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# Chapter 2 Volumetric gas Reservoir Engineering

Download Presentation ## Chapter 2 Volumetric gas Reservoir Engineering

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1. Chapter 2 • Volumetric gas Reservoir Engineering

2. Gas PVT • Gas is one of a few substances whose state, as defined by pressure, volume and temperature (PVT) • One other such substance is saturated steam.

3. The equation of state for gas • Classical ideal gas law • Classical non-ideal gas law • Cubic Equations of State • van der Waals equation of state • Redlich-Kwong equation of state • Soave modification of Redlich-Kwong • Peng-Robinson equation of state • Elliott, Suresh, Donohue equation of state • Non-cubic Equations of State • Dieterici equation of state • Virial Equations of State • Virial Equations of State • The BWRS equation of state • Other Equations of State of Interest • Stiffened equation of state • Ultrarelativistic equation of state • Ideal Bose equation of state

4. The equation of state for an ideal gas (Field units used in the industry) p [=] psia; V[=] ft3; T [=] OR absolute temperature n [=] lbm moles; n=the number of lbm moles, one lbm mole is the molecular weight of the gas expressed in pounds. R = the universal gas constant [=] 10.732 psia∙ ft3 / (lbm mole∙0R) Eq (1.13) results from the combined efforts of Boyle, Charles, Avogadro and Gay Lussac.

5. Note: In this text 1 Darcy = 1.0133×10-8 cm2 or 1 Darcy 10-8 cm2 or 1 Darcy In other book 1 Darcy = ０.986927×10-8 cm2 1 Darcy = ０.986927×10-12 m2 10-12 m2 = 1 1 Darcy 1

6. Non-ideal gas law • Where z = z-factor =gas deviation factor • =supercompressibility factor • = compressibility factor

7. Determination of z-factor • There are three ways to determine z-factor : • (a)Experimental determination • (b)The z-factor correlation of standing and katz • (c)Direct calculation of z-factor

8. (a) Experimental determination • n moles of gas • p=1atm; T=reservoir temperature; => V=V0 • pV=nzRT • z=1 for p=1 atm • =>14.7 V0=nRT • n moles of gas • p>1atm; T=reservoir temperature; => V=V • pV=nzRT • pV=z(14.7 V0) • By varying p and measuring V, the isothermal z(p) function can be • readily by obtained.

9. (b)The z-factor correlation of standing and katz • Requirement: • Knowledge of gas composition or gas gravity • Naturally occurring hydrocarbons: primarily paraffin series CnH2n+2 • Non-hydrocarbon impurities: CO2, N2 and H2S • Gas reservoir: lighter members of the paraffin series, C1 • and C2 > 90% of the volume.

10. The Standing-Katz Correlation • knowing Gas composition (ni) •  Critical pressure (Pci) • Critical temperature (Tci) of each component •  ( Table (1.1) and P.16 )  •  Pseudo critical pressure (Ppc) • Pseudo critical temperature (Tpc) for the mixture •  Pseudo reduced pressure (Ppr) • Pseudo reduced temperature (Tpr) •  Fig.1.6; p.17  z-factor

11. (b’)The z-factor correlation of standing and katz • For the gas composition is not available and the gas gravity (air=1) is available. • The gas gravity (air=1) • ( ) •  fig.1.7 , p18 • Pseudo critical pressure (Ppc) • Pseudo critical temperature (Tpc)

12. (b’)The z-factor correlation of standing and katz •  Pseudo reduced pressure (Ppr) • Pseudo reduced temperature (Tpr) •  Fig1.6 p.17 •  z-factor • The above procedure is valided only if impunity (CO2,N2 and H2S) is less then 5% volume.

13. (c) Direct calculation of z-factor • The Hall-Yarborough equations, developed using the Starling-Carnahan equation of state, are • where Ppr= the pseudo reduced pressure • t=1/Tpr ; Tpr=the pseudo reduced temperature • y=the “reduced” density which can be obtained as the solution of the equation as followed: This non-linear equation can be conveniently solved for y using the simple Newton-Raphson iterative technique.

14. (c) Direct calculation of z-factor • The steps involved in applying thus are: • making an initial estimate of yk, where k is an iteration counter (which in this case is unity, e.q. y1=0.001 • substitute this value in Eq. (1.21);unless the correct value of y has been initially selected, Eq. (1.21) will have some small, non-zero value Fk. • (3) using the first order Taylor series expansion, a better estimate of y can be determined as • where • (4) iterating, using eq. (1.21) and eq. (1.22), until satisfactory convergence is obtained(5) substitution of the correct value of y in eq.(1.20)will give the z-factor. • (5) substituting the correct value of y in eq.(1.20)will give the z-factor.

15. The equation of state for real gas • The equation of Van der Waals(for one lb mole of gas • where a and b are dependent on the nature of the gas. • The principal drawback in attempting to use eq. (1.14) to describe the behavior of real gases encountered in reservoirs is that the maximum pressure for which the equation is applicable is still far below the normal range of reservoir pressures

16. Peng-Robinson equation of state where, ω is the acentric factor of the species R is the universal gas constant.

17. Peng-Robinson equation of state • The Peng-Robinson equation was developed in 1976 in order to satisfy the following goals: • The parameters should be expressible in terms of the critical properties and the acentric factor. • The model should provide reasonable accuracy near the critical point, particularly for calculations of the Compressibility factor and liquid density. • The mixing rules should not employ more than a single binary interaction parameter, which should be independent of temperature pressure and composition. • The equation should be applicable to all calculations of all fluid properties in natural gas processes. • For the most part the Peng-Robinson equation exhibits performance similar to the Soave equation, although it is generally superior in predicting the liquid densities of many materials, especially nonpolar ones. The departure functions of the Peng-Robinson equation are given on a separate article.

