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EXPLORATION GEOPHYSICS

EXPLORATION GEOPHYSICS. THE EXPLORATION TASK. INITIAL DATA GATHANAL AND PROJECT PLANING FOR A FRONTIER TREND. PLAN EXPLORATION APPROACH FOR A MATURE TREND. DEVELOP PLAY PROSPECT FRAMEWORK. NEW DATA GATHERING FOR A FRONTIER TREND. GATHER DATA FOR A MATURE TREND.

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EXPLORATION GEOPHYSICS

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  1. EXPLORATION GEOPHYSICS

  2. THE EXPLORATION TASK INITIAL DATA GATHANAL AND PROJECT PLANING FOR A FRONTIER TREND PLAN EXPLORATION APPROACH FOR A MATURE TREND DEVELOP PLAY PROSPECT FRAMEWORK NEW DATA GATHERING FOR A FRONTIER TREND GATHER DATAFOR A MATURE TREND MAKE PLAY/PROSPECT ASSESSMENT COMMUNICATE ASSESSMENT TO MANAGEMENT PREPARE PRELOCATION REPORT DRILLING

  3. EXPLORATION GEOPHYSICS

  4. Elasticity Source Petroleum related rock mechanics Elsevier, 1992

  5. Elasticity • Definition: The ability to resist and recover from deformations produced by forces. • It is the foundation for all aspects of Rock Mechanics • The simplest type of response is one where there is a linear relation between the external forces and the corresponding deformations.

  6. Stress defines a force field on a material Stress = Force / Area (pounds/sq. in. or psi) s = F / A F Area: A

  7. Stress • In Rock Mechanics the sign convention states that the compressive stresses are positive.  • Consider the cross section area at b, the force acting through this cross section area is F (neglecting weight of the column) and cross sectional area is A’. A’ is smaller than A, therefore stress ’ = F/A’ acting at b is greater than  acting at a

  8. W A’’ c F F b A’ F A a Stress Area Load

  9. Stress • stress depends on the position within the stressed sample. • Consider the force acting through cross section area A’’. It is not normal to the cross section. We can decompose the force into one component Fn normal to the cross section, and one component Fp that is parallel to the section.

  10. Fn Fp F Stress Decomposition of forces

  11. Stress • The quantity  = Fn /A’’ is called the normal stress, while the quantity •  = Fp / A’’ is called the shear stress. • Therefore, there are two types of stresses which may act through a surface, and the magnitude of each depend on the orientation of the surface.

  12. tzx txy tzx tyx tzy tyz General 3D State of Stress in a Reservoir sx, sy, sz Normal stresses xy, yz, zx Shear stresses sx sz sy

  13. Stress •  = xxyxz yxyyz  zx zyz • Stress tensor

  14. Principal Stresses sv Normal stresses on planes where shear stresses are zero s1 > s2 > s3 sh sH

  15. Principal Stresses sv In case of a reservoir, s1 = sv Vertical stress, s2 = sh Minimum horizontal stresses s3 = sH Maximum horizontal stresses sh sH

  16. Types of Stresses • Tectonic Stresses: Due to relative displacement of lithospheric plates Based on the theory of earth’s tectonic plates • Spreading ridge: plates move away from each other • Subduction zone: plates move toward each other and one plate subducts under the other • Transform fault: Plates slide past each other

  17. Types of Stresses • Gravitational Stresses: Due to the weight of the superincumbent rock mass • Thermal Stresses: Due to temperature variation Induced, residual, regional, local, far-field, near-field, paleo ...

  18. Impact of In-situ Stress • Important input during planning stage • Fractures with larger apertures are oriented along the maximum horizontal stress

  19. Natural fractures

  20. Strain (x’, y’, z’) (x, y, z) ShiftedPosition Initial Position

  21. Strain • x’ = x – u • y’ = y – v • z’ = z – w • If the displacements u, v, and w are constants, i.e, they are the same for every particle within the sample, then the displacement is simply a translation of a rigid body.

  22. Strain • Another simple form of displacement is the rotation of a rigid body. • If the relative positions within the sample are changed, so that the new positions cannot be obtained by a rigid translation and/or rotation of the sample, the sample is said to be strained. (figure 8)

  23. P P’ L L’ O O’ Initial Position Shifted Position Strain

  24. Strain • Elongation corresponding to point O and the direction OP is defined as •  = (L – L’)/L   • sign convention is that the elongation is positive for a contraction. • The other type of strain that may occur can be expressed by the change  of the angle between two initially orthogonal directions. (Figure 9)

  25. Q Q  O P O P Initial Position Shifted Position Strain

  26. Strain •  = (1/2)tan is called the shear strain corresponding to point O and the direction OP. We deal with infinitesimal strains. • The elongation (strain) in the x-direction at x can be written as • x = u/x

  27. Strain • The shear strain corresponding to x-direction can be written as • xy = (u/y + v/x)/2 • Strain tensor • Principal strains

  28. Elastic Moduli F X D L D’ L’ Y Schematic showing deformation under load

  29. Elastic Moduli • When force F is applied on its end surfaces, the length of the sample is reduced to L’. • The applied stress is then x = F/A, • The corresponding elongation is  = (L – L’)/L • The linear relation between x and x, can be written as • x = Ex

  30. Elastic Moduli • This equation is known as Hooke’s law • The coefficient E is called Young’s modulus. • Young’s modulus belongs to a group of coefficients called elastic moduli. • It is a measure of the stiffness of the sample, i.e., the sample’s resistance against being compressed by a uniaxial stress.

  31. Elastic Moduli • Another consequence of the applied stress x (Figure 10) is an increase in the width D of the sample. The lateral elongation is y = z = (D – D’)/D. In general D’ > D, thus y and z become negative. • The ratio defined as  = -y/x is another elastic parameter known as Poisson’s ratio. It is a measure of lateral expansion relative to longitudinal contraction.

  32. Elastic Moduli • Bulk modulus K is defined as the ratio of hydrostatic stress p relative to the volumetric strain v. For a hydrostatic stress state we have p = 1 = 2 = 3 while xy = xz = yz = 0. Therefore • K = p/v =  + 2G/3 1 • K is a measure of sample’s resistance against hydrostatic compression. The inverse of K, i.e., 1/K is known as compressibility

  33. Elastic Moduli • Isotropic materials are materials whose response is independent of the orientation of the applied stress. For isotropic materials the general relations between stresses and strains may be written as: • x = ( + 2G) x + y + z • y = x + ( + 2G)y + z • z = x + y + ( + 2G)z • xy = 2Gxy xz = 2Gxzyz = 2Gyz

  34. Elastic Moduli • Expressing strains as function of stresses • Ex = x - (y + z) • Ey = y - (x + z) • Ez = z - (x + y) • Gxy = (1/2)xy • Gxz = (1/2)xz • Gyz = (1/2)yz

  35. Elastic Moduli • In the definition of Young’s modulus and Poisson’s ratio, the stress is uniaxial, i.e., z = y = xy = xz = yz = 0. Therefore •  E = x/x = G (3 + 2G)/ ( + G) 2 •  = -y/x = /(2( + G))   3 • Therefore from equations (1, 2, and 3), knowing any two of the moduli E, , , G and K, we can find other remaining moduli

  36. Elastic Moduli • For rocks,  is typically 0.15 – 0.25. For weak, porous rocks  may approach zero or even become negative. For fluids, the rigidity G vanishes, which according to equation (3) implies  ½. Also for unconsolidated sand,  is close to ½.

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