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