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Prediction of bottom water drive reservoir performance. Prediction of bottom water drive reservoir performance. Determine We from MBE Calculate the values of Sum dp Qt from Unsteady state water drive - PowerPoint PPT Presentation
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Prediction of bottom water drive reservoir performance
Prediction of bottom water drive reservoir performance
1. Determine We from MBE
2. Calculate the values of Sum dp Qt from Unsteady state water drive
3. We can obtained the values of B from the equating We from MBE with B Sum dp Qt
4. The best value of B obtained from drawing the relation between values of B versus Values of Np or time
1. OR
OR B From MBE straight line
FBE
NBNBBwBNPQB
PQBBwBNNBNB
BB
WBWBNN
titwptpt
twptptit
itt
ewptp
This equation of straight line relation between F and E the slope is B
5. Assume three values of Pressure P1,P2 & P3,and Select the values of Production Np ,Wb,..on the basis of the past production history
6. From PVT data Determine Bo,Bt,…..7. Calculate the values of We from MBE for each
pressure8. Plotting pressure Vs water influx calculated by
both MBE and USSE obtained for the three assumed pressures will show the pressure which will result in identical water influx calculation by both method.
9. The intersection of the two We MBE and We fro USSE curves yields the proper pressure
10. Repeat the above steps until abandonment pressure or the time of prediction. Neglecting Cw & Cf
If the pressure less than the saturation pressure
EoNFW
BwBrrBBNBrRBNW
e
wbgsisoiogspope
)()()(
Also, we obtained We From unsteady state and continue the same last steps
Example 1
• A water drive reservoir is an elongated dome with water influx around the entire periphery ( angle =360) of the reservoir. Table below shows the data required. Also given the following N= 34 MMSTB, Pi=2265 Psia Pb= 1135 Psig water viscosity = 0.68 Cp, Porosity = 20 %, Permeability = 100 md, rw=4500 ft, re/rw=4, Ce= 14.x10-6 psi-1. rsi= 192 SCF/STB
Calculate the reservoir pressure at 54 months (considering that Np and Wp at this time are
1.705 MM STB and 160 M STB respectively.Time,
months
P,
Psi
Np
MMSTB
Wp
M STB
Bt
Bbl/STB
Gp,
SCF 10^8
0 2265 0 -
6 2215
12 2115 0.11 5 1.1025 0.2
18 1900
24 1728 0.65 45 1.1056 1.2
30 1600
36 1525 1.075 83 1.1014 2.2
42 1465
48 1405 1.505 130 1.1082 2.8
54 ???? 1.705 160
Solution
• Neglecting (Cf &Cw effect) above Pb
•
FBE
NBNBBwBNPQB
PQBBwBNNBNB
BB
WBWBNN
titwptpt
twptptit
itt
ewptp
As Example step one
• We MBE at 2115= (110,000)(1.1025)+5000-34000000(1.1025-1.1019) = 88500 bbls
• We from Unsteady state
26
3
23
)4500(1041.1468.02.0
10010323.6
10323.6
xxxx
txx
rC
Ktxt
we
D
At P= 2115 PsiTime T’ T’D Qt P Dp DPxQt
182.5 365 5.82 4.70 2215 25 117.5
365 182.5 2.91 3.00 2115 75 225
2505.342
85500)(
5.342
PQ
WB
PQ
MBEe
t
Repeat the last step for all pressures
bblsWe
xWe
PsiPatMBEWeAlso
618000
)1013.11056.1(103445000)1056.1(000,650
1728,6
We from U.S.S.E at P=1728t T’ T’D Qt P Dp DpQt
182.5 730 11.64 6.2 2215 25 155
365 547.5 8.73 5.7 2115 75 422
547.5 365 5.82 4.7 1900 158 743
730 182.5 2.91 3 1728 194 582
3251902
618000)(
1902
PQ
WB
PQ
MBEe
t
We from MBE at P=1525 psi
bblsW
xxW
atPMBe
MBe
1066000)
)1013.11074.1(103483000)1074.1(10075.1)
1525
66
We From U.S.S.Et T’ T’D Qt P Dp DpQt
182.5 1095 17.46 7 2215 25 175
365 912.5 14.55 6.75 2115 75 506
547.5 730 11.64 6.2 1900 158 979
730 547.5 8.73 5.7 1728 194 1106
912.5 365 5.82 4.7 1600 150 705
1095 182.5 2.91 3 1525 102 306
2833777
1066000)(
3777
PQ
WB
PQ
MBEe
t
We from MBE @P=1405 psi
bblsxW
xxW
atPMBe
MBe
61405
66
10565.1)
)1013.11082.1(1034130000)1082.1(10505.1)
We from U.S.S.E @1405 psit T’ T’D Qt P Dp DpQt
182.5 1460 23.28 7.25 2215 25 181
365 1277.5 20.37 7.15 2115 75 536
547.5 1095 17.46 7 1900 158 1106
730 912.5 14.55 6.75 1728 194 1310
912.5 730 11.64 6.2 1600 150 930
1095 547.5 8.73 5.7 1525 102 581
1277.5 365 5.82 4.7 1465 68 320
1460 182.5 2.91 3 1405 60 18
B @ P=1405 psi
3045144
1565000)(
5144
PQ
WB
PQ
MBEe
t
• Draw graph between B and Np or Time
• The best value of constant B from the graph is 300
Prediction of the reservoir performance
• Assume P ; 1340 , 1300 and 1200 > Pb
• From P.V.T dat
• Bt; 1.1087 1.1090 and 1.1099
• Using U.S.S.E
t T’ T’D Qt P Dp DpQt182.5 1642.5 26.19 7.3 2215 25 183
365 1460 23.28 7.25 2115 75 544
547.5 1277.5 20.37 7.15 1900 158 1130
730 1095 17.46 7 1728 194 1358
912.5 912.5 14.55 6.75 1600 150 1012
1095 730 11.64 6.2 1525 102 632
1277.5 547.5 8.73 5.7 1465 68 388
1460 365 5.82 4.7 1405 60 2821642.5 182.5 2.91 3 1340 63 189
1300 83 2491200 133 399
P We U.S.S.E
1340 5718
1300 5778
1200 5928
We from MBE at Assumed pressures
bblsxW
xxW
bblsxW
xxW
bblsxW
xxW
atPMBe
MBe
atPMBe
MBe
atPMBe
MBe
61200
66
61300
66
61340
66
10761.1)
)1013.11099.1(1034160000)1099.1(10705.1)
10790.1)
)1013.11090.1(1034160000)1090.1(10705.1)
10798.1)
)1013.11087.1(1034160000)1087.1(10705.1)
P We, bbl
From U.S.S.E
We,bbl
From MBE
1340 5718 1.798x106
1300 5778 1.790x106
1200 5928 1.761x106
• Plotting pressure Vs water influx calculated by both MBE and USSE obtained for the three assumed pressures will show the pressure which will result in identical water influx calculation by both method.
• The intersection of the two We MBE and We fro USSE curves yields the proper pressure Ans. Pcorrect = 1218 psi
» We = 1.769 x10 6 bbls