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PETE 625 Well Control

Contents. Introduction to course Basic Concepts Liquid Hydrostatics Multimedia Lesson 2. Well ControlNetwork Places - juvkam-wold2

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PETE 625 Well Control

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    1. PETE 625 Well Control Lesson 1 Introduction

    2. Contents Introduction to course Basic Concepts Liquid Hydrostatics Multimedia Lesson 2. Well Control Network Places - juvkam-wold2 – multimedia – Lesson 2 Read: Watson, Chap. 1

    3. 3 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

    4. 4 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

    5. 5 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.

    6. 6 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.

    7. 7 Grading Homework 20% Quiz A 20% Quiz B 20% Project 20% Quiz C 20% See Next Slide for Details

    8. 8 Important Dates (tentative) QUIZ A - Week of October 11 QUIZ B - Week of November 29 Project Presentations: Week of November 29 Quiz C - When ever WCS simulator is complete

    9. 9 Your Instructor Name: Jerome J. Schubert Office: 501K Richardson Phone: 862-1195 e-mail: jschubert@tamu.edu Office Hours: TR 10:00 – 11:30 am

    10. 10 Schedule Week 1 Introduction, Gas Behavior, Fluid Hydrostactics (Ch. 1) Weeks 2&3 Pore Pressure (Ch. 2) Weeks 4&5 Fracture Pressure (Ch. 3) Week 6 SPE ATCE - Houston Weeks 7&8 Kick Detection and Control Methods (Ch. 4) Week 9 Well Control Complications, Special Applications (Ch. 5&6)

    11. 11 Schedule – cont’d Week 10 Well Control Equipment (Ch. 7) Week 11 Offshore Operations (Ch. 8) Week 12 Snubbing & Stripping (Ch. 9) Week 13 Blowout Control (Ch. 10) Week 14 Casing Seat Selection (Ch. 11) Circ. Press. Losses (Appendix A) Surge & Swab Press. (Appendix B) Week 15 Project Presentations

    12. 12 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

    13. 13 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

    14. 14 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

    15. 15 Hydrostatic Pressure, HSP HSP = 0.052 * MW * TVD HSP = Hydrostatic Pressure, psi MW = Mud Weight (density), ppg TVD = Total Vertical Depth, ft

    16. 16 HSP

    17. 17 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

    18. 18 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 = 1 * 1 * TVD ft3 * 62.4 lb/ ft3 * MW ppg/8.33 = 62.4 / 8.33 * MW * TVD

    19. 19 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

    20. 20 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

    21. 21 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

    22. 22 U-Tube Concept

    23. 23 More Terminology Capacity of: casing hole drillpipe Annular capacity Displacement of: Drillpipe Drill collars Rig Pumps Duplex pump Triplex pump KWM, kill weight mud Fluid Weight up

    24. 24 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

    25. 25 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

    26. 26 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

    27. 27 Problem # 3 solution (iii) Plot MW vs. Depth

    28. 28

    29. 29

    30. 30 Addition of Weight Material The amount of barite required to raise the density of one barrel of mud from MW1 to MW2, ppg

    31. 31 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 Vol, bbl Old Mud 42 * MW1 1 Barite WB (WB lbs / 1,490 lb/bbl) Mixture WB + 42 MW1 1 + (WB / 1,490) Density of Mixture = total weight / total volume

    32. 32 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)

    33. 33 Stopping an Influx Increase Pressure at Surface Increase Annular Friction Increase Mud Weight

    34. 34 Stopping an Influx

    35. 35 Stopping an Influx – Soln.1

    36. 36 Stopping an Influx – Soln.2

    37. 37 Stopping an Influx – Soln.3

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