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1 Well Control DRILLING ENGINEERING

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2 4.1 Hydrostatic Pressure in Liquid Columns P = 0.052 D+P o (4.2 b) Where P = Pressure, psig = Density, lb/gal D = Depth, ft

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3 4.2Hydrostatic Pressure in Gas Column (4.4) PV = znRT (4.4) Where, N= number of moles M= Molecular weight of gas m= mass of gas z= gas deviation factor (4.6)

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4 4.3 Hydrostatic Pressure in Complex Fluid Columns (4.7)

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5 Fig 4.3: A Complex Liquid Column

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6 4.3.1Equivalent Density Concept (4.8) Review Examples 4.3 & 4.4

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7 Hydrostatic pressure of the fluid contained within the pore spaces of the sediments depends only on the fluid density. When formation pore pressure is approximately equal to the theoretical hydrostatic pressure for the given vertical depth, formation pressure is said to be normal. Normal pore pressure is usually expressed in terms of the hydrostatic gradient. 6.1 Formation Pore Pressure

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8 Example: Example: Compute the normal formation pressure expected at a depth of 6,000 ft in the Louisiana gulf coast area. Solution: Solution: The normal pressure gradient for the U.S. gulf coast area is 0.465 psi/ft. Thus, the normal formation pore pressure expected at 6,000 ft is; P f = 0.465 psi/ft x 6000 ft = 2,790 psi

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9 4.5 Buoyancy f =Fluid Density s =Steel Density (4.22) (4.21) (4.23)

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10 4.5.1 Determination of Axial Stress Axial Tension in Drill Collar F T Axial Tension in Drill Pipe (4.24 a) (4.24 b)

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11 Fig 4.10: Effect of Hydrostatic Pressure on Axial Forces in Drill string

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12 Calculation of Kill Mud Weights Formation Pressure= P form = 3000 psi Depth = D = 5000 ft Calculate Kill Fluid Density= ρ kill = ? Kill Fluid Density, ρ kill > 11.54 lbm/gal

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13 Example (p291): Example (p291): An offshore Louisiana well to be drilled in 2000 ft of water will penetrate a formation at 10,000ft (sub-sea) having a pore pressure of 6,500 psig. Compute the fracture gradient of the formation assuming the semi submersible used to drill the well will have an 80-ft air gap between the drilling fluid flow line and the sea surface. Compute the vertical overburden gradient assuming a seawater density of 8.5 lbm/gal, an average sediment grain density of 2.6 g/cc, a surface porosity of 0.45 and porosity decline constant of 0.000085/ft. Fracture Resistance

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14 Solution: Solution: At a depth of 10,000 ft subsea, the depth into the sediments is 8,000 ft. Entering Fig 6.50 with a depth of 8,000 ft yields a value of Poisson's ratio of 0.44. The vertical overburden gradient is given by eq 6.6; σ ob = 0.052 (8.5) (2,000) + 0.052 (2.6) (8.33) (8000) 0.052 (2.6 – 1.074) (8.33) (0.45) [ 1-e -0.000085(8000) ] 0.000085 = 884+9,010-1727 = 8167 psig

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15 The horizontal matrix stress, which is the minimum matrix stress, is computed by eq. 6.30 The fracture pressure is given by eq. 6.31 p ff = 1310 + 6500 = 7800 psig Thus, the fracture gradient is

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