Drilling Engineering Fundamentals

Drilling Drilling Engineering Engineering Fundamentals Fundamentals Associate Professor Jorge H.B. Sampaio Jr., PhD Curtin University of Technology Department of Petroleum Engineering [email protected] April 3, 2007

2

Contents 1 Introduction

1–1

1.1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–1 1.2 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–1 1.3 Drilling Rig Types . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–3 1.4 Personnel at Rig Site . . . . . . . . . . . . . . . . . . . . . . . . . 1–5 1.5 Miscellaneous

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–7

2 Rotary Drilling System

2–1

2.1 Power System . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–5 2.1.1 Energy, Work, rk, and Efficie ciency . . . . . . . . . . . . . . . . 2–6 2.2 Hoisting System . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–8 2.2.1 The Derrick . . . . . . . . . . . . . . . . . . . . . . . . . . 2–10 2.2.2 The Drawworks . . . . . . . . . . . . . . . . . . . . . . . . 2–11 2.2.3 The Block & Tackle . . . . . . . . . . . . . . . . . . . . . . 2–12 2.2.4 Load Applied to the the Derric rick . . . . . . . . . . . . . . . . . 2–16 2.3 Dril rilling Fluid Circul culatio tion Syste stem . . . . . . . . . . . . . . . . . . 2–18 2.3.1 Mud Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . 2–20 2.3.2 Solids Control Equipment . . . . . . . . . . . . . . . . . . 2–25 2.3.3 .3.3 Trea reatme tment and Mixi Mixin ng Equip quipm ment . . . . . . . . . . . . . . 2–30 –30 2.4 The Rotar y System . . . . . . . . . . . . . . . . . . . . . . . . . . 2–33 2.4.1 Swivel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–33 2.4. 2.4.2 2 Kelly elly,, Kelly elly Valve alves, s, and and Kelly elly Save Saverr Sub Sub . . . . . . . . . . 2–33 2–33 2.4.3 Rotary Table and Components . . . . . . . . . . . . . . . 2–36 2.5 Well Control System . . . . . . . . . . . . . . . . . . . . . . . . . 2–38 2.6 Well Monitoring System . . . . . . . . . . . . . . . . . . . . . . . 2–43 3 Drillstring Tubulars and Equipment

i

3–1

Curtin University of Technology Department of Petroleum Engineering

Master of Petroleum Well Engineering Drilling Engineering Fundamentals

3.1 Drill Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–1 3.1.1 Drill Pipe Elevator . . . . . . . . . . . . . . . . . . . . . . 3–5 3.2 Drill Collars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–5 3.3 Heavy Wall Drill Pipes . . . . . . . . . . . . . . . . . . . . . . . . 3–6 3.4 Special Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–7 3.4.1 Stabilizers . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–7 3.4.2 Reamers . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–9 3.4.3 Hole–openers . . . . . . . . . . . . . . . . . . . . . . . . . 3–9 3.5 Conn onnecti ection ons s Make–up –up and and Brea reak–o k–out . . . . . . . . . . . . . . . 3–10 –10 3.5. 3.5.1 1 Maxi Maximu mum m Heig Height ht of Tool ool Join Jointt Shou Should lder ers s . . . . . . . . . . 3–11 3–11 3.5.2 Make–up Torque . . . . . . . . . . . . . . . . . . . . . . . 3–13 3.6 Drill Bit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13 3.7 Other Dril rillstrin ring Equipment . . . . . . . . . . . . . . . . . . . . . 3–14 3.7.1 Top Drive . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–14 3.7.2 Bottom Hole Motors . . . . . . . . . . . . . . . . . . . . . 3–15 4 Introduction to Hydraulics

4–1

4.1 Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . . 4–1 4.1. 4.1.1 1 Hydr Hydros osta tati tic c Pres Pressu sure re for Inco Incomp mpre ressi ssib ble Flui Fluids ds . . . . . . 4–2 4–2 4.1. 4.1.2 2 Hydr Hydros osta tati tic c Pres Pressu sure re for Comp Compre ress ssib ible le Flui Fluids ds . . . . . . . 4–4 4–4 4.2 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–7 5 Drillstring Design

5–1

5.1 5.1 Leng Length th of Drill Drill Coll Collar ars s – Neut Neutra rall Point oint Calc Calcul ulat atio ion n . . . . . . . . . 5–1 5–1 5.2 Design Design for for Tensile ensile Force Force,, Torque orque,, Burst, Burst, and Collap Collapse se . . . . . . 5–6 5.2.1 Maximum Tensile Force . . . . . . . . . . . . . . . . . . . 5–6 5.2.2 Maximum Torque . . . . . . . . . . . . . . . . . . . . . . . 5–9 5.2.3 5.2.3 Interna Internall (Burst) (Burst) and Externa Externall (Colla (Collapse pse)) Pressu Pressures res . . . . 5–10 5–10 5.2.4 Dril rillstrin ring Elongation 6 Drilling Hydraulics

. . . . . . . . . . . . . . . . . . . . 5–12 6–1

6.1 Mass and Energy Balance . . . . . . . . . . . . . . . . . . . . . . 6–1 6.1.1 Mass Conserva rvation 6.1.2 Energy Conserva rvation

. . . . . . . . . . . . . . . . . . . . . 6–2 . . . . . . . . . . . . . . . . . . . . 6–3

6.2 Flow Through Bit Nozzles . . . . . . . . . . . . . . . . . . . . . . 6–6



6.2.1 Pressure Drop Acro cross the Bit . . . . . . . . . . . . . . . . 6–6 6.2.2 Hydraulic Power Across the Bit . . . . . . . . . . . . . . . 6–8 6.2.3 Impact Force of the Jets . . . . . . . . . . . . . . . . . . . 6–8 6.3 Required Hydraulic Power . . . . . . . . . . . . . . . . . . . . . . 6–10 6.4 Bit Hydraulics Optimization . . . . . . . . . . . . . . . . . . . . . 6–11 6.4.1 .4.1 Nozzl ozzle e Size ize Sele electio ction n Crit Crite eria ria . . . . . . . . . . . . . . . . 6–13 –13 6.4.2 Graphical Analysis . . . . . . . . . . . . . . . . . . . . . . 6–16 7 Introduction to Drilling Fluids

7–1

7.1 Functions of Drilling Fluids . . . . . . . . . . . . . . . . . . . . . . 7–1 7.2 Types of Drilling Fluid . . . . . . . . . . . . . . . . . . . . . . . . 7–2 7.2.1 Water–Base Fluids . . . . . . . . . . . . . . . . . . . . . . 7–3 7.2.2 Oil–Base Muds . . . . . . . . . . . . . . . . . . . . . . . . 7–6 7.2.3 Synthetic Fluids . . . . . . . . . . . . . . . . . . . . . . . 7–7 7.2.4 Aerated Fluids . . . . . . . . . . . . . . . . . . . . . . . . 7–7 7.3 Laboratory Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–7 7.3.1 Water–Base Mud Tests . . . . . . . . . . . . . . . . . . . 7–7 7.3.2 Oil-Base Mud Testing . . . . . . . . . . . . . . . . . . . . 7–12 7.4 Fluid luid Dens Densit ity y and Visco iscosi sity ty Calc Calcul ula atio tions . . . . . . . . . . . . . . 7–13 –13 7.4.1 Density Calculations . . . . . . . . . . . . . . . . . . . . . 7–14 7.4.2 Density Treatment . . . . . . . . . . . . . . . . . . . . . . 7–15 7.4.3 Initia tial Viscosity Treatment . . . . . . . . . . . . . . . . . . 7–19 8 Rheology and Rheometry

8–1

8.1 Rheo heolog logica ical Cla Classifi ssifica cati tio on of Fluid luids s . . . . . . . . . . . . . . . . . 8–1 8.2 Rheometr y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–4 8.2.1 Visco scosit sity of Newton tonian Fluids . . . . . . . . . . . . . . . . 8–5 8.2. 8.2.2 2 Param aramet eter ers s of Bing Bingha ham– m–Pl Plas asti tic c Mode Modell Flui Fluids ds . . . . . . . 8–5 8–5 8.2. 8.2.3 3 Param aramet eter ers s of Power– wer–La Law w Mode Modell Flui Fluids ds . . . . . . . . . . 8–5 8–5 8.2.4 Gel Strength . . . . . . . . . . . . . . . . . . . . . . . . . 8–6 9 Flow in Pipes and Annuli

9–1

9.1 Laminar Flow in Pipes and Annuli . . . . . . . . . . . . . . . . . . 9–1 9.1.1 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . 9–2 9.1.2 Continuity Equations . . . . . . . . . . . . . . . . . . . . . 9–3



9.1. 9.1.3 3 Newt Newton onia ian n Flow Flow in Pipe Pipes s – Poise oiseui uille lle’’s Equa Equati tion on . . . . . 9–5 9–5 9.1.4 Newtonia Newtonian n Flow Flow in Conce Concentric ntric Annu Annulili – Lamb’s Lamb’s Equation Equation . 9–6 9.1. 9.1.5 5 Slot Slot Appr Appro oxima ximati tion on for Newt Newton onia ian n Flui Fluids ds . . . . . . . . . . 9–9 9–9 9.1.6 9.1.6 Pressu Pressure re Drop Drop Gradie Gradient nt for for Non– Non–Ne Newto wtonia nian n Fluids Fluids . . . . 9–10 9–10 9.2 Turb urbule ulent Flow low in Pipe ipes and and Annuli . . . . . . . . . . . . . . . . . 9–12 –12 9.2. 9.2.1 1 Turb urbulen ulentt Flow Flow of Newt Newton onia ian n Flui Fluids ds in Pipe Pipes s . . . . . . . . 9–12 9–12 9.2.2 9.2.2 Criterio Criterion n for for Lamina Laminarr – Transiti ransition on – Turbule urbulent nt Flow Flow . . . . 9–19 9–19 9.2.3 Other Other Geometri Geometries es – Turbulen urbulentt Flow Flow in Annuli Annuli (Newt (Newtonia onian) n) 9–20 9.2. 9.2.4 4 Turb urbulen ulentt Flow Flow for Non– Non–Ne Newt wton onia ian n Flui Fluids ds . . . . . . . . . 9–22 9–22 10 Drilling Bits

10–1

10.1 Drill Bit Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1 10.1.1 Ro Roller Cone Bit . . . . . . . . . . . . . . . . . . . . . . . . 10–2 10.1.2 Ai Air Drilling Bits . . . . . . . . . . . . . . . . . . . . . . . . 10–8 10.1 0.1.3 Fix Fixe ed Cutte tter Bits its (Dra (Drag g Bits Bits)) . . . . . . . . . . . . . . . . . 10–8 0–8 10.2 Bit Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–15 10.2 0.2.1 PDC PDC Bit Bit Class lassifi ifica cati tio on Syste ystem m . . . . . . . . . . . . . . . 10–1 0–18 10.3 Drill rill Bit Bit Sele Select ctio ion n and Eva Evaluat luatio ion n . . . . . . . . . . . . . . . . . . 10–2 0–20 10.3.1 T To ooth Wear . . . . . . . . . . . . . . . . . . . . . . . . . . 10–21 10.3.2 Be Bearing Wear . . . . . . . . . . . . . . . . . . . . . . . . . 10–21 10.3.3 G Ga age Wear . . . . . . . . . . . . . . . . . . . . . . . . . . 10–22 10.4 10.4 Fac Facto tors rs that that Affe Affect ct the the Rate Rate Of Penet enetra rati tion on . . . . . . . . . . . . . 10–2 10–22 2 10.4.1 Bi Bit Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–22 10.4 0.4.2 For Forma mati tion on Cha Charact racter eris isti tics cs . . . . . . . . . . . . . . . . . . 10–2 0–22 10.4.3 Dri Drillling Fluid Properti rties . . . . . . . . . . . . . . . . . . . 10–23 10.4.4 Op Operating Conditions . . . . . . . . . . . . . . . . . . . . 10–25 A Drill Pipe Dimensions (as in API RP7C)

A–1

List of Figures 1.1 Rig Classification. . . . . . . . . . . . . . . . . . . . . . . . . . . 1–3 2.1 Typical rig components. . . . . . . . . . . . . . . . . . . . . . . . 2–1 2.2 A simplified drillstring. . . . . . . . . . . . . . . . . . . . . . . . . 2–3 2.3 Making a connection.

