1. Prediction of Solution Gas drive reservoir performance

2. Tarner Methods for Predicting Solution gas drive reservoir First of all Recall the MBE for this drive mechanisms @> Pb

3. Instantaneous GOR (I.G.O.R)

4. Oil Saturation

5. First of all we have to draw the past production history versus pressure Also, PVT versus Pressure and So Vs Kg / Ko

7. Select Pressure and Assume Np At Pressure we assume three Np P x Np 1 Np 2 Np 3 Solve N P R P from MBE at each Np We calculate the N P R P from MBE at pressure before Px Represents the amount of gas produced during the time interval required for decreasing the reservoir pressure from Px-I to Px

8. Using the assumed values of Np 1,Np 2 and Np 3 solve the oil saturation for S o1, S o2 and S o3 at each N p 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 R 1,R 2 and R 3 from I.G.O.R equation

9. Calculate the amount of gas produced during the period (P x-1, P x )as follow

10. 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.

11. Example Table below lists reservoir data necessary for depletion drive reservoir PRsBoRpNp,10^6Bg Viscosity ratio Np/N 210013401.48013400 0.00128334.10 180012801.46819363.795 0.00151838.30.0393 150011501.44035848.584 0.00185342.40.0889 12009851.399623011.876 0.00236348.80.1230 10008601.360858013.479 0.00288553.60.1396 7006621.2871301015.236 0.00425062.50.1578 4004651.202 0.00768079.

12. Assume that Swi= 15 % and N = 96.55 x10^6 STB. Calculate Rp and Np at P = 400 psia

13. 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

14. Calculate the amount of gas produced during the period (P 700, P 400 )as follow from MBE

15. Oil Satyration

16. 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

17. We can conclude the results obtained in the following table

18. Np/N 0.17202.02 N178.75 N 0.175203.56 N255 N 0.18205.10 N341.5 N 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

19. 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 2050Initial pressure 1.260------540---- 1690Pb1.2650.0011454059.0 14801.8158251.2380.0013348560.0 14252.1969201.2320.0013847261.0 10504.37417501.1860.0020037165.5 7501.1460.0027327369.2

20. 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:

21. So0.760.750.700.660.6 Kg/Ko0.00610.00820.03550.1000.560 Find by Tarner’s metjhod the fractional oil recovery (Np/N) and R at pressure 750 psia assuming that Swi= 0.2