DRILLING ENGINEERING Well Control 1 4 1 Hydrostatic

DRILLING ENGINEERING Well Control 1

4. 1 Hydrostatic Pressure in Liquid Columns P = 0. 052 D+Po (4. 2 b) Where P = Pressure, psig = Density, lb/gal D = Depth, ft 2

4. 2 Hydrostatic Pressure in Gas Column PV = zn. RT (4. 4) Where, N= number of moles M= Molecular weight of gas m= mass of gas z= gas deviation factor (4. 6) 3

4. 3 Hydrostatic Pressure in Complex Fluid Columns (4. 7) 4

Fig 4. 3: A Complex Liquid Column 5

4. 3. 1 Equivalent Density Concept (4. 8) Review Examples 4. 3 & 4. 4 6

6. 1 Formation Pore Pressure • 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 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. 7

Example: Compute the normal formation pressure expected at a depth of 6, 000 ft in the Louisiana gulf coast area. 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; Pf = 0. 465 psi/ft x 6000 ft = 2, 790 psi 8

4. 5 Buoyancy (4. 22) (4. 21) f=Fluid Density s=Steel Density (4. 23) 9

4. 5. 1 Determination of Axial Stress Axial Tension in Drill Collar FT (4. 24 a) Axial Tension in Drill Pipe (4. 24 b) 10

Fig 4. 10: Effect of Hydrostatic Pressure on Axial Forces in Drill string 11

Calculation of Kill Mud Weights Formation Pressure = Pform = 3000 psi Depth = D = 5000 ft Calculate Kill Fluid Density= ρkill = ? Kill Fluid Density, ρkill > 11. 54 lbm/gal 12

Fracture Resistance Example (p 291): An offshore Louisiana well to be drilled in 2000 ft of water will penetrate a formation at 10, 000 ft (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. 13

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; σob= 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 14

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 pff= 1310 + 6500 = 7800 psig Thus, the fracture gradient is 15