. . . . . . . . . . . . . . . . . . . . . . . . 2–4

2.4 Rig crew setting the slips. . . . . . . . . . . . . . . . . . . . . . . 2–4 2.5 Removing ing one stand of dril rillstri strin ng. . . . . . . . . . . . . . . . . . . 2–5 2.6 Typical hoisting system. . . . . . . . . . . . . . . . . . . . . . . . 2–9 2.7 Stan tand of doubles alon long the mast. . . . . . . . . . . . . . . . . . . 2–10 2.8 Onshore rig drawworks. . . . . . . . . . . . . . . . . . . . . . . . 2–11 2.9 Brak rake belts elts and and magn agnifica ificati tio on lin linkag kage. . . . . . . . . . . . . . . . 2–11 –11 2.10 Drawworks rks schematics. . . . . . . . . . . . . . . . . . . . . . . . 2–12 2.11 .11 Fo Forces rces acti actin ng in the the bloc lock–ta k–tac ckle kle. . . . . . . . . . . . . . . . . . . 2–13 –13 2.12 Derrick floor plan. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–17 2.13 A swivel. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–19 2.14 Rig circulation system. . . . . . . . . . . . . . . . . . . . . . . . . 2–20 2.15 Duplex pumps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22 2.16 Triplex pumps. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22 2.17 Surge dampener. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–24 2.18 Solids control system. . . . . . . . . . . . . . . . . . . . . . . . . 2–25 2.19 Shale shaker configurations.

. . . . . . . . . . . . . . . . . . . . 2–26

2.20 A two–scree reen shale shaker. . . . . . . . . . . . . . . . . . . . . . 2–26 2.21 A vacuum chamber degasser. . . . . . . . . . . . . . . . . . . . . 2–27 2.22 Flow path in a hydr ydrocyc cyclone. . . . . . . . . . . . . . . . . . . . . 2–28 2.23 Solid control equipment. . . . . . . . . . . . . . . . . . . . . . . . 2–28 2.24 Pa Parti rticle size classifica ficatio tion. . . . . . . . . . . . . . . . . . . . . . . 2–29 2.25 In Intern terna al vie view of a cen centrif rifuge. . . . . . . . . . . . . . . . . . . . . . 2–30 v



2.26 Mud cleaner. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–31 2.27 Mud agitator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–31 2.28 Mud gun. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32 2.29 Mud hopper. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32 2.30 Cut views of a swivel. . . . . . . . . . . . . . . . . . . . . . . . . 2–34 2.31 2.31 A squa square re kelly elly and and a hexa hexago gona nall kell kelly y. . . . . . . . . . . . . . . . . 2–35 2–35 2.32 A kelly valve. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–35 2.33 Kelly bushings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–36 2.34 2.34 Master Master bushi bushing ngs s ([a] ([a] and [b]), [b]), and casing casing bushi bushing ng (c). . . . . . . 2–36 2–36 2.35 .35 Ke Kelly lly bush ushing ing and and maste ster bushin shing g. . . . . . . . . . . . . . . . . . 2–37 –37 2.36 2.36 Drillpi Drillpipe pe slip slip (de (detai taill when when set in the master master bushi bushing) ng).. . . . . . . . 2–37 2–37 2.37 2.37 DC slip slips, s, saf safety ety coll collar ar,, and and casi casing ng slip slips. s. . . . . . . . . . . . . . . 2–38 2–38 2.38 A rotary table. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–38 2.39 BOP stacks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–39 2.40 2.40 Ann Annul ular ar BOP’ BOP’s s (a (a and and b) and and an insi inside de BOP BOP (c) (c) . . . . . . . . . . 2–40 2–40 2.41 2.41 BOP BOP:: (a) (a) blin blind d and and pipe pipe rams rams,, (b) (b) shea shearr ram rams. s. . . . . . . . . . . . 2–41 2–41 2.42 2.42 BOP BOP accu accum mulat ulator ors s and and cont contro roll pane panels ls.. . . . . . . . . . . . . . . . 2–42 2–42 2.43 Choke manifold. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–42 2.44 2.44 Wei Weigh ghtt indi indica cato torr (a) (a) and and a dead deadliline ne anch anchor or (b). (b). . . . . . . . . . . 2–44 2–44 2.45 Dril rilling control console. . . . . . . . . . . . . . . . . . . . . . . . 2–44 3.1 Typical rotary drillstrin ring. . . . . . . . . . . . . . . . . . . . . . . . 3–2 3.2 Typical ypical tool tool joint design designs. s. (A) Interna Internall upset DP with with full–hol full–hole e shrink–grip TJ, (B) External upset DP with internal–flush shrink– grip TJ, (C) External upset DP with flash–weld unitized TJ, (D) External–int External–internal ernal upset upset DP with with Hydrill™– Hydrill™–press pressure ure welde welded d TJ. TJ. . 3–3 3.3 3.3 A DP ele elevator ator and and the the link links s to the the hook hook body body.. . . . . . . . . . . . 3–5 3–5 3.4 A spiraled and a slick dril rill collars. . . . . . . . . . . . . . . . . . . 3–6 3.5 Spiraled DC cross–section. . . . . . . . . . . . . . . . . . . . . . 3–6 3.6 A DC elevator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–6 3.8 Heavy wall drill pipes. . . . . . . . . . . . . . . . . . . . . . . . . 3–7 3.9 Some Stabilize Stabilizers: rs: (a) integral, integral, (b) interchang interchangeab eable, le, (c) non–rotat non–rotating, ing, (d) replaceable. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–8 3.10 A roller reamer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–9 3.11 A fixed hole–opener. . . . . . . . . . . . . . . . . . . . . . . . . . 3–9



3.12 Manual tongs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–10 3.13 3.13 Ton Tongs gs in posit positio ion n to make make–u –up p a conn connec ecti tion on.. . . . . . . . . . . . . 3–11 3–11 3.14 A spinner. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–12 3.15 .15 Ton Tong gs posit ositio ion n duri durin ng make–up –up. . . . . . . . . . . . . . . . . . . . 3–12 –12 3.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–12 3.17 An electric rical top driv rive. . . . . . . . . . . . . . . . . . . . . . . . . 3–14 3.18 A bottom hole turbine. . . . . . . . . . . . . . . . . . . . . . . . . 3–15 3.19 A bottom hole PDM. . . . . . . . . . . . . . . . . . . . . . . . . . 3–15 4.1 Stre tress state about a point int in a flui fluid. . . . . . . . . . . . . . . . . . 4–1 4.2 Real gas deviation factor. . . . . . . . . . . . . . . . . . . . . . . 4–5 4.3 Drill rillst stri ring ng sch schemati matics cs for Exam Examp ple 12. . . . . . . . . . . . . . . . 4–9 5.1 5.1 Assu Assump mpti tion on 1 – pres pressu sure re cont contrib ribut utes es to buck buckliling ng.. . . . . . . . . . 5–2 5–2 5.2 Assump Assumptio tion n 2 – pressu pressure re does does not contrib contribute ute to buck bucklin ling. g. . . . . 5–4 6.1 Mass balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–2 6.2 Sche chematic of a circ irculation syst ystem. . . . . . . . . . . . . . . . . . 6–5 6.3 6.3 Long Longit itud udin inal al cut cut of bit bit nozz nozzle les. s. (Cou (Courte rtesy sy SPE) SPE) . . . . . . . . . . 6–6 6–6 6.4 Pressure drop across the bit. . . . . . . . . . . . . . . . . . . . . 6–7 6.5 Jet impact force. . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–9 6.6 Line of maximum hydr ydraulic power. . . . . . . . . . . . . . . . . . 6–16 6.7 Addit dditio ion nal hydr ydrauli aulic c cons constr tra aints ints.. . . . . . . . . . . . . . . . . . . . 6–17 –17 6.8 Idea Ideall surf surfa ace oper peratio ationa nall par parame ameters ters.. . . . . . . . . . . . . . . . 6–18 –18 6.9 Path of optimum hydraulics. . . . . . . . . . . . . . . . . . . . . . 6–19 6.10 .10 Fri Frict ctio ion nal press ressur ure e dro drop lin lines. es. . . . . . . . . . . . . . . . . . . . . 6–20 –20 6.11 Graph for Example 27. . . . . . . . . . . . . . . . . . . . . . . . . 6–21 7.1 A mud balance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–8 7.2 A Marsh funnel. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–8 7.3 A rota rotati tio onal nal visc viscom ome eter ter (rh (rheome ometer) ter).. . . . . . . . . . . . . . . . . 7–9 7.4 A API filter press. . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–9 7.5 A HTHP filter press. . . . . . . . . . . . . . . . . . . . . . . . . . 7–9 7.6 Sand content sieve. . . . . . . . . . . . . . . . . . . . . . . . . . 7–10 7.7 Retor t. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–10 7.8 Methyl blue capacity test kit kit. . . . . . . . . . . . . . . . . . . . . . 7–11



7.9 A pH meter. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–11 7.10 A titration kit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–12 7.11 A perme rmeameter kit. . . . . . . . . . . . . . . . . . . . . . . . . . . 7–12 7.12 An aniline point kit. . . . . . . . . . . . . . . . . . . . . . . . . . . 7–13 7.13 Electric rical stability tester. . . . . . . . . . . . . . . . . . . . . . . . 7–13 7.14 .14 Clay lay perfo rforma rmance nce for visc visco osity sity.. . . . . . . . . . . . . . . . . . . . 7–21 –21 8.1 Typic ypical al graph of Newton tonian ian fluid fluids s. . . . . . . . . . . . . . . . . . . 8–2 8.2 Typic ypical al graph of Bing ingham-p am-pla last stic ic fluid fluids s. . . . . . . . . . . . . . . 8–3 8.3 Typic ypical al graphs phs of power–la r–law w fluid fluids s. . . . . . . . . . . . . . . . . . 8–3 8.4 Arra rrangeme gemen nt of a rota rotati tion ona al visc visco omete meterr. . . . . . . . . . . . . . . 8–4 9.1 Veloc locity profiles of laminar flow. . . . . . . . . . . . . . . . . . . . 9–3 9.2 Veloci locity ty profil rofile e of lam lamina inar flow flow in a slo slot. . . . . . . . . . . . . . . . 9–4 9.3 Slot approximation ion of an annulus. . . . . . . . . . . . . . . . . . . 9–9 9.4 Fluid parti rticle flowing in a pipe. . . . . . . . . . . . . . . . . . . . 9–14 9.5 Stanton chart. rt. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–15 9.6 9.6 Sele Select ctio ion n of the the corr correc ectt pres pressu sure re drop drop value alue.. . . . . . . . . . . . 9–19 9–19 10.1 Ty Typical roller cone bits. . . . . . . . . . . . . . . . . . . . . . . . . 10–2 10.2 Cut view of a roller ler con cone bits. . . . . . . . . . . . . . . . . . . . . 10–3 10.3 Cut vie view of a non– on–sea sealed led bea bearin ring bit. it.

. . . . . . . . . . . . . . . 10–4 0–4

10.4 A sealed bearing bit. . . . . . . . . . . . . . . . . . . . . . . . . . 10–5 10.5 Cut vie view of a roll rolle er bea bearin ring con cone. . . . . . . . . . . . . . . . . . . 10–5 0–5 10.6 Cut vie view of a jour journ nal beari earin ng cone cone.. . . . . . . . . . . . . . . . . . 10–6 0–6 10.7 Geometry of bit cones. . . . . . . . . . . . . . . . . . . . . . . . . 10–7 10.8 Cone offsets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–8 10.9 Air drilling bits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–9 10.10 .10Stee teel blade drag bits. ts. . . . . . . . . . . . . . . . . . . . . . . . . . 10–10 10.11A diamond bit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–10 10.1 10.12 2Sche Schema mati tic c and and nome nomenc ncla latu ture re of diam diamon ond d bit. bit. . . . . . . . . . . . 10–1 10–11 1 10.13PDC bits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–13 10.1 10.14 4Sche Schema mati tic c and and nome nomenc ncla latu ture re of a PDC PDC bit. bit. . . . . . . . . . . . . 10–1 10–14 4 10.15Nozzles in a PDC bit.

