1. Introduction

PETE 625Well ControlLesson 1Introduction

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

Catalog DescriptionPETE 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

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

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.

References - GeneralApplied 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.

Grading Homework20% Quiz A20% Quiz B20% Project20% Quiz C 20%

See Next Slide for Details

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

Your Instructor Name:Jerome J. Schubert

Office:501K Richardson

Phone:862-1195

e-mail:[email protected]

Office Hours: TR 10:00 11:30 am

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

Schedule contdWeek 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

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

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

How do we prevent kicks?We must maintain the pressure in the wellbore greater than formation pressureBut,We must not allow the pressure in the wellbore to exceed the fracture pressureThis is done by controlling the HSP of the drilling fluid, and isolating weak formations with casingHSP = HydroStatic Pressure

Hydrostatic Pressure, HSPHSP = 0.052 * MW * TVD

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

HSPTVD10 ppg mudHSP =HSP=HSP

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

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

Solution, cont.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 * TVDWF = PA

Solution, cont. 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

TerminologyPressurePressure gradientFormation pressure (Pore)Overburden pressureFracture pressurePump pressure (system pressure loss)SPP, KRP, Slow circulating pressure, kill rate pressureSurge & swab pressureSIDPP & SICPBHP

U-Tube ConceptHSP = 5,200 psi5,6005,6005,600400400600600HSP = 5,200 psiMud HSP =4,800 psiMud HSP =4,800 psiInflux HSP =200 psiInflux HSP =200 psi

More TerminologyCapacity of:casingholedrillpipeAnnular capacityDisplacement of:DrillpipeDrill collarsRig PumpsDuplex pumpTriplex pumpKWM, kill weight mudFluid Weight up

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

Problem # 3The 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

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

Problem # 3 solution (iii) Plot MW vs. DepthDepth

05,00010,00015,00020,00025,00030,000Mud Density, ppg0 5 10 15 208.33

11.0

16.0

Problem # 3 solution(iv) Plot mud gradient vs. Depth

Depth Gradient ft psi/ft0-6,000 0.4336,000-14,000 0.572 14,000-TD 0.832Depth

05,00010,00015,00020,00025,00030,000Mud Gradient, psi/ft0 0.2 0.4 0.6 0.8 0.9 0.433

0.572

0.832

Problem # 3 solution(iv) Plot HSP vs. Depth

ft psi@ 6,000 2,599 @14,000 7,175 @ 30,000 20,487Depth

05,00010,00015,00020,00025,00030,000Mud Pressure, kpsi8 5 10 15 20 2,599 psi 7,175 psi 20,487 psi

Addition of Weight MaterialThe amount of bariterequired to raise thedensity of one barrelof mud from MW1 to MW2, ppg

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 Mud42 * MW11Barite WB(WB lbs / 1,490 lb/bbl)

Mixture WB + 42 MW1 1 + (WB / 1,490)

Density of Mixture = total weight / total volume

Problem # 4New Density = Weight / VolumeMW2 = (WB+42 MW1 lbs) / {[1+(WB/1,490)bbl]*42 gal/bbl}42 MW2 [1+(WB/1,490)] = WB+42 MW1 lbsWB [(MW2 / 35.4) -1] = 42 MW1 42 MW2 WB(MW2 - 35.4) = (42 * 35.4) * (MW1 - MW2)

Stopping an InfluxIncrease Pressure at Surface

Increase Annular Friction

Increase Mud Weight

Stopping an InfluxPressureDepthMud Hydrostatic Pressure

Stopping an Influx Soln.1PressureDepthMud Hydrostatic Pressure

Stopping an Influx Soln.2PressureDepthMud Hydrostatic Pressure

Stopping an Influx Soln.3PressureDepthMud Hydrostatic Pressure