This document provides an overview of drilling equipment used on oil rigs. It discusses various components of the rig including rig pumps, solids control equipment, air drilling systems, rotary systems, swivels, well control systems, well monitoring systems, and offshore drilling equipment. It includes examples of calculating pump factors and pump rates. Schematics and diagrams are provided to illustrate blowout preventers, casing strings, capacities, and other drilling equipment. The document is intended to teach students about the equipment used in well drilling as part of a PETE 411 course.

1. 1
PETE 411
Well Drilling
Lesson 3
The Rig - Drilling Equipment

2. 2
Contents
 Rig Pumps
 Solids Control Equipment
 Air Drilling
 The Rotary System
 The Swivel
 The Well Control System
 The Well Monitoring System
 Offshore Drilling

3. 3
Assignments:
READ: ADE Ch. 1, to p.30
HW #1: ADE 1.1, 1.2, 1.3
Due Monday, Sept. 9, 2002
at the beginning of class
NOTE: Answers in book are not
always correct...

4. 4
Schematic of
Rig
Circulating
System for
liquid
drilling fluid

5. 5

6. 6
Example 1.3
Compute the pump factor in units of
barrels per stroke for a double-acting
duplex pump having:
6.5-inch liners (dL)
2.5 inch rods (dr)
18-inch strokes (LS)
and a volumetric efficiency of 90%. (EV)

7. 7
Solution:
The pump factor for a duplex pump can
be determined using Equation 1.10
( )
( )( ) ( ) ( )[ ]
stroke/in1991
5.25.629.018
2
dd2EL
2
F
3
22
2
r
2
LVSp
=
−
π
=
−
π
=

8. 8
Recall:
There are 231 in.3
inan U. S. gallon and
42 U.S. gallons in a U.S. barrel. Thus
converting to the desired field units
yields:
1991 in.3
/stroke * gal/231 in.3
* bbl/42 gal.
= 0.2052 bbl/stroke.
Thus: Pump Factor = 0.2052 bbl/stroke

9. 9
Pump Factor = 3 * π/4 dL
2
LS EV/(231 * 42)

10. 10
Example:
Pump Factor for Triplex Pump
( )( )
bbl/stroke09442.0
.in
42231
bbl
stroke
in
90.0126
4
3
ELd
4
3
3
2
VS
2
L
=
∗





π
=
∗




π
∗=

11. 11
Example: Pump Rate
= Pump Factor * Strokes/min
= 0.09442
= 7.554 bbl/min
= 317.3 gal/min
min
stks
80
stk
bbl
∗
Pump Rate = 317 gal/min

12. 12
Hydrocyclone
• desander
• desilter
* No moving parts
* Low cost* Pressure drop
* Diameter

13. 13
Decanting Centrifuge Use?

14. 14
l
l
Use?

15. 15
Fig. 1.33
Schematic
of Rotary
System

16. 16
Fig. 1.34
Cutaway
View of
Swivel
ROTATING
* Seals
* Bearings

17. 17
Fig. 1.38
Cutaway View and
Dimensions for
Example Tool Joint
PIN
BOX
TJ
Shoulder

18. 18
Fig. 1.39
Stabilizer
* Keeps pipe in center of hole
* Aids in drilling straight hole
* Prolongs bit life

19. 19
Fig. 1.41
Kick Detection
During Drilling
Operations
1
2
3
KICK
GAIN IN PIT
VOLUME EQUAL
TO KICK VOLUME

20. 20
Fig. 1.46
Remote
Control
Panel for
operating
Blowout
PreventersCHOKE
What to do if KICK occurs?

21. 21
Fig. 1.44 Annular Blowout Preventer
DP
TJ
DC
OH
Press

22. 22
Ram Blowout Preventer

23. 23
Ram Blowout Preventer - cont’d
SHEAR / BLIND
RAM ASSEMBLY

24. 24
Fig. 1.48 Rotating Blowout Preventer

25. 25
Fig. 1.51
High-
Pressure
Circulating
System for
Well
Control
Operations
Keep BHP const.
Kick

26. 26
Fig. 1.56
Subsurface
Well
Monitoring
System
MWD

27. 27
Fig. 1.57 Spread Mooring Systems
OFFSHORE DRILLING

28. 28
Fig. 1.58
Schematic of
Equipment
for Marine
Drilling

29. 29
Fig. 1.59 Operation of a
Heave Compensator

30. 30
Fig. 1.60
Subsea
BOP
Stack

31. 31
Fig. 1.63 Subsea Equipment
Installation Procedure

32. 32
Typical Casing Strings
Water Level
Depth
Seafloor Below ML
Conductor pile 36” 30” 200’
Conductor Casing 26” 20” 1000’
Surface Casing 17 1/2” 13 3/8” 4000’

33. 33
Some Typical Casing Strings
Depth
Below ML
Conductor pile 36” 30” 200’
Conductor Casing 26” 20” 1000’
Surface Casing 17 1/2” 13 3/8” 4000’
Hole Csg. Depth

34. 34
Capacity = Area * Length

35. 35
What is the capacity of 10,000 ft of 5”
OD, 19.50 lb/ft drillpipe?
Capacity = Area * Length
Area = π/4 d2
= π/4 * 4.2762
= 14.36 in2
Length = 10,000 ft = 120,000 in
Capacity = 14.36 *120,000 in3
/(231*42 in3
/bbl)
Capacity = 177.6 bbls

36. 36
What is the displacement of 10,000 ft
of 5” OD, 19.50 lb/ft drillpipe?
Capacity = Area * Length
Area = π /4 (od2
- id2
) = π /4 * (52
- 4.2762
)
= 5.275 in2
Length = 10,000 ft = 120,000 in
Displ. = 5.276 * 120,000 in3
/(231 * 42 in3
/bbl)
Displacement = ? Bbls