. . . . . . . . . . . . . . . . . . . . . . . . 10–14

10.1 10.16 6Back Back rake rake and and sid side e rak rake e angl angles es in PDC PDC bit bits. s. . . . . . . . . . . . . 10–1 10–14 4



10.1 10.17 7IADC IADC roll roller er cone cone bit bit clas classi sific ficat atio ion n chart chart.. . . . . . . . . . . . . . . 10–1 10–16 6 10.1 10.18 8Tooth ooth wear wear diag diagra ram m for mill milled ed toot tooth h bits bits.. . . . . . . . . . . . . . . 10–2 10–21 1 10.19 10.19Correlatio Correlation n between between rock strength strength and threshold threshold WOB. WOB.

. . . . . 10–23 10–23

10.2 10.20 0Variat ariatio ion n of of ROP ROP with with diff differ eren entt flui fluid d prop properti erties es.. . . . . . . . . . . 10–2 10–24 4 10.2 10.21 1Eff Effect ect of diff differ eren enti tial al pres pressu sure re in the the ROP ROP. . . . . . . . . . . . . . . 10–2 10–25 5 10.22Exponential relationship between of differential pressure and ROP.10–26 10.23 10.23Effect Effect of WOB WOB (a) and rotary rotary spee speed d (b) (b) in the ROP ROP. . . . . . . . . 10–26 10–26



List of Tables 2.1 Heating values of fuels. . . . . . . . . . . . . . . . . . . . . . . . 2–6 2.2 Bloc lock–ta k–tac ckle kle effici fficie ency ncy ($\e ($\eta ta=0 =0.9 .96 6$). $). . . . . . . . . . . . . . . . 2–14 –14 10.1 IADC cod codes for rol roller cone bits. . . . . . . . . . . . . . . . . . . . 10–17 10.2 Range for IADC bit profile file. . . . . . . . . . . . . . . . . . . . . . . 10–19 10.3 Rang ange for IAD IADC bit hydr ydraulic ulic desi desig gn. . . . . . . . . . . . . . . . . 10–1 0–19 10.4 10.4 Rang Range e for IADC IADC cutt cutter er size size and and dens densit ity y. . . . . . . . . . . . . . . 10–2 10–20 0 A.1 New Drill rill Pipe Dimensional Data . . . . . . . . . . . . . . . . . . A–2 A.2 New New Drill Drill Pipe Pipe Tors Torsion ional al and Tensile ensile Data. Data. Courtes Courtesy y API API . . . . . A–3 A.3 New Drill Drill Pipe Collaps Collapse e and Internal Internal Pressure Pressure Data. Data. Courtesy APIA–4 APIA–4 A.4 Premium Premium Drill Pipe Torsional orsional and Tensile ensile Data. Data. Courtesy Courtesy API . . A–5 A.5 Premium Premium Drill Pipe Pipe Collapse Collapse and and Internal Internal Pressure Pressure Data. Data. CourCourtesy API . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A–6

xi





Chapter 1 Introduction 1.1 1. 1

Obje Object ctiv ives es

The purpose of this text is to give students an introduction to the principles and some recommended procedures practiced in drilling engineering. All chapters in general general contain a theoretica theoreticall introductio introduction, n, examples examples,, and exercises exercises.. ReferReferences for further readings are given at the end of the text. Necessary equations and procedures to solve the exercises are presented throughout the text.

1.2 General When a drilling project project is commenced, commenced, two goals govern govern its aspects. aspects. The first is to build the well according to its purpose and in a safe manner (i.e, avoiding personal injuries and avoiding technical problems). The second is to complete it with minimum cost. Thereto the overall costs of the well during its lifetime in conjunction with the field development aspects shall be minimized. The overall cost minimization, or optimization , may influence the location from where the well is drilled (e.g., an extended reach onshore or above reservoir offshore), the drilling technology applied (e.g., conventional or slim–hole drilling, overbalanced or underbalanced, vertical or horizontal, etc), and which evaluation procedures are run to gather subsurface information to optimize future wells. On the other hand, the optimization is influenced by logistics, environmental regulations, etc. To build a hole, different drilling technologies have been invented: • Percussion ercussion drilling drilling – Cable drilling – Drillstring

* With mud CHAPTER 1 Introduction to Drilling

→ “Pennsylvanian “Pennsylvanian drilling” → Quick percussion drilling Page 1–1


* Without mud


→ “Canadian drilling”

• Rotating Rotating drilling drilling – Full cross-section drilling

* Surface driven · Rotary bit · Rotary nozzle nozzle * Subsurface driven · Turbine drilling drilling · Positive Positive displacement displacement motor drilling · Electro Electro motor motor drilling drilling – Annular drilling

* Diamond coring * Shot drilling • Special Special technique techniques s – Abrasive jet drilling – Cavitating jet drilling – Electric arc and plasma drilling – Electric beam drilling – Electric disintegration drilling – Explosive drilling – Flame jet drilling – Implosion drilling – Laser drilling – REAM drilling – Replaceable cutterhead drilling – Rocket exhaust drilling – Spark drilling – Subterrene drilling – Terra drilling – Thermal-mechanical drilling – Thermocorer drilling

Throughout this text, rotary drilling technology is discussed exclusively. CHAPTER 1 Introduction to Drilling

Page 1–2



Figure 1.1: Rig Classification.

1.3 1. 3

Drill Dr illin ing g Ri Rig g Types ypes

The diagram in Figure 1.1 shows a general classification of rotary drilling rigs. Several pictures of the different types of rigs are presented in Figures (a) to (l) below.

(a) Jackknife rig.

CHAPTER 1 Introduction to Drilling

(b) Por table mast.

Page 1–3


(c) A cantilever rig on a barge.

(e) A tender assisted platform. rm.

(g) A Jack–Up rig



(d) A self–contained platform.

(f) A submersible platform. rm.

(h) Semi–submersible vessel.

Page 1–4


(i) A drill–ship

(k) (k) Cais Caisso son n vess vessel el (als (also o call called ed spar spar– – buoy).

1.4 1. 4


(j) A tension–leg platform.

(l) Diagram of a spar–buoy.

Per erso sonn nnel el at Ri Rig g Si Site te

This This secti section on descr describ ibes es the the crew crew requ requir irem emen ents ts and and tasks tasks of some some indi individ vidua uall crew crew members at the rig site. Peopl eople e dire directl ctly y invo involv lved ed in drill drillin ing g a well well are are empl emplo oyed yed eith either er by the the oper operat atin ing g company, the drilling contractor, or one of the service and supply companies. The operating company is the owner of the lease/block and principal user of the services provided by the drilling contractor and the different service companies. To drill an oil or gas well, the operating company (or simply called operator) acquires the right from the land owner under which the prospective reservoir CHAPTER 1 Introduction to Drilling

Page 1–5



may may exist, to drill and produce produce from it. Usually Usually,, when a well has to be drilled, drilled, an auction is run by the operator and various drilling contractors are invited to place place their their bid. bid. Since Since drillin drilling g contra contracto ctors rs are compani companies es that that perf perform orm the actual drilling of the well, their main job is to drill a hole to the depth/location and specification specifications s set by the operator operator.. Along Along with hiring a drilling contractor contractor,, the operator usually employs various service and supply companies to perform logging, cementing, or any other special operations, including maintaining the drilling fluid in its planed condition. Most drilling crews consist of a tool pusher, a driller, a derrickman, a mud logger, and two or three rotary helpers (also called floormen or roughnecks). Along with this basic crew configuration the operator sends usually a representative, called company man to the rig. For offshore operations the crews usually consist of many more employees. Tool Pusher: The tool pusher supervises all drilling operations and is the leading man of the drilling contractor on location. Along with this supervision duties, he has to coordinate company and contractor affairs. Two or three crews operate 24/7, and it is a responsibility of the Tool Pusher to supervise and coordinate these crews. Company Company Man: The company man is in direct charge of all company’s activities on the rig site. He is responsible for the drilling strategy as well as the supplies supplies and services in need. need. His decisions decisions directly effect the progress progress of the well. Driller: The driller operates the drilling machinery on the rig floor and is the overall overall supervisor supervisor of all floormen. floormen. He reports directly to the tool pusher and is the person who is most closely involved in the drilling process. He operates, from his position at the control console, the rig floor brakes, switches, levers, and all other related controls that influence the drilling para parame mete ters rs.. In case of a kick kick he is the first first person person to take take actio action n by moving the bit off bottom and closing the BOP. Derrick Man: The derrickman works on the so–called monkeyboard, a small platform up in the derrick, usually about 90 ft above the rotary table. When a connection is made or during tripping operations he is handling and guiding guiding the upper end of the pipe. During During drilling operations operations the derrickman is responsible for maintaining and repairing the pumps and other equipment as well as keeping tabs on the drilling fluid. Floormen: During tripping, the rotary helpers are responsible for handling the lower end of the drill pipe as well as operating tongs and wrenches to make or break up a connection connection.. During other other times, they also maintain equipment, keep it clean, do painting and in general help where ever help is needed. Mud Engineer, Mud Logger: The service company who provides the mud almost always sends a mud engineer and a mud logger to the rig site. They CHAPTER 1 Introduction to Drilling

Page 1–6



are constantly responsible for logging what is happening in the hole as well as maintaining the proper mud conditions.

1.5 1. 5

Misc Mi scel ella lane neou ous s

According to a wells final depth, it can be classified into:

< 2000 Shallow well: 2000 m Conve Conventi ntiona onall well: well: 2 000 m – 3500 3500 m Deep well: 3500 m – 5000 m > 5 000m Ultra deep well: 000m With the help of advanced technologies in MWD/LWD and extended reach drilling drilling technique techniques, s, horizontal horizontal departures of more than10000 than10000 m are possible today (e.g.,Wytch Farm).


Page 1–7




Page 1–8



Chapter 2 Rotary Drilling System The most common common drilling rigs in use today are rotary drilling rigs. rigs. Their Their main tasks are to create rotation of the drillstring and facilities to advance and lift the drillstring, casings, and special equipment into and out of the hole drilled. The main components of a rotary drilling rig can be seen in Figure 2.1.

