PETE 411 Well Drilling Lesson 10 Drilling Hydraulics (cont’d) 1 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 2 READ: ADE, Ch. 4 to p. 135 HW #5: ADE # 4.3, 4.4, 4.5, 4.6 due September 27, 2002 3 Fh Buckling of Tubulars Fh - Fb l Neutral Point l Slender pipe suspended in wellbore Partially buckled slender pipe Neutral Point Fb 4 Buckling of Tubulars • Long slender columns, like DP, l Neutral Point have low resistance to bending and tend to fail by buckling if... • Force at bottom (Fb) causes neutral point to move up • What is the effect of buoyancy on buckling? • What is NEUTRAL POINT? Neutral Point Fb 5 What is NEUTRAL POINT? • One definition of NEUTRAL POINT is the point above which there is no tendency towards buckling l • Resistance to buckling is indicated, in part, by: Neutral Point The Moment of Inertia Neutral Point I d 64 4 n d4 in 4 6 Consider the following: 19.5 #/ft drillpipe Depth = 10,000 ft. Mud wt. = 15 #/gal. DPHYD = 0.052 (MW) (Depth) = 0.052 * 15 * 10,000 DPHYD = 7,800 psi Axial tensile stress in pipe at bottom = - 7,800 psi What is the axial force at bottom? 7 What is the axial force at bottom? Cross-sectional area of pipe = (19.5 / 490) * (144/1) = 5.73 in2 Axial compressive force = pA lbf 2 7,800 5 . 73 in 2 in = 44,700 lbf. Can this cause the pipe to buckle? 8 Axial Tension: FT = W1 - F2 FT FT = w x - P2 (AO - Ai ) At surface, FT = 19.5 * 10,000 - 7,800 (5.73) = 195,000 - 44,694 = 150,306 lbf. At bottom, FT = 19.5 * 0 - 7,800 (5.73) = - 44,694 lbf Same as before! 9 F2 Stability Force: Ai FS = Aipi - AO pO FS = (Ai - AO) p (if pi = pO) At surface, FS = - 5.73 * 0 = 0 At bottom, FS = (-5.73) (7,800) = - 44,694 lbs THE NEUTRAL POINT is where FS = FT Therefore, Neutral point is at bottom! PIPE WILL NOT BUCKLE!! 10 Compression 44,770 Tension 150,306 0 FS FT Zero Axial Stress Neutral Point Depth of Zero Axial Stress Point = 150,306 = 7,708 11 ft 19.5 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) LDC FBIT WDC LDC LDC FBIT f WDC 1 s lbf lbf / ft lbf lbf / ft FBIT * S.F. f WDC 1 s 13 State of stress in pipe at the neutral point?14 At the Neutral Point: The axial stress is equal to the average of the radial and tangential stresses. r t Z 2 15 Stability Force: 0 FT FS FT FS = Ai Pi - Ao Po If FS > axial tension then the pipe may buckle. If FS < axial tension then the pipe will NOT buckle. 16 At the neutral point: FS = axial load To locate the neutral point: Plot FS vs. depth on “axial load (FT ) vs. depth plot” The neutral point is located where the lines intersect. 17 NOTE: If pi = po = p, then Fs = or, d 4 2 o di 2 p AS Fs = - AS p 18 Axial Load with FBIT = 68,000 lbf 19 Stability Analysis with FBIT = 68,00020 lbf Nonstatic Well Conditions FLUID FLOW Physical Laws Rheological Models Equations of State 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 q v 2 2.448 d q v 2 2 2.448 d 2 d1 WHERE v = average velocity, ft/s q = flow rate, gal/min d = internal diameter of pipe, in. d2 = internal diameter of outer pipe or borehole, in. d1 =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: E 2 E1 p2V2 p1V1 1 2 2 g D2 D1 v 2 v1 2 W Q In the wellbore, in many cases { Q = 0 (heat) = constant27 In practical field units this equation simplifies to: p2 p1 0.052 D2 D1 8.074 *10 4 v22 v12 DPp Dp f where p1 and p2 v1 and v2 Dpp Dpf D1 and D2 are pressures in psi is density in lbm/gal. are velocities in ft/sec. is pressure added by pump between points 1 and 2 in psi is frictional pressure loss in psi are depths in ft. 28 Determine the pressure at the bottom of the drill collars, if Dp f 1,400 psi 400 gal/min. 12 lbm/gal. D2 10,000 ft. (bottom of drill collars) D1 0 (mud pits) q IDDC 2.5 in. Dp p 3,000 psi 29 Velocity in drill collars q v2 2 2.448 d (gal/min) 2 (in ) 400 v2 26 .14 ft/sec 2 2.448 * (2.5) Velocity in mud pits, v1 0 30 p 2 p1 0.052 (D2 D1 ) 8.074 * 10 -4 ( v v ) DPp DPf 2 2 2 1 p 2 0 0.052 * 12 (10,000 - 0) - 8.074 * 10 -4 * 12 (26.14 2 0 2 ) 3,000 1,400 0 6,240 6.6 3,000 1,400 Pressure at bottom of drill collars = 7,833 psig NOTE: KE in collars 0 May be ignored in many cases31 p2 p1 0.052 ( D2 D1 ) 8.074 *10 -4 (v v ) DPp DPf 2 2 2 1 32 Fluid Flow Through Nozzle Assume: D2 D1 DPp 0 v1 0 v 2 vn DPf 0 4 p 2 p1 8.074 * 10 v and vn 2 n Dp 4 8.074 * 10 33 If DPf 0 Equation may be written as vn c d Dp 4 8.074 * 10 c d 0.95 This accounts for all the losses in the nozzle. Example: vn 0.95 1,000 305 ft/sec 4 8.074 * 10 * 12 34 35 For multiple nozzles in // Vn is the same for each nozzle even if the dn varies! This follows since Dp is the same across each nozzle. vn c d Dp 8.074 * 10 4 Δp bit q & vn 3.117 A t 8.311 * 10 -5 q2 2 d C A 2 t 36 Hydraulic Horsepower of pump putting out 400 gpm at 3,000 psi = ? Power rate of doing work F* s/t Dp * A q A PH qDp In field units: qDp HHP 1714 400 * 3,000 700hp 1714 37 What is Hydraulic Impact Force developed by bit? Consider: CD 0.95 q 400 gal/min 12 lb/gal Δp n 1,169 psi 38 Impact = rate of change of momentum q vn Dmv m Fj Dv Dt 32.17 * 60 Dt CD 0.95 Fj 0.01823 c d q Dp q 400 gal/min 12 lb/gal Δp n 1,169 psi Fj 0.01823 * 0.95 * 400 12 * 1,169 820 lbf 39