PETE 411 Well 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

HW #4 ADE #1.18, 1.19, 1.24 Due Monday, Sept 23, 2002 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

Drilling Hydraulics Applications 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

Fig. 4-2. The Well Fluid System ppore < pmud < pfrac Well Control Fig. 4-2. The Well Fluid System

Forces Acting on a Fluid Element 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

Incompressible fluids If p0 = 0 (usually the case except during well control or cementing procedures) then,

Compressible Fluids …………… (1) …………… (2) …… (3) But, …… (4) from (3) p = pressure of gas, psia V = gas volume, gal Z = gas deviation factor n = moles of gas R = universal gas constant = 80.3 T = temperature, R r = density, lbm/gal M = gas molecular wt. m = mass of gas …………… (1) …………… (2) …… (3) But, …… (4) from (3)

Compressible Fluids r = density, lbm/gal p = pressure of gas, psia V = gas volume, gal Z = gas deviation factor n = moles of gas R = universal gas constant, = 80.3 T = temperature, oR r = density, lbm/gal M = gas molecular wt. m = mass of gas, lbm

Compressible Fluids From Eqs. (2) and (4): 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?

(i) What is pressure at 10,000 ft? 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?

(iv) What is psurf if p10,000 = 8,000 psia? Example (iv) What is psurf if p10,000 = 8,000 psia?

Fig. 4-3. A Complex Liquid Column

Fig. 4-4. Viewing the Well as a Manometer Pa = ? Fig. 4-4. Viewing the Well as a Manometer

Figure 4.4

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!

For steel, immersed in mud, the buoyancy factor is: 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

Simple Example - Empty Wellbore Drillpipe weight = 19.5 lbf/ft 10,000 ft 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

Example - 15 lb/gal Mud in Wellbore Drillpipe weight = 19.5 lbf/ft 10,000 ft - 41,100 153,900 195,000 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 Anywhere in the Drill Collars: Axial Tension = Wt. - Pressure Force - Bit Wt.

Anywhere in the Drill Pipe: Axial Tension = Wts 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?

Example A1 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’

Cross-sectional area of collars, Example Cross-sectional area of collars, A2

Example 1. At 10,600 ft. (bottom of drill collars) 4 1. At 10,600 ft. (bottom of drill collars) Compressive force = pA = 357,200 lbf [ axial tension = - 357,200 lbf ] 3 2 1

Example 2. At 10,000 ft+ (top of collars) FT = W2 - F2 - Fb 4 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 Fb = FBIT = 0 3 2 1

Example 3. At 10,000 ft - (bottom of drillpipe) FT = W1+W2+F1-F2-Fb 4 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 3 2 1

Example 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 3 2 1

Fig. 4-11. Axial tensions as a function of depth for Example 4.9

Example - Summary 1. At 10,600 ft FT = -357,200 lbf [compression] 3. At 10,000 - ft FT = +23,500 lbf [tension] 4. At Surface FT = +218,500 lbf [tension]