Figure 2.1: Typical rig components. Since the rig rate (rental cost of the rig) is one of the most influencing cost factors to the total cost of a well, careful selection of the proper type and capacity pacity is vital for a successful successful drilling project. project. CHAPTER 2 Rotary Drilling Systems

Page 2–1



For all rigs, the depth of the planned well determines basic rig requirements like hoisting capacity, power system, circulation system (mud pressure, mud stream, stream, mud cleaning), cleaning), and the pressure control control system. The selection selection of the most cost–efficient rig involves both quantitative and qualitative considerations. The most important rig systems are: 1. Power Power system, system, 2. Hoisting Hoisting system, system, 3. Drilling Drilling fluid circulation circulation system, system, 4. Rotary system, system, 5. Derrick Derrick and substructure substructure,, 6. Well Well control control system, 7. Well Well monitoring monitoring system. system. The proper way to calculate the various requirements is discussed below. The qualitative aspects involve technical design, appropriate expertise and training of the drilling crew, contractors track record, and logistics handling. For offshore rigs, factors like water depth, expected sea, winds, and currents conditions, and location (supply time) have to be considered. It should be understood that rig rates are not only influenced by the rig type but they are also strongly dependent on by the current market situation (oil price, drilling activity, activity, rig availabili availability ty,, location, location, etc). Therefo Therefore, re, for the rig selection, basic rig requirements are determined first. Then drilling contractors are are cont contac acte ted d for off offers ers of a prop propos osed ed spud spud date date (dat (date e at which which drill drillin ing g oper operat atio ion n commences commences)) and alternative alternative spud dates. dates. This flexibility flexibility to schedule schedule the spud date may reduce rig rates considerably. Before describe the various rig systems listed above, it is important to understa derstand nd the drillin drilling g proces process. s. In rotary rotary drillin drilling, g, the rock is destro destroye yed d by the action action of rotation rotation and axial force applied applied to a drilling drilling bit. The bit acts on the soil destroying the rock, whose cuttings must be removed from the bottom of the borehole in order to continue drilling. The drilling bit is located at the end of a drill string which is composed of drill pipes (also (also called joints or singles), drill collars , and other specialized drilling tools connected end to end by threads to the total length of the drill string, which roughly roughly corresponds corresponds to the current depth of the borehole. borehole. Drill collars are thick thick walled tubes responsible for applying the axial force at the bit. Rotation at the bit is usually usually obtained obtained by rotating rotating the whole drill string from the surface. surface. (See Figure 2.2.) The lower portion of the drill string, composed of drill collars and specialized drilling tools, are called bottom hole assembly (BHA). A large variety of bit models models and designs designs are avail availab able le in indust industry ry.. The choice choice of the right bit, CHAPTER 2 Rotary Drilling Systems

Page 2–2



Figure 2.2: A simplified drillstring.

based on the characteristics of the formations to be drilled, and the right parameters (weight on bit and rotary speed) are the two most basic problems the drilling drilling engineer engineer faces faces during drilling planning planning and drilling drilling operation operation.. The cuttings created by the bit action are lifted to the surface by the drilling fluid, which is continuously pumped from the surface to the bottom through inside of the hollow drill string. At the bit, the drilling fluid is forced through nozzles in a fluid jet action which removes the cuttings from under the bit. The fluid returns to the surface carrying the cuttings, through the annular space between the drill string and the borehole borehole.. The carrying capacity capacity of the drilling drilling fluid is an important characteristics of the drilling fluid. Other important characteristics are the capacity to prevent formation fluids from entering in the borehole, and the capacity pacity to maintain the stability stability of the borehole borehole wall. At the surface, surface, the cuttings are separated separated from the drilling fluid by several several solid removal removal equipmen equipment. t. The drilling fluid accumulates in a series of tanks where it receives the necessary treatment. From the last tank in this series, the drilling mud is picked up by the system of pumps and pumped again down the hole. As drilling progresses, new joints are added to the top of the drill string increa creasi sing ng its its leng length th,, in an oper operat atio ion n call called ed connection . The The diag diagra ram m in Figu Figure re 2.3 2.3 depicts the process of adding a new joint to the drill string. During the drilling of the length of the kelly, a new joint is picked from the pipe rack and stabbed stabbed into the mousehole mousehole using rig r ig lift equipmen equipment. t. At the kelly kelly down, the kelly is pulled out of the hole. A pipe slips (see figure 2.4) is used to transfe transferr the weight weight of the drillstring drillstring from the hook to the master bushing. bushing. The CHAPTER 2 Rotary Drilling Systems

Page 2–3



Figure 2.3: Making a connection.

Figure 2.4: Rig crew setting the slips. connection at the first tool joint is broken and the kelly is swang and stabbed onto the joint in the “mousehole.” The new joint is stabbed on and connected to the top of the drillstring. The drillstring is picked up to remove the slips and the drillstring is lowered until the kelly bushing fits the master bushing. Then drilling is re–initiated. As the bit gets dull, a round trip is performed to bring the dull bit to the surface and replace it by a new one. A round trip is performed also to change the BHA. The drillstring drillstring is also removed removed to run a casing string. string. The operation operation is done by removing stands of two (“doubles”), three (“thribbles”) or even four (“fourbles”) joints connected, and stacking them upright in the rig. During trips, the kelly and swivel swivel is stabbed into the “rathole". “rathole".”” The diagram diagram in Figure Figure 2.5 depicts the process of removing a stand of the drillstring. The process repeats until the whole whole drillstring is out of the hole. hole. Then the drill string is run again again into CHAPTER 2 Rotary Drilling Systems

Page 2–4



Figure 2.5: Removing one stand of drillstring. the hole and drilling continue continues. s. The process process to run the drillstring into the hole is exactly the reverse of that shown in Figure 2.5. Someti Sometime mes s the drillstr drillstring ing is not comple completel tely y run out of the hole. hole. It is just just lifted up to the top of the open-hole section and then lowered back again while contin continuou uously sly circul circulati ating ng with with drillin drilling g mud. mud. Such a trip, trip, calle called d wipe wiperr trip trip , is carri carried ed out to clean the hole from remaining cuttings that may have settled along the open–hole section.

2.1 2. 1

Power wer Syst System em

The power system of a rotary drilling rig has to supply power to items 2 to 7 in the list above. In addition, the system must provide power for pumps in general, rig light, air compressors, etc. Since the largest power consumers on a rotary drilling rig are the hoisting, the circulation system, and the rotary system, these componen components ts determine mainly mainly the total power power requirements requirements.. During typical typical drilling operations, the hoisting and the rotary systems are not operated at the same time. Therefore the same engines can be used to perform both functions. Drilling rig power systems are classified as direct drive type and electric type . In both cases cases,, the source sources s of energy energy are diesel diesel fueled fueled engine engines. s. In the direct drive type, internal combustion engines supply mechanical power to the rig. rig. Most Most rigs rigs use use one one to thre three e engi engine nes s to powe powerr the the dra drawwor wworks ks and and rota rotary ry tab table. le. Power is usually transmitted to the elements by gears, chains, belts, clutches, and and torque torque conve converters rters.. The engines engines are usuall usually y rated rated between between 400 hp and CHAPTER 2 Rotary Drilling Systems

Page 2–5



800 800 hp. hp. The power power is used used primaril primarily y to turn the drill string, string, pump pump the drilling drilling fluid, and raise the drillstring. Engines Engines also power power generato generators rs that supply supply the electri electricit city y used used on and around around the rig. Usuall Usually y there there are two genera generator tor sets in the rig. rig. The The rig can run with with one one of these these units units but it woul would d run close close to maximum output at night. The second provides for back–up and allows allows for other options. These engines are generally rated at 300 hp to 350 hp. Rigs may also employ one or two engines to power the drilling fluid pumps. Total output varies from 300 hp to 800 hp. In the electric type, several diesel engines are used to generate electricity (DC and AC at various voltage levels) that are transmitted to the various rig systems. DC electric motors are compact and powerful, and can operate in a wide range of power and torque. There is considerable flexibility of equipment placement, allowing better space utilization and weight distribution. This is extreme extremely ly important important in offshore offshore rigs. As guideline guideline,, power power requireme requirements nts for most onshore rigs are between 1,000 to 3,000 hp. Offshore rigs in general use much more power. The performance of a rig power system is characterized by the output horsepower power,, torque, torque, and fuel consumption consumption for various engine speeds. speeds. These three parameters are related by the efficiency of each system.

2.1.1

Energy Energy,, Work, and Efficiency Efficiency

The The energy energy consum consumed ed by the engine engines s comes comes from from burnin burning g fuels. fuels. Table able 2.1 presents the heating values for some types of fuels used in internal combustion engines. The engine transforms the chemical energy of the fuel into work. No engine can transform transform totally the chemical chemical energy into work. Most of the energy that enter enters s the engine engine is lost lost as heat. The thermal thermal efficienc efficiency y E t of a machine is defined as the ratio of the work W generated to the chemical energy consumed Q: W E t = . Q Evidently, in order to perform this calculation, we must use the same units both to the work and to the chemical energy. Important conversion factors are:

1 BTU = 778. 778.17 lbf /ft, ft,

Table 2.1: Heating values of fuels. Fuel Fuel Type ype Heat Heatin ing g Value alue Dens Densit ity y (BTU (BTU/l /lbm bm)) (lbm (lbm/g /gal al)) Diesel 19000 7.2 Gasoline 20000 6.6 Butane (liquid) 21000 4.7 Methane (gas) 24000 –

CHAPTER 2 Rotary Drilling Systems

Page 2–6



1 cal = 4. 4.1868 Joule = 4. 4.1868 N m, m, 1 BTU = 252 cal. cal. Engines are normally rated by the power P they can deliver at a given working regime. Power if defined as the rate work is performed, that is work per unit of time. If Q˙ is the rate of chemical energy consumed by the machine (chemical energy per unit of time), we can rewrite the expression for the thermal efficiency as: P E t = . Q˙ To calculate Q˙ we need to know the type of fuel and the rate of fuel consumpti sumption on in mass mass per per unit unit time. time. (Consu (Consumpt mption ion of gaseou gaseous s fuels fuels is normall normally y given in mass per unit time, but consumption for liquid fuels is normally given in volume per unit time. In the latter, we need to know the density of the fluid.) A system produces mechanical work when the sole result of the process could could be the raisin raising g of a weigh weightt (most (most time time limite limited d by its efficie efficiency ncy). ). In this case, the work W done by the system is given by

W = F h , where F is the weight and h is the height. Since power is the rate the work is produced, if we take the time derivative of the work we obtain power:

P =

dW dh = F = F v , dt dt

where P is power, and v the velocity (assuming F constant). constant). When a rotating rotating machine is operating (an internal combustion engine or an electrical motor, for example), we cannot measure its power, but we can measure its rotating speed (normal (normally ly in RPM) RPM) and the torque torque at the shaft. shaft. This This is normally normally performe performed d in a machine called dynamometer . The expressio expression n relating power power to angular velocity and torque is: P = ω T , where ω is the angular angular velocity (in radians radians per unit of time) and T is the torque. A comm common on unit unit of power power is the the hp (hors (horse e powe power). r). One One hp is the the powe powerr required to raise a weight of 33,000 lbf by one foot in one minute:

1 hp = 33, 33, 000

lbf ft lbf ft = 550 . min s

For T in ft lbf and N in RPM we have:







r ad//s 1 hp π rad N [RPM] [RPM] T [ft T [ft lbf] 30 RPM 550 lbf ft/ ft/s that is

P [hp] P [hp] = CHAPTER 2 Rotary Drilling Systems



= P [hp] P [hp] . .

N [RPM] N [RPM] T T [ft [ft lbf] . 5252 Page 2–7



When the rig is operated at environments with non–standard temperatures (85◦ F) or at high altitudes, the mechanical horsepower requirements have to be corrected. The correction should follow the American Petroleum Institute (API) standard 7B-llC: 1. Deduction of 3% of the standard brake horsepower for each 1000 ft of altitude above mean sea level.

1 0◦ F rise or 2. Deduction Deduction of 1% of the standard standard brake horsepow horsepower er for each 10 fall in temperature above or below 85 ◦ F.

Example 1: A diesel engine gives gives an output output torque of 1740 ft lbf at an engine speed of 1200 RPM. If the fuel consumption rate was 31.5 gal/hr, what is the output power and the overall efficiency of the engine.

Solution: The power delivered at the given regime is:

P =

1200 RPM 1740 ft lbf = 397. 397.5 hp 5252

×

Diesel is consumed at 31.5 gal/hr. From Table 2.1 we have:

gal Q˙ = 31. 31.5 hr

BTU 19000 = 4, 309, 309, 200 BTU/ BTU/hr × 7.2 lbm × gal lbm

Converting to hp, results in:

BTU Q˙ = 4, 309, 309, 200 hr

778.17 lbf ft 1 hr 1 hp = 1693. 1693.6 hp × 778. × × BTU 3600 s 550 lbf ft/ ft/s

The thermal efficiency is:

E t =

2.2 2. 2

P 397. 397.5 = = 23. 23.5% 1693. 1693.6 Q˙

Hois Hoisti ting ng Syst System em

The hoisting system is used to raise, lower, and suspend equipment in the well (e.g., (e.g., drillstring, casing, casing, etc). The hoisting hoisting equipme equipment nt itself consists consists of: (See Figure 2.6.) • derrick derrick (not shown), shown), CHAPTER 2 Rotary Drilling Systems

Page 2–8



Figure 2.6: Typical hoisting system. • draw draw works, works, • fast line, line, • crown crown block, block, • traveling traveling block, block, • dead line, line, • deal line line anchor anchor, • storage storage reel, reel, • hook. The drilling line (wire rope) is usually braided steel cable varying from 1 inch to 1 3 /4 inches in diameter. It is wound around a reel or drum in the drawworks. Power (torque and rotation) is transmitted to the drawworks, allowing the drilling line in or out. The hoisting systems is composed by the derrick, the drawworks, and the block-tackle system. CHAPTER 2 Rotary Drilling Systems

Page 2–9



Figure 2.7: Stand of doubles along the mast.

