PETE 625 Well Control Lesson 1 Introduction
Contents Introduction to course Basic Concepts Liquid Hydrostatics Multimedia Lesson 2. Well Control Network Places - juvkam-wold2 – multimedia – Lesson 2
Read: Watson, Chap. 1
Catalog Description PETE 625. Well Control. (3.0). Credit 3. Theory of pressure control in drilling operations and during well kicks; abnormal pressure detection and fracture gradient determination; casing setting depth selection and advanced casing design; theory supplemented on well control simulators. Prerequisite: PETE 411 3
Textbook Advanced Well Control, by David Watson, Terry Brittenham and Preston Moore. SPE Textbook Series, 2003 Class Notes and Homework Assignments can be found at: http//pumpjack.tamu.edu/~schubert 4
References – Well Control Kicks and Blowout Control, by Neal Adams and Larry Kuhlman. 2nd Editions. PennWell Publishing Company, Tulsa, OK, 1994. Blowout Prevention, by W.C. Goins, Jr. and Riley Sheffield. Practical Drilling Technology, Volume 1, 2nd Edition. Gulf Publishing Company, Houston, 1983. Advanced Blowout and Well Control, by Robert D. Grace. Gulf Publishing Company, Houston, 1994. IADC Deepwater Well Control Guidelines, Published by the International Association of Drilling Contractors, Houston, TX, 1998. Guide to Blowout Prevention, WCS Well Control School, Harvey, LA, 2000. 5
References - General Applied Drilling Engineering, by Adam T. Bourgoyne Jr., Martin E. Chenevert, Keith K. Millheim and F.S. Young Jr., Society of Petroleum Engineers, Richardson, TX, 1991. Drilling Engineering, A complete Well Planning Approach, by Neal Adams and Tommie Carrier. PennWell Publishing Company, Tulsa, OK, 1985 Practical Well Planning and Drilling Manual, by Steve Devereux. PennWell Publishing Company, Tulsa, OK, 1998.
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Grading Homework Quiz A Quiz B Project Quiz C
20% 20% 20% 20% 20%
See Next Slide for Details 7
Important Dates (tentative) QUIZ A - Week of October 11 QUIZ B - Week of November 29 Project Presentations: November 29
Week of
Quiz C - When ever WCS simulator is complete
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Your Instructor Name:
Jerome J. Schubert
Office:
501K Richardson
Phone:
862-1195
e-mail:
[email protected]
Office Hours: TR 10:00 – 11:30 am 9
Schedule Week 1
Introduction, Gas Behavior, Fluid Hydrostactics (Ch. 1)
Weeks 2&3 Weeks 4&5 Week 6 Weeks 7&8
Pore Pressure
(Ch. 2)
Fracture Pressure
(Ch. 3)
Week 9
Well Control Complications, Special Applications (Ch. 5&6)
SPE ATCE - Houston Kick Detection and Control Methods (Ch. 4)
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Schedule – cont’d Week 10
Well Control Equipment
(Ch. 7)
Week 11 Week 12 Week 13 Week 14
Offshore Operations
(Ch. 8)
Snubbing & Stripping
(Ch. 9)
Week 15
Project Presentations
Blowout Control
(Ch. 10)
Casing Seat Selection (Ch. 11) Circ. Press. Losses (Appendix A) Surge & Swab Press. (Appendix B)
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Definitions What is a Kick? An unscheduled entry of formation fluids into the wellbore of sufficient quantity to require shutting in the well
What is a Blowout? Loss of control of a kick 12
Why does a kick occur? Pressure in the wellbore is less than the pressure in the formation Permeability of the formation is great enough to allow flow A fluid that can flow is present in the formation 13
How do we prevent kicks? We must maintain the pressure in the wellbore greater than formation pressure But, We must not allow the pressure in the wellbore to exceed the fracture pressure This is done by controlling the HSP of the drilling fluid, and isolating weak formations with casing HSP = HydroStatic Pressure
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Hydrostatic Pressure, HSP HSP = 0.052 * MW * TVD HSP = Hydrostatic Pressure, psi MW = Mud Weight (density), ppg TVD = Total Vertical Depth, ft 15
HSP 10 ppg mud
TVD
HSP
=
HSP
=
HSP 16
Problem # 1 Derive the HSP equation Calculate the HSP for each of the following: 10,000 ft of 12.0 ppg mud 12,000 ft of 10.5 ppg mud 15,000 ft of 15.0 ppg mud 17
