1. PETE 411 Well 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. 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

4. 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

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

7. Forces Acting on a Fluid Element
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. Incompressible fluids
If p0 = 0 (usually the case except during well control or cementing procedures)
then,

12. 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)

13. 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

14. Compressible Fluids
From Eqs. (2) and (4):
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. (i) What is pressure at 10,000 ft?
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. (iv) What is psurf if p10,000 = 8,000 psia?
Example
(iv) What is psurf if p10,000 = 8,000 psia?

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

20. Fig. 4-4. Viewing the Well as a Manometer
Pa = ?
Fig 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. 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 * = 77,100 lbs

25. Axial Forces in Drillstring
Fb = bit weight

26. Simple Example - Empty Wellbore
Drillpipe weight = 19.5 lbf/ft ,000 ft
0 lbf
195,000 lbf
OD = in
ID = in
DEPTH, ft
A = in2
AXIAL TENSION, lbf
W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf

27. Example - 15 lb/gal Mud in Wellbore
Drillpipe weight = 19.5 lbf/ft ,000 ft
- 41,100
153,900
195,000 lbf
OD = in
ID = in
DEPTH, ft
A = in2
AXIAL TENSION, lbf
F = P * A
= 7,800 * 5.265
= 41,100 lbf
Pressure at bottom = * 15 * 10,000 = 7,800 psi
W = 195, ,100 = 153,900 lbf

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

29. Anywhere in the Drill Pipe: Axial Tension = Wts
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. Example A1 Pressure at top of collars = 0.052 (15) 10,000 = 7,800 psi
Pressure at bottom of collars = (15) 10,600
= 8,268 psi
Cross-sectional area of pipe,
10,000’
10,600’

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

33. 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

34. Example 2. At 10,000 ft+ (top of collars) FT = W2 - F2 - Fb4
2. At 10,000 ft+ (top of collars)
FT = W2 - F2 - Fb
= 147 lbm/ft * 600 ft - 357,200
= 88, ,200
= -269,000 lbf
Fb = FBIT = 0 3 2 1

35. Example 3. At 10,000 ft - (bottom of drillpipe) FT = W1+W2+F1-F2-Fb4
3. At 10,000 ft - (bottom of drillpipe)
FT = W1+W2+F1-F2-Fb
= 88, lbf/in2 * 37.5in ,200
= 88, , ,200
= + 23,500 lbf 3 2 1

36. Example 4. At Surface FT = W1 + W2 + F1 - F2 - Fb
= 19.5 * 10, ,500
= 218,500 lbf
Also: FT = WAIR * BF = 283,200 *
= 218,345 lbf 3 2 1

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

38. 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]