18. Application of the real gas equation of state • Equation of state of a real gas • This is a PVT relationship to relate surface to reservoir volumes of hydrocarbon. • the gas expansion factor E, • Real gas equation for n moles of gas at standard conditions •  • Real gas equation for n moles of gas at reservoir conditions •  • > • > surface volume/reservoir volume • [=] SCF/ft3 or STB/bbl

19. Example • Reservoir condition: P=2000psia; T=1800F=(180+459.6)=639.60R; z=0.865 > surface volume/reservoir or SCF/ft3 or STB/bbl

20. (2) Real gas density • where n=moles; M=molecular weight) • at any p and T • For gas • For air

21. (2) Real gas density • At standard conditions zair = zgas = 1 • in general • (a) If is known, then or , (b) If the gas composition is known, then where

22. (3)Isothermal compressibility of a real gas since p.24, fig.1.9

23. Exercise 1.1 - Problem • Exercise1.1 Gas pressure gradient in the reservoir • (1) Calculate the density of the gas, at standard conditions, whose composition is listed in the table 1-1. • (2) what is the gas pressure gradient in the reservoir at 2000psia and 1800F(z=0.865)

24. Exercise 1.1 -- solution -1 • (1) Molecular weight of the gas since • or from • At standard condition

25. Exercise 1.1 -- solution -2 • (2) gas in the reservoir conditions

26. Exercise 1.1 -- solution -3

27. Fluid Pressure Regimes The total pressure at any depth = weight of the formation rock + weight of fluids (oil, gas or water) ~ 1 psi/ft * depth(ft)

28. Fluid Pressure Regimes • Density of sandstone

30. Overburden pressure • Overburden pressure (OP) = Fluid pressure (FP) + Grain or matrix pressure (GP) • OP=FP + GP • In non-isolated reservoir PW (wellbore pressure) = FP • In isolated reservoir PW (wellbore pressure) = FP + GP’ where GP’<=GP

31. Normal hydrostatic pressure • In a perfectly normal case , the water pressure at any depth • Assume :(1) Continuity of water pressure to the surface • (2) Salinity of water does not vary with depth. • [=] psia • psi/ft for pure water • psi/ft for saline water

32. Abnormal hydrostatic pressure ( No continuity of water to the surface) • [=] psia • Normal hydrostatic pressure c = 0 • Abnormal (hydrostatic) pressure c > 0 → Overpressure (Abnormal high pressure) • c < 0 → Underpressure (Abnormal low pressure)

33. Conditions causing abnormal fluid pressures • Conditions causing abnormal fluid pressures in enclosed water bearing sands include • Temperature change ΔT = +1℉ → ΔP = +125 psi in a sealed fresh water system • Geological changes – uplifting; surface erosion • Osmosis between waters having different salinity, the sealing shale acting as the semi permeable membrane in this ionic exchange; if the water within the seal is more saline than the surrounding water the osmosis will cause the abnormal high pressure and vice versa.

34. Are the water bearing sands abnormally pressured ? • If so, what effect does this have on the extent of any hydrocarbon accumulations?

35. Hydrocarbon pressure regimes • In hydrocarbon pressure regimes • psi/ft • psi/ft • psi/ft

36. = + = P 0 . 45 * D 15 [ ] psia in water zone w = = + = P ( at D 5600 ft ) 0 . 45 * 5600 15 2535 psia w = = + = P ( at D 5500 ft or at OWC ) 0 . 45 * 5500 15 2490 psia w = = P ( at D 5500 ft or at OWC ) 2490 o = + 0 . 35 * D C o = = or C 2490 - 0.35 * 5500 565 o Þ = + P 0.35 * D 565 in oil zone o = = + = P ( at D 5200 ft ) 0 . 35 * 5200 5 65 2385 psia o = = + = P ( at D 5200 ft ) 0 . 45 * 5200 15 2355 psia w = = + = P ( at D 5000 ft ) 0 . 35 * 5000 565 2315 psia o = = + = P ( at D 5000 ft ) 0 . 45 * 5000 15 2265 psia w Pressure Kick – Oil and Water P(psia) Pw=2265 Po=2315 5000 oil 5200 Pw=2355 Po=2385 OWC 5500 D=5500ft Pw=Po=2490 5600 Pw=2535 water Depth(ft)

37. pressure kick-gas and water P(psia) Pw=2265 Pg=2450 5000 Gas Pw=2355 Pg=2466 5200 GWC D=5500ft 5500 Pw=Pg=2490 5600 Pw=2535 water Depth(ft)

38. pressure kick-gas, oil and water Pg=2396 P(psia) Pw=2265 5000 Gas Pw=2355 5200 Pg=2412 Pw=2400 GOC 5300 Po =Pg=2420 D=5300ft Pw=2445 5400 Po=2455 oil OWC 5500 Pw= Po=2490 D=5500ft 5600 Pw=2535 water Depth(ft)

39. Pressure Kick • Assumes a normal hydrostatic pressure regime Pω= 0.45 × D + 15 • In water zone • at 5000 ft Pω(at5000) = 5000 × 0.45 + 15 = 2265 psia • at OWC (5500 ft) Pω(at OWC) = 5500 × 0.45 + 15 = 2490 psia

40. Pressure Kick • In oil zone Po = 0.35 x D + C • at D = 5500 ft , Po = 2490 psi • → C = 2490 – 0.35 × 5500 = 565 psia • → Po = 0.35 × D + 565 • at GOC (5200 ft) Po (at GOC) = 0.35 × 5200 + 565 = 2385 psia

41. Pressure Kick • In gas zone Pg = 0.08 D + 1969 (psia) • at 5000 ft Pg = 0.08 × 5000 + 1969 = 2369 psia