Bài giảng Well drilling engineering - Chapter 5: Drilling Hydraulics (Part 2) - Đỗ Quang Khánh

1 Well Drilling Engineering  Drilling Hydraulics (cont’d) Dr. DO QUANG KHANH 2 10. Drilling Hydraulics (cont’d) Effect of Buoyancy on Buckling The Concept of Stability Force Stability Analysis Mass Balance Energy Balance Flow Through Nozzles Hydraulic Horsepower Hydraulic Impact Force 3 READ:ADE, Ch. 4 HW #: ADE # 4.14, 4.15, 4.16, 4.21 4 Buckling of Tubulars Slender pipe suspended in wellbore Partially buckled slender pipe Neutral Point Neutral Point F h - F b F h F b 5 Buckling of Tubulars Neutral Point Neutral Point Long slender columns, like DP, have low resistance to bending and tend to fail by buckling if... Force at bottom (F b ) causes neutral point to move up What is the effect of buoyancy on buckling? What is NEUTRAL POINT? F b 6 What is NEUTRAL POINT? Neutral Point Neutral Point One definition of NEUTRAL POINT is the point above which there is no tendency towards buckling Resistance to buckling is indicated, in part, by: The Moment of Inertia 7 Consider the following : 19.5 #/ft drillpipe Depth = 10,000 ft. Mud wt. = 15 #/gal. D P HYD = 0.052 (MW) (Depth) = 0.052 * 15 * 10,000 D P HYD = 7,800 psi Axial tensile stress in pipe at bottom = - 7,800 psi What is the axial force at bottom? 8 Axial Tension: F T = W 1 - F 2 F T = w x - P 2 (A O - A i ) At surface, F T = 19.5 * 10,000 - 7,800 (5.73) = 195,000 - 44,700 = 150,300 lbf. At bottom, F T = 19.5 * 0 - 7,800 (5.73) = - 44,700 lbf Same as before! F T F 2 9 Stability Force: F S = A i p i - A O p O F S = (A i - A O ) p (if p i = p O ) At surface, F S = - 5.73 * 0 = 0 At bottom, F S = ( - 5.73) (7,800) = - 44,700 lbs THE NEUTRAL POINT is where F S = F T Therefore, Neutral point is at bottom! PIPE WILL NOT BUCKLE!! A i 10 Compression Tension 44,700 0 150,300 F S F T Zero Axial Stress Neutral Point Depth of Zero Axial Stress Point = 11 Length of Drill Collars Neutral Point Neutral Point 12 Length of Drill Collars In Air: In Liquid: In Liquid with S.F.: (e.g., S.F =1.3) 13 State of stress in pipe at the neutral point s t s Z s r s r s Z s t Steel Elemental Volume 14 At the Neutral Point: The axial stress is equal to the average of the radial and tangential stresses. 15 Stability Force: F S = A i p i - A o p o If F S > axial tension then the pipe may buckle. If F S < axial tension then the pipe will NOT buckle. F S F T 0 F T 16 At the neutral point: F S = axial load To locate the neutral point: Plot F S vs. depth on “axial load ( F T ) vs. depth plot” The neutral point is located where the lines intersect. 17 NOTE: If p i = p o = p, then F s = or, F s = - A S p A S 18 Axial Load with F BIT = 68,000 lbf 19 Stability Analysis withF BIT = 68,000 lbf 20 Nonstatic Well Conditions Physical Laws Rheological Models Equations of State FLUID FLOW 21 Physical Laws Conservation of mass Conservation of energy Conservation of momentum 22 Rheological Models Newtonian Bingham Plastic Power – Law API Power-Law 23 Equations of State Incompressible fluid Slightly compressible fluid Ideal gas Real gas 24 Average Fluid Velocity  Pipe Flow Annular Flow WHERE v = average velocity, ft/s q = flow rate, gal/min d = internal diameter of pipe, in. d 2 = internal diameter of outer pipe or borehole, in. d 1 =external diameter of inner pipe, in . 25 26 Law of Conservation of Energy States that as a fluid flows from point 1 to point 2: In the wellbore, in many cases Q = 0 (heat) r = constant { 27 In practical field units  this equation simplifies to: p 1 and p 2 are pressures in psi r is density in lbm/gal. v 1 and v 2 are velocities in ft/sec. D p p is pressure added by pump between points 1 and 2 in psi D p f is frictional pressure loss in psi D 1 and D 2 are depths in ft. where 28 Determine the pressure at the bottom of the drill collars, if (bottom of drill collars) (mud pits) 29 Velocity in drill collars Velocity in mud pits, v 1 30 Pressure at bottom of drill collars = 7,833 psig NOTE: KE in collars May be ignored in many cases 31 32 Fluid Flow Through Nozzle Assume: 33 If This accounts for all the losses in the nozzle. Example: 34 35 For multiple nozzles in // V n is the same for each nozzle even if the d n varies! This follows since D p is the same across each nozzle. & 36 Hydraulic Horsepower of pump putting out 400 gpm at 3,000 psi = ? Power In field units: 37 What is Hydraulic Impact Force developed by bit? Consider: 38 Impact = rate of change of momentum