Solution to Problem # 1 Consider a column of fluid: Cross-sectional area = 1 sq.ft. Height = TVD ft Density = MW
Weight of the fluid = Vol * Density 3
3
= 1 * 1 * TVD ft * 62.4 lb/ ft * MW ppg/8.33 = 62.4 / 8.33 * MW * TVD 18
Solution, con’t. This weight is equally distributed over an area of 1 sq.ft. or 144 sq.in. Therefore, Pressure = Weight / area = 62.4 MW * TVD/(8.33*144) HSP = 0.052 * MW * TVD
W
F = PA 19
Solution, con’t. HSP = 0.052 * MW * TVD HSP1 = 0.052 * 12 * 10,000 = 6,240 psi HSP2 = 0.052 * 10.5 * 12,000 = 6,552 psi HSP3 = 0.052 * 15.0 * 15,000 = 11,700 psi 20
Terminology Pressure Pressure gradient Formation pressure (Pore) Overburden pressure Fracture pressure Pump pressure (system pressure loss)
SPP, KRP, Slow circulating pressure, kill rate pressure Surge & swab pressure SIDPP & SICP BHP
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U-Tube Concept 400 600
HSP = 5,200 psi
400
600
Mud HSP =4,800 psi
HSP = 5,200 psi Mud HSP =4,800 psi
Influx HSP =200 psi
Influx HSP =200 psi 5,600
5,600
5,600
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More Terminology Capacity of:
Rig Pumps
casing
Duplex pump
hole
Triplex pump
drillpipe
Annular capacity
KWM, kill weight mud Fluid Weight up
Displacement of: Drillpipe Drill collars 23
Problem # 2 Calculate the mud gradient for 15.0 ppg mud G15 = 0.052 * MW = 0.052 * 15 = 0.780 psi/ft Calculate the HSP of 15,000’ of 15 ppg mud HSP = 0.780 * 15,000 = 11,700 psi
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Problem # 3 The top 6,000 ft in a wellbore is filled with fresh water, the next 8,000 with 11 ppg mud, and the bottom 16,000 ft is filled with 16 ppg mud. (i) What is the BHP? (ii) What is the pressure 1/2 way to bottom? (iii) Plot the mud density vs. depth (iv) Plot the mud gradient vs. depth (v) Plot the pressure vs. depth 25
Problem # 3 solution (i) BHP = 0.052 * [(8.33 * 6,000)
+ (11 * 8,000) + (16 * 16,000)] = 20,487 psi
(ii) Pressure 1/2 way down (at 15,000 ft) = 0.052 * [(8.33 * 6,000)
+ (11 * 8,000) + (16 * 1,000)] = 8,007 psi
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Problem # 3 solution (iii) Plot MW vs. Depth
Mud Density, ppg 0 5 10 15 20 0 5,000 D 10,000 e 15,000 p t 20,000 h 25,000 30,000
8.33
11.0
16.0
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Problem # 3 solution (iv) Plot mud gradient vs. Depth Depth Gradient ft psi/ft 0-6,000 0.433 6,000-14,000 0.572 14,000-TD 0.832
Mud Gradient, psi/ft 0 0.2 0.4 0.6 0.8 0.9 0 5,000 D 10,000 e 15,000 p t 20,000 h 25,000 30,000
0.433
0.572
0.832
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Problem # 3 solution (iv) Plot HSP vs. Depth ft
psi
@ 6,000 2,599 @14,000 7,175 @ 30,000 20,487
0
Mud Pressure, kpsi 8 5 10 15 20
5,000
2,599 psi
10,000
7,175 psi
D 15,000 e p 20,000 t 25,000 h 30,000
20,487 psi
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Addition of Weight Material The amount of barite required to raise the density of one barrel of mud from MW1 to MW2, ppg
1,490 ( MW2 − MW1 ) WB = ( 35.4 − MW2 ) where WB = Barite Required, lb/bbl MW1 = Old Mud Density, ppg MW2 = New Mud Density, ppg 1,490 = Wt. of 1 bbl of Barite, lbs 35.4 = Wt. of 1 gal of Barite, lbs 30
Problem # 4, Derive Barite Eq. Consider one bbl of mud of density, MW1, add WB lbs of barite to increase the mud density to MW2. Wt, lb Old Mud Barite Mixture
Vol, bbl
42 * MW1 1 WB (WB lbs / 1,490 lb/bbl) WB + 42 MW1
1 + (WB / 1,490)
Density of Mixture = total weight / total volume
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Problem # 4 New Density = Weight / Volume MW2
= (WB+42 MW1 lbs) / {[1+(WB/1,490)bbl]*42 gal/bbl}
42 MW2 [1+(WB/1,490)] = WB+42 MW1
lbs
WB [(MW2 / 35.4) -1] = 42 MW1 – 42 MW2 WB(MW2 - 35.4) = (42 * 35.4) * (MW1 - MW2)
1,490 ( MW2 − MW1 ) WB = ( 35.4 − MW2 ) 32
Stopping an Influx 1. Increase Pressure at Surface 2. Increase Annular Friction 3. Increase Mud Weight 33
Depth
Stopping an Influx
Mud Hydrostatic Pressure
Pressure
34
Depth
Stopping an Influx – Soln.1
Mud Hydrostatic Pressure Pressure
35
Depth
Stopping an Influx – Soln.2
Mud Hydrostatic Pressure Pressure
36
Depth
Stopping an Influx – Soln.3
Mud Hydrostatic Pressure Pressure
37