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1、1,1.6-1,Wellbore Hydraulics, Pressure Drop Calculations,2,1.6-2 Wellbore Hydraulics,Hydrostatics Buoyancy Pipe Tension vs. Depth Effect of Mud Pressure Laminar and Turbulent Flow Pressure Drop Calculations Bingham Plastic Model API Power-Law Model,3,1.6-3 Buoyancy Force = weight of fluid displaced (
2、Archimedes, 250 BC),Figure 4-9. Hydraulic forces acting on a submerged body,4,1.6-4 Effective (buoyed) Weight,Buoyancy Factor,Valid for a solid body or an open-ended pipe!,We = buoyed weight W = weight in air Fb = buoyancy force V = volume of body rf = fluid density rs = body density,5,1.6-5 Axial F
3、orces in Drillstring,Fb = bit weight F1 & F1 are pressure forces,6,1.6-6 Axial Tension in Drill String,Example A drill string consists of 10,000 ft of 19.5 #/ft drillpipe and 600 ft of 147 #/ft drill collars suspended off bottom in 15#/gal mud (Fb = bit weight = 0). What is the axial tension in the
4、drillstring as a function of depth?,7,1.6-7 Example,Pressure at top of collars = 0.052 (15) 10,000 = 7,800 psi Pressure at bottom of collars = 0.052 (15) 10,600 = 8,268 psi Cross-sectional area of pipe,A1,10,000,10,600,8,Cross-sectional area of collars,A2,A1,1.6-8 Example contd,9,1. At 10,600 ft. (b
5、ottom of drill collars) Compressive force = p A = 357,200 lbf axial tension = - 357,200 lbf ,4,3,2,1,1.6-9 Example - contd,10,1.6-10 Example - contd,2. At 10,000 ft+ (top of collars) FT = W2 - F2 - Fb = 147 lbm/ft * 600 ft - 357,200 = 88,200 - 357,200 = -269,000 lbf,4,3,2,1,Fb = FBIT = 0,11,3. At 10
6、,000 ft - (bottom of drillpipe) FT = W1+W2+F1-F2-Fb = 88,200 + 7800 lbf/in2 * 37.5in2 - 357,200 = 88,200 + 292,500 - 357,200 = + 23,500 lbf,4,3,2,1,1.6-11 Example - contd,12,4. At Surface FT = W1 + W2 + F1 - F2 - Fb = 19.5 * 10,000 + 88,200 + 292,500 - 357,200 - 0 = 218,500 lbf Alternatively: FT = WAIR * BF = 283,200 * 0.7710 = 218,345 lbf,4,3,2,1,1.6-12 Example - contd,13,Fig. 4-11. Axial tensions as a function of depth for Example 4.9,1.6-13,14,1.6-14 Axial Load with FBIT = 68,000 lbf,