2.2.1 2.2.1 The The De Derr rric ick k The derrick or mast is a steel tower. 1 The purpose of the derrick is to provide height to raise and lower the drillstring (and also casing) out and into the borehole. Derricks are rated by the API according to their height and their ability to withstand withstand wind and compressive compressive loads. API has publishe published d standards standards for the particu particular lar specifica specificatio tions. ns. The The higher higher the derrick derrick is, the longer longer stands stands it can handle, which in turn reduces the tripping time. Derricks are designed to handle two, three, or four joints. The The derric derrick k stands stands above above the derrick derrick floor floor. The The derric derrick k floor floor is the stage stage where several surface drilling operations occur. At the derrick floor are located the drawworks, drawworks, the driller’s driller’s console, console, the driller’s driller’s house house (or “doghouse”), “doghouse”), the rotary table, the drilling fluid manifold, and several other tools to operate the drillstring. The space below the derrick floor is the substructure. The height of the the subs substru tructu cture re shou should ld be enou enough gh to acco accomm mmod odat ate e the the well well cont contro roll equi equipm pmen ent. t. (See Figure 2.1.) At about 3 /4 of the height of the derrick is located a platform called called “monkey “monkey board”. board”. This platform platform is used to operate operate the drillstring drillstring stands during trip operations. During drillstring trips, the stands are kept stood in in the mast, held by “fingers” in the derrick rack near the monkey board, as shown in Figure 2.7.

1

If the tower tower is jacked jacked up, it is called called mast. If the tower tower is erected erected on the site, site, it is called called derrick.


Page 2–10



Figure 2.8: Onshore rig drawworks.

Figure 2.9: Brake belts and magnification linkage.

2.2.2 2.2.2 The The Dr Draw aww works orks The drawworks provides hoisting and braking power required to handle the heav heavy y equipm equipment ents s in the boreh borehole ole.. It is is composed composed of a wire wire rope rope drum, drum, mechanica mechanicall and hydraulic hydraulic brakes, brakes, the transmissi transmission, on, and the cathead cathead (small (small winches operated by hand or remotely to provide hoisting and pulling power to operate small loads and tools in the derrick area). Figure 2.8 shows a typical onshore rig drawworks. The reeling–in of the drilling line is powered by an electric motor or Diesel engine, engine, and the reeling–out reeling–out is powered powered by gravity gravity.. To control control the reeling reeling out, mechanical brakes and auxiliary hydraulic or magnetic brakes are used, which dissipates the energy required to reduce the speed and/or stop the downward movement of the suspended equipment. (See Figure 2.9.) CHAPTER 2 Rotary Drilling Systems

Page 2–11



Figure 2.10: Drawworks schematics. The drawworks take power from Diesel engines or electrical motors, and an assembly of gears and clutches reduces the rotary speed to power the drum and the various catheads. A schematic of the internal mechanisms of a drawworks is shown in Figure 2.10. As shown in the schematics, the drum surface has a helical groove to accommodate the drilling line without causing excessive stress and stain. This also helps the drilling line to lay neatly neatly when reeled in.

2.2.3 2.2.3

The Block Block & Tackle ackle

The drawworks, drawworks, although although very powerful powerful,, cannot cannot provide provide the pull required required to raise raise the the heav heavy y drill drillst strin ring. g. The The requ requir ired ed pull pull is obta obtain ined ed with with a syste system m of pull pulley eys. s. The drilling line coming from the drawworks, called fast line , goes over a pulley system mounted at the top of the derrick, called the crown block , and traveling block . The down to another pulley system called the traveling The asse assemb mbly ly of of crown block, traveling block and drilling line is called block-tackle . The number of lines n of a tackle is twice the number of (active) pulleys in the traveling block. The last line of the tackle is called dead line and and is anchored to the derrick floor, close to one of its legs. legs. Below Below and connected connected to the traveling traveling block block is a hook to which drilling equipment can be hung. As the drilling line is reeled in or out of the drawwork drawworks, s, the traveling traveling block block rises and lowers lowers along along the derrick. derrick. This raises and lowers the equipment in the well. The block-tackle system provides a mechanical advantage to the drawworks, and reduces the total load applied to the derrick. We will be interested in calculating the fast line force F f f (provided by the drawworks) required to raise a weight W in the hook, and the total load CHAPTER 2 Rotary Drilling Systems

Page 2–12



Figure 2.11: Forces acting in the block–tackle. applied to the rig and its distribution on the derrick floor.

2.2.3.1 2.2.3.1

Mechanic Mechanical al advan advantage tage and Efficienc Efficiency y

The mechanical advantage AM of the block–tackle is defined as the ratio of the load W in the hook to the tensile force force on the fast line F f f :

AM =

W . F f f

For an ideal, frictionless system, the tension in the drilling line is the same throughout the system, so that W = n F f f . (See Figure Figure 2.11.) 2.11.) Therefo Therefore, re, the ideal mechanical advantage is equal to the number of lines strung through the traveling traveling block: (AM )ideal = n = n . In a real pulley, however, the tensile forces in the cable or rope in a pulley are not identical. identical. If F i and F o are the input and output tensile forces of the rope in the pulley, the efficiency η of a real pulley is given by the following ratio:

η =

F o . F i

We will assume that all pulleys in the hoisting system have the same efficiency, and we want to calculate the mechanical advantage of a real pulley system. system. If F f f is the force in the fast line, the force F 1 in the line over the first pulley (in the crown block) is given by

F 1 = ηF η F f f . CHAPTER 2 Rotary Drilling Systems

Page 2–13



The force in the line over the second pulley (in the traveling block) is

F 2 = η = ηF F 1 = η 2 F f f . Using the same reasoning over and over, the force in the i th line is

F i = η i F f f . The total load W acting in the hook is equal to the sum of the forces in each line of the traveling block. This means that the load W is given by

W = F 1 + F 2 +

· · · + F n = (η + η2 + · · · + ηn)F f f .

It can be easily shown that the expression between parenthesis can be written as η η n+1 . 1 η Therefore we have: η η n+1 W = F f f . 1 η Consequently, the real mechanical advantage is given by:

− − − −

W η η n+1 AM = = . F f f 1 η

− −

The overall efficiency E of the system of pulleys is defined as the ratio of the real mechanical advantage to the ideal mechanical advantage:

AM η η n+1 E = = . (AM )ideal n(1 η )

−

−

(2.1)

If the efficiency of the pulleys η is known, Block–tackle overall efficiency E can be calcul calculate ated d using using Expres Expressio sion n 2.1. 2.1. A typica typicall value value for the efficie efficiency ncy of ball–bearing pulleys is η = 0.96. Table able 2.2 shows shows the calculated calculated and industry industry average overall efficiency for the usual number of lines. Table 2.2: Block–tackle efficiency ($\eta=0.96$). E ave n E ave 6 0.869 0.874 8 0.836 0.841 10 0.804 0.810 12 0.775 0.770 16 0.746 0.740 Therefore, if E is is known, the fast line force F f f required to rise a load W is given by W F f f = (2.2) nE CHAPTER 2 Rotary Drilling Systems

Page 2–14


2.2.3. 2.2 .3.2 2


Hook Hook Power ower

For an ideal block–tackle system, the input power (provided by the drawworks) is equal to the output or hook power (available to move the borehole equipments). In this case, the power delivered by the drawworks is equal to the force in the fast line F f f times the velocity of the fast line v f , and the power developed at the hook is equal to the force in the hook W times the velocity of the traveling block v b . That is P d = F = F f f vf = W vb = P = P h . Since for the ideal case n F f f = W , we have that

vb =

vf , n

that is, the velocity of the block is n times slower than the velocity of the fast line, and this is valid also for the real case. Considering the Equation (2.2)

F f f =

W n E

(2.3)

2.2 which represents the real relationship between the force in the fast line and the weight in the hook, and multiplying both sides by v f we obtain:

F f f vf = P d =

W vf W vb P h = = , n E E E

P h , E which which repr repres esen ents ts the the real real rela relatio tions nshi hip p betw betwee een n the the powe powerr deliv deliver ered ed by the the draw draw-works and the power available in the hook, where E is the overall efficiency of the block–tackle system. P d =

Example 2: A rig must hoist a load of 300,00 300,000 0 lbf. The drawwo drawworks rks can provide a maximum input input power power to the block–tac block–tackle kle system of as 500 hp. hp. Eight Eight lines are strung between the crown block and traveling block. Calculate (1) the tension in the fast line when upward motion is impending, (2) the maximum hook horsepower, (3) the maximum hoisting speed.

Solution: Using E = 0.841 (average efficiency for n = 8) we have:

W 300, 300, 000 lbf = n E 8 0.841

(1)

F f f =

(2)

P d = 500 hp =


×

→

P h P h = E 0.841

F f f = 44, 44, 590 lbf

→

P h = 421 hp Page 2–15


(3)



550 lbf ft/ ft/s P h = 421 hp 1 hp vb =

2.2.4 2.2.4



231, 231, 550 lbf ft/ ft/s 300, 300, 000 lbf


= 231, 231, 550 lbf ft/ ft/s = W vb = 300, 300, 000 lbf

→

× vb

ft/s = 46. 46.3 ft/ ft/min vb = 0.772 ft/

Load Load Applied Applied to the Derric Derrick k

The total load applied to the derrick, F D is equal to the load in the hook plus the force acting in the dead line plus the force acting in the fast line:

F D = W + F f f + F d . The worst scenario for the force in the fast line is that for the real case. From Section 2.2.3.1 the force in the fast line is:

F f f =

W n E

(2.4)

2.2 For the dead line, however, the worst scenario (largest force) is that of ideal case. In this case, the force in the dead line is:

F d =

W . n

Therefore, the total load applied to the derrick is:

F D = W +

W W (n + 1)E 1) E + + 1 + = W . n E n n E

The total load F D , however, is not evenly distributed over all legs of the derrick. derrick. In a conven conventiona tionall derrick, derrick, the drawworks drawworks is usually usually located between between two of the legs of the derrick. (See Figure 2.12.) The dead line, however must be anchored close to one of the remaining two legs. 2 From this configuration the load in each leg is:

Leg A :

W W n + 4 + = W , 4 n 4n

W , 4 W W nE + + 2 Legs C and D : + = W . 4 2nE 4nE Leg B :

2

The side of the derrick opposite to the drawworks is called V–gate. This area must be kept free to allow pipe handling. Therefore, the dead line cannot be anchored between legs A and B.


Page 2–16



Figure 2.12: Derrick floor plan. Evidently, the less loaded leg is leg B. We can determine under which conditions the load in leg A is greater then the load in legs C and D:

n + 4 nE + + 2 W > W 4n 4nE

→

E > 0. 0 .5 .

Since the efficiency E is usually greater than 0.5, leg A will be the most loaded leg, and very likely it will be the first to fail in the event of an excessive load load is applie applied d to the hook. hook. If a derric derrick k is designed designed to support support a maxim maximum um Lmax nominal load Lmax, each leg can support 4 . Therefor Therefore, e, the maximum maximum hook hook load that the derrick can support for a given line arrangement is

Lmax n + 4 = W max max 4 4n

→

W max max =

n L max . n + 4

The equivalent derrick load, F DE DE , is defined as four times the load in the most loaded loaded leg. For For the derrick configuratio configuration n above, above, the equivalen equivalentt derrick derrick load is n + 4 F DE W . DE = n The equivalent derrick load (which depends on the number of lines) must be less than the nominal capacity of the derrick. The derrick efficiency factor , E D is defined as the ratio of the total load applied to the derrick to the equivalent derrick load:

F D E D = = F DE DE


(n+1)E +1)E +1 +1 W n E n+4 W n

=

(n + 1)E 1) E + + 1 . (n + 4)E 4) E

Page 2–17



Example 3: For For the data of Example 2, calculate calculate (1) the actual actual derrick load, load, (2) the equivalent derrick load, and (3) the derrick efficient factor.

