PETE 411 Well Drilling

# PETE 411 Well Drilling

## PETE 411 Well Drilling

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##### Presentation Transcript

1. PETE 411Well Drilling Lesson 9 Drilling Hydraulics - Hydrostatics

2. 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

3. 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

4. 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

5. 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

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

7. Forces Acting on a Fluid Element F1 = F2 = F3 = FWV = specific wt. of the fluid

8. Pressures in a fluid column • At equilibrium, S F = 0 0 = F1 + F2 + F3 (p = rgh)

9. Incompressible Fluids Integrating,

10. Incompressible Fluids In field units, 1’ x 1’ x 1’ cube

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

12. 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)

13. 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

14. From Eqs. (2) and (4): Compressible Fluids Integrating, Assumptions?

15. 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?

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

17. Example cont’d (ii) What is density at surface? (iii) What is density at 10,000 ft?

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

19. Fig. 4-3. A Complex Liquid Column

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

21. Figure 4.4

22. Buoyancy Force = weight of fluid displaced (Archimedes, 250 BC) Figure 4-9. Hydraulic forces acting on a foreign body

23. Effective (buoyed) Weight Buoyancy Factor Valid for a solid body or an open-ended pipe!

24. 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

25. Axial Forces in Drillstring Fb = bit weight

26. 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

27. 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

28. Anywhere in the Drill Collars:Axial Tension = Wt. - Pressure Force - Bit Wt.

29. Anywhere in the Drill Pipe:Axial Tension = Wts. - Pressure Forces - Bit Wt. FT

30. 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?

31. 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’

32. A1 Example Cross-sectional area of collars, A2

33. 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

34. 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

35. 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

36. 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

37. 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]