.590
1.6 164,830 0.46 1.548 0.717 1.394 0.524 0.893 0.601
1.8 185,440 0.879 1.715 0.794 1.259 0.534 0.905 0.610
2.0 206,040 0.906 1.875 0.869 1.151 0.543 0.920 0.613
2.5 257,550 0.969 2.256 1.046 0.956 0.562 0.937 0.628
3.0 309,060 1.000 2.619 1.214 0.823 0.575 0.949 0.637
4.0 412,080 1.000 3.336 1.545 0.647 0.597 0.963 0.653
5.0 515,100 1.000 4.053 1.877 0.533 0.611 0.973 0.660
6.0 618,120 1.000 4.770 2.208 0.453 0.622 0.980 0.664
(10) (11) (12) (13) (14) (15) (16) (17) (18)
(N
p
)
newly
WOR
S
E
D
N
p
W
p
M i
w
Eq. Eq. Eq. Eq. Eq. Eq. Fig. Eq.
14-69 14-70 14-9 14-86 14-87 14-62 14-42 14-76
0.3056 1.5* 0.4173 47,027 0 0.400 0.800 0.91 244.9
0.2545 2.03 0.4387 56,223 9,604 0.430 0.860 0.94 252.9
0.2182 2.55 0.4533 63,716 21,223 0.450 0.900 0.96 258.3
0.1910 3.12 0.4680 70,816 33,303 0.480 0.960 0.98 263.7
0.1697 3.63 0.480 77,138 46,326 0.500 1.000 1.0 269.1
0.1528 4.24 0.484 81,400 61,812 0.510 1.020 1.02 274.5
0.1223 5.56 0.504 93,518 98,780 0.542 1.084 1.08 287.9
0.000 22.3
†
0.516 100,078 142,418 0.560 1.120 1.12 301.4
0.0000 31.2 0.5373 104,209 240,481 0.600 1.200 1.20 322.9
0.0000 43.2 0.5467 106,032 341,314 0.625 1.250 1.25 336.4
0.0000 58.8 0.5520 107,060 443,100 0.635 1.270 1.27 341.8
kS
rw w
@
2
S
w2
984 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 984
(19) (20) (21) (22) (23) (24)
t = t = (t) Q
o
Q
w
=
(i
w
)
avg
W
inj
(20)(19) days Eq. 14-55 (11)(22)
334.4 90.7 136.1
248.9 20,600 82.7 417.1 78.6 159.6
255.6 20,610 80.6 497.7 69.6 177.5
261.0 20,600 79.0 576.5 61.5 191.9
266.4 20,610 77.3 663.8 56.5 205.1
271.8 20,600 75.9 729.7 51.5 218.4
282.2 51,510 183.0 912.7 43.1 239.6
295.6 103,020 349.0 1261.7 12.8 285.4
312.7 103,020 330.0 1591.5 9.9 308.9
330.6 103,020 312.0 1903.5 7.6 328.3
339.1 103,020 305.0 2208.3 5.7 335.2
*Equation 14-70.
†
Equation 14-20.
To illustrate the use of Equation 14-70 in calculating the WOR
s
values
of column 11, the value of the surface water–oil ratio when W
inj
/W
iBT
reaches 2 bbl/bbl is calculated below:
Step 1. Calculate the coefficient E, which remains constant for all the
values of W
inj
/W
iBT
:
Step 2. Calculate the parameter λ:
Step 3. Calculate the incremental oil produced from the newly swept
area when (W
inj
/W
iBT
) = 2 from Equation 14-69:
Step 4. Calculate WOR
s
from Equation 14-70:
WOR STB STB
s
=
−
(
)
−−
(
)
=
0 920 1 0 1528
1 0 920 1 0 1528
120
100
424
..
..
.
.
./
∆N E bbl bbl
p
newly
(
)
==
(
)
(
)
=λ 1 1115 0 13745 0 1528.. . /
λ=
=
=0 2749 0 2749
1
2
0 13745...
W
W
iBT
inj
E
SS
ES S
wf wi
ABT wBT wi
=
−
−
(
)
=
−
−
(
)
=
0 469 0 1
0 717 0 563 0 1
1 1115
..
.. .
.
Principles of Waterflooding 985
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 985
Figure 14-47 documents results of Examples 14-15 and 14-16 graphically.
Figure 14-47. Performance curves for Example 14-16.
Water Fingering and Tonguing
In thick, dipping formations containing heavy viscous oil, water tends
to advance as a “tongue” at the bottom of the pay zone. Similarly, dis-
placement of oil with a gas will result in the gas attempting to overrun
the oil due to gravity differences unless stopped by a shale barrier within
the formation or by a low overall effective vertical permeability. In lin-
ear laboratory experiments, it was observed that the fluid interface
remains horizontal and independent of fluid velocity when the viscosities
of the two phases are equal. If the oil and water have different viscosi-
ties, the original horizontal interface will become tilted.
In a dipping reservoir, Dake (1978) developed a gravity segregation
model that allows the calculation of the critical injection rate I
crit
that is
required to propagate a stable displacement. The condition for stable dis-
placement is that the angle between the fluid interface and the direction
of flow should remain constant throughout the displacement as shown in
Figure 14-48. Dake introduced the two parameters, the Dimensionless
Gravity Number “G” and the End-point Mobility Ratio M*, that can be
used to define the stability of displacement. These two parameters are
defined by the following relationships:
986 Reservoir Engineering Handbook
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 986
Figure 14-48. Stable and unstable displacement in gravity segregated displace-
ment: (a) stable: G > m – 1, M > 1, and < ; (b) stable: G > M – 1, M < 1, > ;
and (c) unstable: G < M – 1. (Courtesy of Elsevier.)
1. Dimensionless gravity number. The dimensionless gravity number
G is given by:
where k = absolute permeability, md
k
rw
= relative permeability to water as evaluated at S
or
A = cross-sectional area
w
= water density, lb/ft
3
= dip angle
2. End-point mobility ratio. The end-point mobility ratio M* is defined by:
M
kS
kS
rw or
ro wi
o
o
*
@
@
=
(
)
µ
µ
14 - 88
G
kk A
i
rw w o
ww
=
×−
(
)
(
)
(
)
−
7 853 10
6
. ρρ θ
µ
sin
14 - 87
Principles of Waterflooding 987
Reservoir Eng Hndbk Ch 14 2001-10-25 17:38 Page 987