Solution: (1) The actual derrick load is given by

F D =

(n + 1)E 1) E + + 1 (8 + 1) 0.841 + 1 W = n E 8 0.841

× ×

300, 000 = 382, 382, 000 lbf lbf × 300,

(2) The equivalent derrick load is given by

F DE DE =

n + 4 8+4 W = n 8

300, 000 = 450, 450, 000lbf × 300,

(3) The derrick efficiency factor is

E D =

F D 382, 382, 000 = = 85% F DE 450, 450, 000 DE

2.3 Drillin Drilling g Fluid Fluid Cir Circul culati ation on System System The drilling fluid plays several functions in the drilling process. The most important are: 1. clean the rock fragment fragments s from beneath the bit and carry them to surface, surface, 2. exert exert sufficient sufficient hydrostatic hydrostatic pressure against against the formation formation to prevent prevent formation fluids from flowing into the well, 3. maintain stability of the the borehole walls, 4. cool and lubricate lubricate the drillstring drillstring and bit. Drilling fluid is forced to circulate in the hole at various pressures and flow rates. Drillin Drilling g fluid fluid is stored stored in steel steel tanks tanks locate located d beside beside the rig. Powerf owerful ul pumps pumps force the drilling fluid through surface high pressure connections to a set of valves called pump manifold , located located at the derrick floor. floor. From the manifold, manifold, the fluid goes up the rig within a pipe called standpipe to to approximately 1/3 of the height of the mast. From there the drilling fluid flows through a flexible high pressure hose to the top of the drillstring. The flexible hose allows the fluid to flow continuously as the drillstring moves up and down during normal drilling operations. The fluid enters in the drillstring through a special piece of equipment called swivel (Figure 2.13) located at the top of the kelly. The swivel permits rotating the drillstring while the fluid is pumped through the drillstring. 3 The drilling fluid 3

See Section 2.4.1 for details.


Page 2–18



Figure 2.13: A swivel. then then flows flows down down the the rota rotati ting ng dril drills lstr trin ing g and and jets jets out out thro throug ugh h nozz nozzle les s in the the dril drilll bit bit at the bottom bottom of the hole. The drilling fluid picks the rock cuttings generated generated by the drill bit action on the formation. formation. The drilling fluid then flows up the borehole through the annular space between the rotating drillstring and borehole wall. At the top of the well (and above the tank level, the drilling fluid flows through the flow line to a series of screens called the shale shaker . The shale shale shaker shaker is designed designed to separate separate the cuttings cuttings from the drilling mud. mud. Other devices devices are also used to clean the drilling fluid before it flows back into the drilling fluid pits. Figure 2.14 depicts the process described above. The principal components of the mud circulation system are: 1. pits or or tanks, tanks, 2. pumps, pumps, 3. flow line, line, 4. solids and contaminants removal removal equipment, equipment, 5. treatment treatment and mixing equipmen equipment, t, 6. surface piping piping and valves, valves, 7. the drillstrin drillstring. g. The tanks (3 or 4 – settling tank, mixing tank(s), suction tank) are made of steel steel sheet. sheet. They They contai contain n a safe safe excess xcess (neither (neither to big nor to small) small) of the CHAPTER 2 Rotary Drilling Systems

Page 2–19



Figure 2.14: Rig circulation system. total volume of the borehole. In the case of loss of circulation, this excess will provide the well with drilling fluid while the corrective measures are taken. The number of active tanks depends on the current depth of the hole (bypasses allow to isolate one or more tanks.) The tanks will allow enough retaining time so that much of the solids brought from the hole can be removed from the fluid.

2.3. 2.3.1 1

Mud Mud Pump Pumps s

The great majority of the pumps used in drilling operations are reciprocating positive displacement pumps (PDP). Advantages of the reciprocating PDP when compared to centrifugal pumps are: • ability ability to pump pump fluids fluids with with high high abrasi abrasive ve solids solids conten contents ts and with large large solid particles, • easy to operate operate and maintai maintain, n, • sturdy and and reliable, reliable, • ability ability to operate in a wide range range of pressure and flow rate. rate. CHAPTER 2 Rotary Drilling Systems

Page 2–20



Centrifugal pumps are very sensitive to abrasive solid contents mud, and do not offer a wide range of operation compared to PDP. PDP are composed of two major parts, namely: Power end: receives power power from engines and transform transform the rotating movement into reciprocating movement. Fluid end: converts the reciprocating power into pressure and flow rate.

The efficiency E m of the the power power end, end, that that is the efficien efficiency cy with which rotating rotating mechanical power is transformed in reciprocating mechanical power is of the order of 90%. 90%. The efficie efficiency ncy E v of the fluid end (also called volumetric efficiency), that is, the efficiency that the reciprocating mechanical power is transformed into hydraulic power, can be as high as 100%. Rigs normally have two or three PDPs. During drilling of shallow portions of the hole, when the diameter is large, the two PDPs are connected in parallel to provide the highest flow rate necessary to clean the borehole. As the borehole deepens, less flow rate and higher pressure are required. In this case, normally only only one one PDP PDP is used used whil while e the the othe otherr is in stan standb dby y or in pre preventi entiv ve main mainte tena nanc nce e. The great flexibility in the pressure and flow rate is obtained with the possibility of changing changing the diameters diameters of the pair piston–lin piston–liner er.. The flow rate depends depends on the following parameters: • stroke stroke length length L S (normally fixed), • liner diameter diameter dL , • rod diame diameter ter d R (for duplex PDP only), • pump pump speed speed N (normally given in strokes/minute), • volumetric volumetric efficien efficiency cy E V V of the pump. In addition, the pump factor F p is defined as the total volume displaced by the pump in one stroke. There are two types of PDP: double-action duplex pump, and single-action triplex pump. Triplex PDPs, due to several advantages, advantages, (less bulky, bulky, less pressure fluctuation, cheaper to buy and to maintain, etc,) has taking place of the duplex PDPs in both onshore and offshore rigs. 2.3.1. 2.3 .1.1 1

Duplex Duplex PDP

The The dupl duple ex mud mud pump pump cons consis ists ts of two two doub double le–a –acti ction on cylin cylinde ders rs (see (see Figu Figure re 2.16 2.16-a). This This means means that that drillin drilling g mud mud is pumpe pumped d with with the forwa forward rd and backw backward ard movement of the barrel. CHAPTER 2 Rotary Drilling Systems

Page 2–21



(a) Piston scheme (double action).

(b) A duplex unit.

Figure 2.15: Duplex pumps.

(a) Piston scheme (single action).

(b) A Triplex unit.

Figure 2.16: Triplex pumps. For a duplex pump (2 double–action cylinders) the pump factor is given by: π 2 F p = 2dL2 dR LS E V V . 2



−



A typical duplex duplex pump is shown shown in Figure 2.16-b 2.16-b.. 2.3.1. 2.3 .1.2 2

Triple riplex x PDP

The triplex mud pump consists of three single–action cylinders (see Figure ?? a). This means that drilling mud is pumped only in the forward movement of the barrel. For a triplex pump the pump factor is given by:

F p =

3π 2 d LS E V V . 4 L

A typical triplex pump is shown in Figure ?? -b. 2.3.1. 2.3 .1.3 3

Pump Pump Flow Flow Rate Rate

For both types of PDP, the flow rate is calculated from:

q = N = N F p . CHAPTER 2 Rotary Drilling Systems

Page 2–22



For N in strokes per minute (spm), dL , d R , and L S in inches, F p p in in3, and q in gallons per minute (gpm) we have:

q = =

1 N F p . 231

Note that in this particular formulation, the volumetric efficiency of the pump is included in the pump factor. 2.3.1. 2.3 .1.4 4

Pump Pump Power ower

Pumps Pumps convert convert mechanica mechanicall power power into hydraulic hydraulic power. power. From the definition definition of power we can write: P = F v . In its motion, the piston exerts a force on the fluid that is equal to the pressure ∆ p times the area A of the piston, and the velocity v is differential in the piston ∆ p equal to the flow rate q divided divided by the area A , that is

P H (∆ p A) H = (∆ p

q = ∆ p q . A

(2.5)

∆ p in psi, and q in For P H in gpm we have: H in hp, ∆ p P H H =

∆ p q . 1714. 1714.29

(2.6)

Example 4: Compute the pump pump factor in gallons per stroke and and in barrels per stroke for a triplex pump having 5.5 in liners and 16 in stroke length, with a volumetric efficiency of 90%. At N = 76spm, the pressure differential between the input and the output of the pump is 2400 psi. Calculate the hydraulic power transferred to the fluid, and the required mechanical power of the pump if E m is 78%.

Solution: The pump factor (triplex pump) in in 3 per stroke is:

F p =

3π 4

90% = 1026 in3 × 5.52 × 16 × 90%

Converting to gallons per stroke and to barrels per stroke gives:

F p = 1026

1 1 = 4.44gps = 4. 4.44 × = 0.1058bps × 231 42

The flow rate at N = 76spm is:

q = N F p = 78spm CHAPTER 2 Rotary Drilling Systems

337.44gpm × 4.44gps = 337. Page 2–23



The hydraulic power transferred to the fluid is:

P H H =

2400psi 334. 334.44gpm = 468hp 1714. 1714.29

×

To calculate the mechanical power required by the pump we must consider the efficiencies: 1 1 P = P = 468hp = 667hp 90% 78%

×

2.3.1.5 2.3.1.5

×

Surge Surge Dampener Dampeners s

Due to the reciprocating action of the PDPs, the output flow rate of the pump presents a “pulsation” (caused by the changing speed of the pistons as they move along the liners). This pulsation is detrimental to the surface and downhole equipment equipment (particularly with MWD pulse telemetry system). system). To decrease decrease the pulsation, surge dampeners are are used at the output of each pump. A flexible diaphrag diaphragm m creates creates a chamber chamber filled with nitrogen nitrogen at high pressure. pressure. The fluctuation of pressure is compensated by a change in the volume of the chamber. The schematic of a typical surge dampener is shown in Figure 2.17. A relief valve located in the pump discharge line prevents line rupture in case the pump is started against a closed valve.

Figure 2.17: Surge dampener.


Page 2–24


2.3.2 2.3.2


Solids Solids Contr Control ol Equipm Equipment ent

The purpose of the solids control equipment is to reduce to a minimum the amount amount of inert solids and gases in the drilling fluid. They They are: 1. Shale shakers, shakers, 2. Degassers Degassers,, 3. Desanders (hydrocyclones), 4. Desilters (hydrocyclones), 5. Centrifuges Centrifuges,, 6. Mud cleane cleaners. rs. Figure 2.18 shows a sketch of a typical solids control system (for unweighted fluid). fluid). Fine particles particles of inactive inactive solids solids are continuo continuously usly added to the fluid during drilling. drilling. These solids solids increase the density density of the fluid and also the friction pressure pressure drop, drop, but do not contribute contribute to the carrying capacity of the fluid. The amount of inert solids must be kept as low as possible.

Figure 2.18: Solids control system.


Page 2–25



Figure 2.19: Shale shaker configurations.

Figure 2.20: A two–screen shale shaker. 2.3.2. 2.3 .2.1 1

Shale Shale Shaker Shakers s

The The shal shale e shak shaker er remo remove ves s the the coar coarse se soli solids ds (cut (cutti ting ngs) s) gene genera rate ted d durin during g drill drillin ing. g. It is locate located d at the end of the flow line. line. It constitut constitutes es of one or more more vibra vibratin ting g screens in the range of 10 to 150 mesh over which the mud passes before it is fed to the mud pits. (See Figure 2.19.) The screens are vibrated by eccentric heavy cylinders connected to electric motors. motors. The vibration vibration promotes promotes an efficient separation separation without without loss of fluid. Figure 2.20 shows a typical two–screen shale shaker. 2.3.2. 2.3 .2.2 2

Degass Degassers ers

Gases that might enter the fluid must also be removed. Even when the fluid is overbalanced, the gas contained in the rock cut by the bit will enter the fluid and must be removed. The degasser removes gas from the gas cut fluid by creating a vacuum in a vacuum vacuum chamber. chamber. The fluid flows down an inclined flat surface as a thin layer. layer. The vacuum vacuum enlarges enlarges and coalesce the bubb bubbles. les. Degassed Degassed CHAPTER 2 Rotary Drilling Systems

Page 2–26



Figure 2.21: A vacuum chamber degasser.

fluid is draw from chamber by a fluid jet located at the discharge line. A typical degasser diagram is shown in Figure 2.21.

