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Chapter 3 Material Balance Applied to Oil Reservoirs

Chapter 3 Material Balance Applied to Oil Reservoirs. § 3.1 Introduction -The Schilthuis material balance equation - Basic tools of reservoir engineering => Interpreting and predicting reservoir performance. -Material balance 1. zero dimension – this chapter

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Chapter 3 Material Balance Applied to Oil Reservoirs

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  1. Chapter 3 Material Balance Applied to Oil Reservoirs § 3.1 Introduction -The Schilthuis material balance equation - Basic tools of reservoir engineering => Interpreting and predicting reservoir performance. -Material balance 1. zero dimension – this chapter 2. multi-dimension (multi-phase) – reservoir simulation

  2. § 3.2 General form of the material balance equation for a hydrocarbon reservoir Underground withdrawal (RB) = Expansion of oil and original dissolved gas (RB)………(A) + Expansion of gascap gas (RB) ……………… ………(B) + Reduction in HCPV due to connate water expansion and decrease in the pore volume (RB)……………………… …….……(C)

  3. Expansion of oil & originally dissolved gas

  4. Expansion of the gascap gas Expansion of the gascap gas =gascap gas (at p) –gascap (at pi)

  5. Change in the HCPV due to the connate water expansion & pore volume reduction

  6. Underground withdrawal

  7. The general expression for the material balance as

  8. where

  9. No initial gascap, negligible water influx • With water influx eq(3.12) becomes • Eq.(3.12) having a combination drive-all possible sources of energy.

  10. reducing the M.B to a compact form to • quantify reservoir performance • determining the main producing • characteristics, • for example, GOR; water cut • determining the pressure decline in the • reservoir • - estimating the primary recovery factor • - Solution gas drive • - Gascap drive • Natural water • drive • - Compaction drive In terms of § 3.4 Reservoir Drive Mechanisms Reservoir drive mechanism

  11. § 3.5 Solution gas drive (a) above the B.P. pressure (b) below the B.P. pressure

  12. Above the B.P. pressure- no initial gascap, m=0- no water flux, We=0 ; no water production, Wp=0- Rs=Rsi=Rp from eq.(3.7)

  13. Exercise3.1 Solution gas drive, undersaturated oil reservoir Determine R.F. Solution: FromTable2.4(p.65) Eq(3.18)

  14. Table 2.4 Field PVT P(psia) Bo (Rb/STB) Rs(SCF/STB) Bg( Rb/SCF) 4000 (pi) 1.2417 510 3500 1.2480 510 3300 (pb) 1.2511 510 0.00087 3000 1.2222 450 0.00096 • 1.2022 401 0.00107 • 1.1822 352 0.00119 • 1.1633 304 0.00137 1800 1.1450 257 0.00161 • 1.1287 214 0.00196 1200 1.1115 167 0.00249 • 1.0940 122 0.00339 • 1.0763 78 0.00519 300 1.0583 35 0.01066

  15. Bo as Function of Pressure

  16. Rs as Function of Pressure

  17. Bg and E as Function of Pressure

  18. Producing Gas-oil Ratio (R) as Function of Pressure

  19. Below B.P. pressure (Saturation oil)P<Pb =>gas liberated from saturated oil

  20. Exercise3.2 Solution gas drive; below bubble point pressure Reservoir-described in exercise 3.1 Pabandon = 900psia(1) R.F = f(Rp)? Conclusion?(2) Sg(free gas) = F(Pabandon)?Solution:(1) From eq(3.7)

  21. Eq(3.7) becomes Conclusion:

  22. liberated gas in the reservoir total amount of gas gas produced at surface gas still dissolved in the oil = − − (2) the overall gas balance

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