PETE 625Well Control

Lesson 1Introduction

Contents

Introduction to course

Basic Concepts

Liquid Hydrostatics

Multimedia Lesson 2. Well ControlNetwork Places - juvkam-wold2 – multimedia – Lesson 2

Read: Watson, Chap. 1

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

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

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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.

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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 20% Quiz A 20% Quiz B 20% Project 20% Quiz C 20%

See Next Slide for Details

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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

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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

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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)

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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

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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

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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

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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 pressureThis 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

16

HSP

TVD

10 ppg mud

HSP = HSP = HSP

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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

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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

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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

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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

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Terminology

PressurePressure gradientFormation pressure

(Pore)Overburden

pressureFracture pressurePump pressure

(system pressure loss)

SPP, KRP, Slow circulating pressure, kill rate pressure

Surge & swab pressure

SIDPP & SICPBHP

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U-Tube Concept

HSP = 5,200 psi

5,600 5,600 5,600

400

400600 600

HSP = 5,200 psi

Mud HSP =4,800 psi

Mud HSP =4,800 psi

Influx HSP =200 psi

Influx HSP =200 psi

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More Terminology

Capacity of:casing

hole

drillpipe

Annular capacity

Displacement of:Drillpipe

Drill collars

Rig PumpsDuplex pump

Triplex pump

KWM, kill weight mud

Fluid Weight up

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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

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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

Depth

0

5,000

10,000

15,000

20,000

25,000

30,000

Mud Density, ppg0 5 10 15 20

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

Depth

0

5,000

10,000

15,000

20,000

25,000

30,000

Mud Gradient, psi/ft0 0.2 0.4 0.6 0.8 0.9

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

Depth

0

5,000

10,000

15,000

20,000

25,000

30,000

Mud Pressure, kpsi8 5 10 15 20

2,599 psi

7,175 psi

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

lbsBarite,ofgal1ofWt.35.4

lbsBarite,ofbbl1ofWt.1,490

ppgDensity,MudNewMW

ppgDensity,MudOldMW

lb/bblRequired,BariteW

where

2

1

B

2

12B MW35.4

MWMW1,490W

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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 1Barite WB (WB lbs / 1,490 lb/bbl)

Mixture 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)

2

12B MW35.4

MWMW1,490W

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Stopping an Influx

1. Increase Pressure at Surface

2. Increase Annular Friction

3. Increase Mud Weight

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Stopping an Influx

Pressure

Dep

th

Mud Hydrostatic Pressure

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Stopping an Influx – Soln.1

Pressure

Dep

th

Mud Hydrostatic Pressure

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Stopping an Influx – Soln.2

Pressure

Dep

th

Mud Hydrostatic Pressure

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Stopping an Influx – Soln.3

Pressure

Dep

th

Mud Hydrostatic Pressure