2.3.2.3 2.3.2.3

Hydrocy Hydrocyclo clones nes (Desan (Desanders ders and Desilter Desilters) s)

Hydrocyclon Hydrocyclones es are simple devices devices with no internal internal moving moving parts. The drilling fluid enters the device through a tangential opening in the cylindrical section, impelled by a centrifugal pump. The centrifugal force generated by the whirling motion pushes the solid particles towards the internal wall of the inverted cone. As the whirling flux moves downwards the rotating speed increased and the diameters diameters decreases decreases.. The fluid free of solid particles particles is “squeeze “squeezed” d” out of the flow and swirls upwards in a vortex motion, leaving the hydrocyclone from the upper exit. The solids leave the hydrocyclone from the apex of the cone (underflow). For maximum efficiency, efficiency, the discharge from the apex exit of hydrocyclone should be in a spray in the shape of a hollow cone rather than a rope shape. Figure 2.22 shows the fluid/solids paths in a hydrocyclone. Hydrocyclones are classified according to the size of the particles removed as desanders (cut point in the 40–45 µm size range) or desilters (cut point in the 10–20 µm size range). At the cut point of a hydrocyclone 50% of the particles of that size is discarded. The desander is a set of two or three 8in or 10in hydrocyclones, and are positioned after the shale shaker and the degasser (if used). used). The desilte desilterr is a set of eight to twelve twelve 4in or 5in hydroc hydrocycl yclone ones. s. It removes particles that can not be removed by the desander. Figures 2.23 shows a desander desander (a), and a desilter desilter (b). Note the size and number number of hydrocyclo hydrocyclones nes in each case. A typical typical drilling drilling solid particle distribution distribution and particle size range classification are shown in the diagram in Figure 2.24. CHAPTER 2 Rotary Drilling Systems

Page 2–27



Figure 2.22: Flow path in a hydrocyclone.

(a) Desander.

(b) Desilter.

Figure 2.23: Solid control equipment.


Page 2–28



Figure 2.24: Particle size classification. 2.3.2. 2.3 .2.4 4

Centrif Centrifuge uges s

The centrifuge is a solids control equipment which separates particles even smaller, which can not be removed by the hydrocyclones. It consists of a rotating cone–shape drum, with a screw conveyor. (See Figure 2.25.) Drilling fluid is fed through the hollow conveyor. The drum rotates at a high speed and creates a centrifugal centrifugal force force that causes causes the heavier heavier solids to decant. decant. The screw rotates in the the same same dire direct ctio ion n of the the drum drum but at a slig slight ht slo slower wer spee speed, d, push pushin ing g the the soli solids ds toward the discharge line. The colloidal suspension exits the drum through the overflow ports. The drums are enclosed in an external, non–rotating casing not shown in the figure.

2.3.2. 2.3 .2.5 5

Mud Cle Clean aner ers s

Inert solids in weighted fluid (drilling fluid with weight material like barite, iron oxide, etc) can not be treated with hydrocyclones alone because the particle sizes of the weighting material are within the operational range of desanders and desilters. 4 This is shown in the diagram in Figure 2.24, which includes the particle size distribution of typical industrial barite used in drilling fluids. A mud cleaner is a desilter desilter unit in which the underflow underflow is further processed processed by a fine vibrating screen, mounted directly under the cones. The mud cleaner sepa separa rate tes s the the low low dens densit ity y inert inert solid solids s (und (undes esir irab able le)) from from the the high high dens densit ity y weig weight ht-4

Weighting material are relatively expensive additives, which must be saved.


Page 2–29



Figure 2.25: Internal view of a centrifuge. ing particles. (See Figure 2.26-a.) Hydrocyclon Hydrocyclones es discriminate discriminate light particles from heavy heavy particles. particles. Bentonite Bentonite are lighter than formation solids because they are of colloidal size (although of the same density). density). Barite particles particles are smaller smaller than formation formation solids because because they are denser. The desilter desilter removes removes the barite and the formation solids particles in the underflow, leaving only a clean mud with bentonite particles in a colloidal suspension in the overflow overflow. The thick slurry in the underflow underflow goes goes to the fine screen, which which sepa separa rate te the the larg large e (low (low dens densit ity) y) partic particle les s (form (format atio ion n soli solids ds)) from from the the smal smalll (high density) barite particles, thus conserving weighting agent and the liquid phase but at the same time returning many fine solids to the active system. The thick barite rich slurry is treated with dilution and mixed with the clean mud (colloidal (colloidal bentonite). bentonite). The resulting resulting mud is treated to the right density and viscosity and re–circula re–circulates tes in the hole. A diagram diagram of a mud cleaner is shown in Figure 2.26-b. 2.26-b. Mud cleaners are used mainly with oil– and synthetic–base fluids where the liquid discharge from the cone cannot be discharged, either for environmental or economic economic reasons. It may also be used with weighted weighted water–base water–base fluids fluids to conserve barite and the liquid phase.

2.3.3

Treatment reatment and Mixing Mixing Equipment Equipment

Drilling fluid is usually a suspension of clay (sodium bentonite) in water. Higher density fluids can be obtained by adding finely granulated (fine sand to silt size – see Figure 2.24) barite (BaSO4). Various chemicals or additives are also used in different situations. The drilling fluid continuous phase is usually water (freshwate (freshwaterr or brine) called called water–ba water–base se fluids. When the continuous continuous phase is oil (emulsion (emulsion of water in oil) it is called called oil–base oil–base fluid. The basic drilling drilling fluids CHAPTER 2 Rotary Drilling Systems

Page 2–30


(a) Unit of a mud cleaner


(b) Principle of the mud cleaner.

Figure 2.26: Mud cleaner.

(a)

Figure 2.27: Mud agitator. physical physical properties properties are density, density, viscosity viscosity,, and filtrate. filtrate. Fresh water density is 8.37 pounds pounds per gallon gallon (ppg). Bentonite Bentonite adds viscosity viscosity to the fluids and also increa increases ses the density density to about about 9 to 10 ppg. ppg. Highe Higherr densi density ty (15 to 20 ppg) ppg) is obtained with barite, iron oxide, or any other dense fine ground material. Water base fluids are normally made at the rig site (oil base mud and synthetic thetic fluids are normally manufa manufacture ctured d in a drilling fluid plant). Special Special treatment and mixing equipment exists for this purpose. Tank agitators, mud guns, mixing hoppers, and other equipment are used for these purposes. Tank agitators or blenders (Figure 2.27-a) are located in the mud tanks to homog homogen enize ize the fluid in the tank. tank. They They help help to keep the variou various s suspen suspended ded material homogeneously distributed in the tank by forcing toroidal and whirl motions of the fluid in the tank. (See Figure 2.27-b.) Mud guns are mounted in gimbals at the side of the tanks, which allow aimCHAPTER 2 Rotary Drilling Systems

Page 2–31



Figure 2.28: Mud gun.

(a)

(b)

Figure 2.29: Mud hopper.

ing a mud jet to any point in the tank. They help to homogenize the properties of two tanks, and spread liquid additives in a large area of the tank (from a pre-mixed tank). (See Figure 2.28.) Centrifugal pumps power the mud guns. The mixing hopper (see Figure 2.29) allows adding powder substances and additives in the mud system. The hopper is connected to a Venturi pipe. Mud is circulated by centrifugal pumps and passes in the Venturi at high speed, sucking the substance into the system. CHAPTER 2 Rotary Drilling Systems

Page 2–32


2.4 2. 4


The The Rota Rotary ry Syst System em

The rotary system is the set of equipments necessary to promote the rotation of the bit. The bit must be mechanically and hydraulically connected to the rig. This connectio connection n is made by the drillstring. drillstring. The purpose purpose of the drillstring drillstring is to transmit transmit axial force, force, torque, torque, and drilling drilling fluid (hydrau (hydraulic lic power) to the bit. The basic drillstring is composed of the following components: • Swivel, Swivel, • Kelly Kelly and accessories accessories,, • Rotary table table and component components, s, • Drillstring Drillstring tubulars tubulars (drill pipe, drill collars, collars, etc.), • Drill bit. Several other components and equipment can be connected to the drillstring to perform several tasks and to lend to the drillstring special features.

2.4. 2.4.1 1

Swiv Swivel el

The swivel is suspended by the hook of the traveling block and allows the drillstring to rotate as drilling fluid is pumped pumped to within within the drillstring. drillstring. Without Without the swivel, drilling fluid could not be pumped downhole, or the drillstring could not rotate. The swivel also supports the axial load of the drillstring. See Figure 2.30 for cuts of a swivel showing the internal parts. A flexible hose connects to the gooseneck which is hydraulically coupled to the top of the swivel stem by a stuffing box. The stem shoulder rest on a large thrust tapered roller bearing, which transmits the drillstring weight to the swivel body, and then to the bail. The thread connector of the swivel is cut left–hand so that it will not tend to disconnect when the drillstring is rotated by the kelly or by the top drive.

2.4.2

Kelly Kell y, K Kelly elly Valves, and Kelly Kelly Saver Sub

Below and connected to the swivel is a long four-sided (square) or six-sided (hexagon) steel bar with a hole drilled through the middle for a fluid path called kelly . The purpose purpose of the kelly kelly is to transmi transmitt rotary rotary motion motion and torque torque to the drillstring (and consequently to the drill bit), while allowing the drillstring to be lowered or raised during rotation. The square or hexagonal section of the kelly allows it to be gripped and turned by the kelly bushing and rotary table (see Sectio Section n 2.4.3) 2.4.3).. The kelly kelly bushi bushing ng has an inside inside profile profile matchi matching ng the kelly’ kelly’s s outside profile (either square or hexagonal), but with slightly larger dimensions CHAPTER 2 Rotary Drilling Systems

Page 2–33



(a) Lateral cut.

(b) Front cut.

Figure 2.30: Cut views of a swivel. so that the kelly can freely move move up and down inside it. The overal overalll length of the kelly varies from 40 ft to 54 ft. It is common (and advisable) to include two special valves at both ends of the kelly, called kelly valves . (The upper kelly valve has left–hand threads.) The kelly valve consists of a ball valve which allows free passage of drilling fluids withou withoutt pressu pressure re loss. loss. This This is a safety safety device device that can be closed closed to prevent prevent flow from inside the drillstring drillstring during critical operations operations like kick control. control. It also isolates the drillstring from the surface equipment and allows disconnecting the kelly during critical operations. A kelly saver sub is simply a short length pipe with has male threads on one one end end and and femal female e on the other. other. It is screwed screwed onto the bottom bottom of the lowe lowerr kelly kelly valve or topdrive topdrive and onto the rest of the drillstring. drillstring. When the hole must must be deepened, and pipe added to the drillstring, the threads are unscrewed between the kelly saver sub and the rest of the drillstring, as opposed to between the kelly valve valve or topdrive topdrive and the saver saver sub. This means that the connection connection between the kelly or topdrive and the saver sub rarely is used, and suffers minimal wear and tear, whereas the lower connection is used in almost all cases and suffers suffers the most wear and tear. tear. The saver saver sub is expenda expendable ble and does not represent represent a major investmen investment. t. Howev However er,, the kelly or topdrive component component threads are spared by use of a saver sub, and those components represent a significant significant capital cost and considerabl considerable e downtime downtime when replaced. replaced. It is imporimportant that both lower kelly valve and kelly saver sub be of the same diameter of CHAPTER 2 Rotary Drilling Systems

Page 2–34



Figure 2.31: A square kelly and a hexagonal kelly.

