Prediction of Solution Gas drive reservoir performance

Tarner Methods for Predicting Solution gas drive reservoir

• First of all Recall the MBE for this drive mechanisms

• @> Pb

tit

gsiptp

oio

op

BB

BrRBNN

PbP

BB

BNN

)(

Instantaneous GOR (I.G.O.R)

sg

o

g

o

o

g rB

B

K

KR

Oil Saturation

)1/(

)(

wioi

op

s

SNB

BNNSo

VolumePore

VolumeoilSo

• First of all we have to draw the past production history versus pressure

• Also, PVT versus Pressure and

• So Vs Kg / Ko

g

gsogsisoiopp

g

gsitptitpp

B

BrBNpBrrBBNRN

B

BrBNBBNRN

)(

)()(

Select Pressure and Assume Np• At Pressure we assume three Np

• Px Np1 Np2 Np3 • Solve NPRP from MBE at each Np• We calculate the NPRP from MBE at pressure before

Px

•

G

where

NpRpRpNpNG

NpRpRpNpNG

NpRpRpNpNG

pxPxp

pxPxp

pxPxp

13333

12222

11111

)()(@

)()(@

)()(@

Represents the amount of gas produced during the time interval required for decreasing the reservoir pressure from Px-I to Px

• Using the assumed values of Np1,Np2 and Np3 solve the oil saturation for So1, So2 and So3 at each Np assumed respectively.

• Using the calculated values of oil saturation determine relative permeability ratio (kg/Ko)1, (kg/Ko)2& (kg/Ko)3 at each saturation from the relative permeability ratio saturation curve.

• Calculate R1,R2 and R3 from I.G.O.R equation

• Calculate the amount of gas produced during the period (Px-1, Px )as follow

)2

)((

)2

)((

)2

)((

13133

12122

11111

xpxp

xpxp

xpxp

RRNNG

RRNNG

RRNNG

• The values of amount of gas produced during interval pressure calculated by both methods ( MBE and IGOR) are then plotted on a graph Vs Np

• The plot of the two lines will intersect. This intersection is only place where one value of Np will satisfy both equation.

Example • Table below lists reservoir data necessary for

depletion drive reservoir

P Rs Bo Rp Np,10^6 Bg Viscosity ratio

Np/N

2100 1340 1.480 1340 0 0.001283 34.1 0

1800 1280 1.468 1936 3.795 0.001518 38.3 0.0393

1500 1150 1.440 3584 8.584 0.001853 42.4 0.0889

1200 985 1.399 6230 11.876 0.002363 48.8 0.1230

1000 860 1.360 8580 13.479 0.002885 53.6 0.1396

700 662 1.287 13010 15.236 0.004250 62.5 0.1578

400 465 1.202 0.007680 79.

• Assume that Swi= 15 % and N = 96.55 x10^6 STB .

• Calculate Rp and Np at P = 400 psia

Solution • As Np = 0.1578 N at P= 700 psia, then assume. • Np = 0.17 N, Np = 0.175 N and Np = 0.18 N•

NNpRp

xNxNNpRp

NNpRpNNp

NNpRpNNp

NNpRpNNp

NNpNNpRp

xNxNNpRp

P

P

P

P

P

P

P

26.68900425.0

00425.0662287.11578.000425.0)6621340(480.1287.1

36.89418.0@

82.892175.0@

28.89117.0@

17.0@28.891007680.0

007680.0465202.117.0007680.0)4651340(480.1202.1

700@

700@

400@

400@

400@

400@

400@

• Calculate the amount of gas produced during the period (P700, P400)as follow from MBE

NNpRpRpNpNNG

NNpRpRpNpNNG

NNNNNG

pxPxp

pxPxp

p

1.205)()(18.0@

56.203)()(175.0@

02.20226.68928.89117.0@

13333

12222

11

Oil Satyration

566.0

570.0

573.0)15.01)(480.1

202.1)(17.01(

)1/(

)(

3

2

1

o

o

o

wioi

opo

S

S

S

SNB

BNNS

• From data given and plot the relation between relative permeability ratio versus oil saturation

• From this figure we can obtained the values of (kg/Ko)1,=1.28 (kg/Ko)2= 1.314

and

(kg/Ko)3=1.395

•

NG

NG

NNG

xxR

GOR

GOR

GOR

5.3413

8.2552

73.1782

13010162911578.017.01

16291465007680.0

202.17928.11

We can conclude the results obtained in the following table

Np/N

0.17 202.02 N 178.75 N

0.175 203.56 N 255 N

0.18 205.10 N 341.5 N

GORGMBEG

The results: Npc=0.1716 N , DG= 202 N : Npc= 16.56x!0^6 STB

Using DG relation ship of DG 202N= (0.1716N-0.1578N)x(13010+Rc)/2 then

Rc= 16265.3 SCF/STB

Problem 1 • Given the following data for an oil reservoir

P,

psia

Np, MMSTB

R, SCF/STB

Bo,

BBL/STB

Bg

,BBL/SCF

R,

SCF/STB

Viscosity ratio

2050 Initial pressure

1.260 ------ 540 ----

1690 Pb 1.265 0.00114 540 59.0

1480 1.815 825 1.238 0.00133 485 60.0

1425 2.196 920 1.232 0.00138 472 61.0

1050 4.374 1750 1.186 0.00200 371 65.5

750 1.146 0.00273 273 69.2

• Determine : (neglecting Cw &Cf)– The fractional oil recover at the bubble point

presssure.– The original oil in place – The reservoir drive mechanism– The STB of oil that would have been recovered at

1050 psia if all the gas produced except 170 SCF/STB on a cumulative basis had injected back into the reservoir

– Using the following permeability relationship:

So 0.76 0.75 0.70 0.66 0.6

Kg/Ko 0.0061 0.0082 0.0355 0.100 0.560

Find by Tarner’s metjhod the fractional oil recovery

(Np/N) and R at pressure 750 psia assuming that

Swi= 0.2