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## PETE 411 Well Drilling

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**PETE 411Well Drilling**Lesson 9 Drilling Hydraulics - Hydrostatics**Drilling Hydraulics - Hydrostatics**• Hydrostatic Pressure in Liquid Columns • Hydrostatic Pressure in Gas Columns • Hydrostatic Pressure in Complex Columns • Forces on Submerged Body • Effective (buoyed) Weight of Submerged Body • Axial Tension in Drill String sA = FA/A**Read:Applied Drilling Engineering, Ch.4 (Drilling**Hydraulics) to p. 125 HW #4 ADE #1.18, 1.19, 1.24 Due Monday, Sept 23, 2002**WHY?**Drilling Hydraulics Applications • Calculation of subsurface hydrostatic pressures that may tend to burst or collapse well tubulars or fracture exposed formations • Several aspects of blowout prevention • Displacement of cement slurries and resulting stresses in the drillstring**Drilling Hydraulics Applications cont’d**• Bit nozzle size selection for optimum hydraulics • Surge or swab pressures due to vertical pipe movement • Carrying capacity of drilling fluids**ppore < pmud < pfrac**Well Control Fig. 4-2. The Well Fluid System**Forces Acting on a Fluid Element**F1 = F2 = F3 = FWV = specific wt. of the fluid**Pressures in a fluid column**• At equilibrium, S F = 0 0 = F1 + F2 + F3 (p = rgh)**Incompressible Fluids**Integrating,**Incompressible Fluids**In field units, 1’ x 1’ x 1’ cube**p0**D Incompressible fluids p If p0 = 0 (usually the case except during well control or cementing procedures) then,**T = temperature, R**r = density, lbm/gal M = gas molecular wt. m = mass of gas Compressible Fluids …………… (1) …………… (2) …… (3) But, p= pressure of gas, psia V= gas volume, gal Z = gas deviation factor n = moles of gas R = universal gas constant = 80.3 …… (4) from (3)**Compressible Fluids**T = temperature, oR r = density, lbm/gal M = gas molecular wt. m = mass of gas, lbm p= pressure of gas, psia V= gas volume, gal Z = gas deviation factor n = moles of gas R = universal gas constant, = 80.3**From Eqs. (2) and (4):**Compressible Fluids Integrating, Assumptions?**Example**Column of Methane (M = 16) Pressure at surface = 1,000 psia Z=1, T=140 F (i) What is pressure at 10,000 ft? (ii) What is density at surface? (iii) What is density at 10,000 ft? (iv) What is psurf if p10,000 = 8,000 psia?**Example (i)**(i) What is pressure at 10,000 ft?**Example cont’d**(ii) What is density at surface? (iii) What is density at 10,000 ft?**Example**(iv) What is psurf if p10,000 = 8,000 psia?**Pa = ?**Fig. 4-4. Viewing the Well as a Manometer**Buoyancy Force = weight of fluid displaced (Archimedes, 250**BC) Figure 4-9. Hydraulic forces acting on a foreign body**Effective (buoyed) Weight**Buoyancy Factor Valid for a solid body or an open-ended pipe!**Example**For steel, immersed in mud, the buoyancy factor is: A drillstring weighs 100,000 lbs in air. Buoyed weight = 100,000 * 0.771 = 77,100 lbs**Axial Forces in Drillstring**Fb = bit weight**Drillpipe weight = 19.5 lbf/ft 10,000 ft**Simple Example - Empty Wellbore 0 lbf 195,000 lbf OD = 5.000 in ID = 4.276 in DEPTH, ft A = 5.265 in2 AXIAL TENSION, lbf W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf**Drillpipe weight = 19.5 lbf/ft 10,000 ft**Example - 15 lb/gal Mud in Wellbore - 41,100 0 153,900 195,000 lbf OD = 5.000 in ID = 4.276 in DEPTH, ft A = 5.265 in2 AXIAL TENSION, lbf F = P * A = 7,800 * 5.265 = 41,100 lbf Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psi W = 195,000 - 41,100 = 153,900 lbf**Anywhere in the Drill Collars:Axial Tension = Wt. - Pressure**Force - Bit Wt.**Anywhere in the Drill Pipe:Axial Tension = Wts. - Pressure**Forces - Bit Wt. FT**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 drillstring as a function of depth?**A1**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, 10,000’ 10,600’**A1**Example Cross-sectional area of collars, A2**4**Example 3 2 1. At 10,600 ft. (bottom of drill collars) Compressive force = pA = 357,200 lbf [ axial tension = - 357,200 lbf ] 1**4**Example Fb = FBIT = 0 3 2 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 1**4**Example 3 2 3.At 10,000 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 1**4**Example 3 2 4. At Surface FT = W1 + W2 + F1 - F2 - Fb = 19.5 * 10,000 + 23,500 = 218,500 lbf Also: FT = WAIR * BF = 283,200 * 0.7710 = 218,345 lbf 1**Fig. 4-11. Axial tensions as a function of depth for Example**4.9**Example - Summary**1. At 10,600 ftFT = -357,200 lbf [compression] 2. At 10,000 + ftFT = -269,000 lbf [compression] 3. At 10,000 - ftFT = +23,500 lbf [tension] 4. At SurfaceFT = +218,500 lbf [tension]