Figure 2.32: A kelly valve.


Page 2–35



the drill pipe tool-joints to allow stripping into the hole during control operations.

2.4.3 2.4.3

Rotary Rotary Table able a and nd Compon Component ents s

Torque and is transmitted to the kelly by the kelly bushing . The kelly kelly bushin bushing g has an inside profile matching the kelly’s outside profile (either square or hexagonal), but with slightly larger dimensions so that the kelly can freely move up and down inside it (see Figure 2.33).

Figure 2.33: Kelly bushings.

Figure 2.34: Master bushings ([a] and [b]), and casing bushing (c). The kelly bushing fits in the master bushing , which, in turn, attach to the rotary table. table. It connects to the master bushing bushing either by pins of by a squared CHAPTER 2 Rotary Drilling Systems

Page 2–36



Figure 2.35: Kelly bushing and master bushing. link. A master casing bushing is is used to handle casings. Figure 2.34 shows two master bushing and one master casing bushing. The master bushing transmit torque and rotation from the rotary table to the kelly bushing. Figure 2.35 shows a kelly bushing, master bushing, and rotary table assembly. The master bushing (and also the master casing bushing) has a tapered interna internall hole hole as shown shown in the schema schematics tics in Figure Figure 2.36-a 2.36-a.. The purpose purpose of the tapered tapered hole is to receive receive the pipe slips (see Figure 2.36-b). 2.36-b). During pipe pipe connection or drillstring trip operations, this tapered hole receives either the drill pipe slips, or the drill collar slips, or the casing slips, which grips the tubular and frees the hook from its weight. Because of the slick shape of most drill collars, a safety clamp is always

(a)

(b)

Figure 2.36: Drillpipe slip (detail when set in the master bushing). CHAPTER 2 Rotary Drilling Systems

Page 2–37


(a)


(b)

(c)

Figure 2.37: DC slips, safety collar, and casing slips.

Figure 2.38: A rotary table. used used above above the drill collar collar slips (mandat (mandatory!) ory!) If the drill collars collars slides slides in the slips, the safety clamp works as a stop to force the slips to grip the drill collar. A drill collar slips, a safety collar, and a casing slips are shown in Figure 2.37. The rotary table (Figure 2.38) receives power from the power system (either mechanica mechanicall or electric.) A gearbo gearbox x allows allows several several combinations combinations of torque torque and speed.

2.5 2. 5

Wel elll Cont Contrrol Syst System em

The functions of the well control system are to detect, stop, and remove any undesired entrance of formation fluids into the borehole. An undesired entrance of formation fluid into the borehole borehole is called called kick and may occur due to several CHAPTER 2 Rotary Drilling Systems

Page 2–38


(a) A fixed rig BOP.


(b) A floating rig BOP.

Figure 2.39: BOP stacks. reasons (high pressure formations, insufficient drilling fluid density, drillstring swab swab,, loss loss of circula circulatio tion, n, format formation ion fracture fracture,, etc). etc). If the undesir undesired ed entra entrance nce of fluid feedbacks and the fluid continuously enters the borehole reaching the surface, it is called blowout . Blowouts (in particular gas blowouts) are extremely dangerous and put the crew, the rig, the drilling operation, and the reservoir at risk. The well control system must detect, control, and remove the undesired entrance of fluids into the borehole. The system is composed of sensors (flow rate, surface volume, annular and drillstring pressure etc,) capable to detect an increase of flow or volume in the fluid system, the blowout preventer (BOP), the circulating pressure control manifold (choke manifold), and the kill and choke lines. The BOP is a set of pack–offs capable of shutting the annular space between the surface casing and the drillstring. Because of the diversity in shape of the annular, annular, several different device types exist and they are normally nor mally assembled bled togeth together er (and (and in variou various s configu configurat ration ions) s) called called BOP stack stack (see (see Figu Figure re 2.39 2.39.. The BOP stack is located under the rotary table in land and fixed marine rigs, and on the bottom of the sea in mobile and floating rigs. The various types of BOP devices are: Annular BOP: The purp purpos ose e of the the annul nnula ar BOP is to shut shut the the annular lar in fron frontt of any kind of drillstring equipment (except stabilizers) or even without drillstring. The active element is an elastomeric ribbed donut that is squeezed around around the drillstring drillstring by an hydraulic hydraulic ram (see Figure Figure 2.40-a and -b). It is located located at the top of the BOP stack. Controllin Controlling g the pressure applied applied to the ram, it is possible to strip the drillstring in and out while keeping the CHAPTER 2 Rotary Drilling Systems

Page 2–39



(a)

(b)

(c)

Figure 2.40: Annular BOP’s (a and b) and an inside BOP (c) annular closed (requires the use of an inside-BOP, shown in Figure 2.40c, which should be connected immediately to the drillstring when a kick is identified). The inside bop acts as a check valve, allowing fluid be pumped down the drillstring, but blocking back flow.

Blind ram: The blind rams (normally one at the top of all other rams) allows shuttin shutting g the boreho borehole le with with no drillstring drillstring element element in front front of it. (See (See FigFigure 2.41-a, upper ram.) If the blind ram is applied to a drillpipe, the pipe will but no seal is obtained. Pipe rams: The pipe rams allows shutting the annular in front a compatible drill pipe (not in front of tool joints.) Normally two rams are used (a special CHAPTER 2 Rotary Drilling Systems

Page 2–40



spool between the two is used where the kill and choke line is connected. (See Figure 2.41-a, lower ram.) The use of two pipe rams also permit to snub the drillstring during the well control operation. Shear rams: The shear ram (normally one below the blind ram or below all other other rams) can shear a drill drill pipe pipe and provid provide e seal. seal. (See (See Figure Figure 2.412.41b.) This This is a last last resour resource ce when when all other other rams and annu annular lar had failed. failed. Circulation through the drillstring is lost and, if the shear ram is the lower one, the drillstring falls into the borehole.

(a)

(b)

Figure 2.41: BOP: (a) blind and pipe rams, (b) shear rams. All these safety devices are hydraulically actuated by a pneumatic–hydraulic system (actuators and accumulators), which can operate completely independent of the power system of the rig. Two control panels are normally used, one at the rig floor, and a remote one away from the risky area. The accumulators are steel bottles lined with a elastomeric bladers forming two separated compartments. One compartment is filled with oil, which powers the BOP. The other compartment is filled with air or nitrogen at high pressure. The pressure pressure of the gas pressurizes pressurizes the oil across across the elastomeric elastomeric liner. liner. Rig power power,, during during ordinary ordinary operation operation,, keeps keeps the gas in the accumulators accumulators under pressure. The accumulators should be able to provide hydraulic power to close and open all elements of the BOP stack a number of times without external power. Choke Manifold

During a kick control operation, some of the BOP stack devices are actuated to close the annulus and divert the the returning fluid to the choke line . The choke line directs the returning fluid to a manifold of valves and chokes called choke manifold , which allows to control the flow pressure at the top of the annular adjusting the flow area open to flow. The choke manifold also direct the flow to a flare (in case of a gas kick), or to the pits (if mud) or to special tanks (if oil). CHAPTER 2 Rotary Drilling Systems

Page 2–41



Figure 2.42: BOP accumulators and control panels.

Figure 2.43: Choke manifold.


Page 2–42


2.6 2. 6


Wel elll Moni Monito tori ring ng Syst System em

Several sensors, gauges, meters, indicators, alarms, and recorders exist in the rig to provide all data required to control (safely, efficiently, and reliably) of all operations under way in the rig. Among the most important parameters are: • weight weight on bit (WOB) (WOB) and hook load, load, • rate of penetrati penetration on (ROP), (ROP), • rotary speed, speed, • torque, torque, • circulating circulating (pump) (pump) pressure, pressure, • flow rate rate (in and and out), • drilling drilling fluid gain/los gain/loss, s, • mud temper temperature ature,, • mud density density,, • total hydroca hydrocarbon rbon gas in the drilling drilling fluid. Accurate and reliable indication of hook load and weight on bit are essential for the efficient control of rate of penetration, bit life, borehole cleaning, and borehole direction. The weight indicator works in conjunction with the deadline anchor using either tension or compression hydraulic load cells. The deadline anchor senses the tension in the deadline and hydraulically actuates the weight indicator. Most weight weight indicators indicators have have two hands hands and two scales. The inner scale shows the hook load and the outer one shows the weight-on–bit. To obtain obtain the weight–on– weight–on–bit, bit, the driller perform perform the following following steps: with the bit out of the bottom, the drillstring is put to rotate and the weight of the dril drills lstr trin ing g is obse observ rved ed in the the cent centra rall scal scale; e; using using the the knob knob at the the rim rim of the the weig weight ht indicator, the outer scale is adjusted so that the zero of the outer scale aligns with the longer hand. The driller lowers the drillstring slowly observing the long hand. hand. When the bit touches touches the bottom, part of the weight weight of the drillstring drillstring is transfe transferred rred from the hook to the bit (the weight–on weight–on–bit.) –bit.) The amount amount of weight transferred corresponds to the decrease of hook load, indicated by the long pointer (turning counterclockwise). All modern rigs have control consoles that shows all pertinent parameters in analog and or digital displays. All parameters and operations may be recorded in physical (paper) or magnetic media for post analysis. Some automated operations like constant weight–on–bit and constant torque are possible in most rigs. CHAPTER 2 Rotary Drilling Systems

Page 2–43



(a)

(b)

Figure 2.44: Weight indicator (a) and a deadline anchor (b).

Figure 2.45: Drilling control console.


Page 2–44



Chapter 3 Drillstring Tubulars and Equipment The purpose of the drillstring is to transmit mechanical power (torque and rotation), hydraulic power (pressure and flow rate), and weight to the bit. The drillstring is composed mainly of the following elements: • Drill pipes, pipes, • Heavy Heavy wall drill drill pipes, pipes, • Drill collars, collars, • Several Several special elements elements and tools. tools. Figure Figure 3.1 shows shows an schematic schematic of a typical rotary drillstring. drillstring.

3.1 Dr Dril illl Pi Pip pes Below the kelly assembly (upper kelly valve + kelly +lower kelly valve + kelly saver sub) is a length of drill pipes (DP). Drill Pipe is a primary and important drillstring member. Since the drill pipes are generally compose the upper and longest portion of the drillstring, they must be light and strong. The drill pipe body is a seamless pipe with outside diameter (OD) varying from 2 3 /8 in to 6 5/8 in. The outside diameter and the wall thickness t determine the linear weight of the drill pipe. The inside diameter (ID) is equal to OD minus 2t. Drill pipes are made of high grade steel (there are also drill pipes made of aluminum aluminum,, carbon carbon fiber, fiber, etc). API has standardized standardized four four steel grades: grades: E–75, E–75, X–95, X–95, G–105, G–105, and S–135. S–135. The figures represent represent the minimum yield strength strength Y s (in ksi) of the the steel. steel. Drill Drill pipes pipes are specifie specified d with with the follo followin wing g basic basic parameters: CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–1



Figure 3.1: Typical rotary drillstring. 1. Length Length range: range: Range I: 18ft to 22ft, Range II: 27ft to 30ft (most common), common), Range III: 38ft to 45 ft,

2. Nominal Nominal linear weight: weight: in general general 2 or 3 linear linear weight or wall thickness thickness for each standard OD 3. Wall Wall upset: EU (external (external upset), IU (internal upset), upset), and IEU (internal & external upset). The wall upset is a length of extra thickness at both ends of the drill pipe body to provide a smooth transition between the pipe body and the tool joint, in order to reduce the stress concentration, 4. Tool joint OD, ID, ID, and tong length, 5. Steel grade: grade: (D-55), (D-55), E-75, X-95, G-105, G-105, S-135, 6. Connectio Connection n size and and type: type: from 23 /8 in to 51/2 in, type IF (internal flush), EF (external flush), FH (full hole), XH (extra hole), SH (slim hole), DS (double streamline), and NC (numbered connection), The API RP-7G contains the specification of all API standard drill pipes approved for oil and gas drilling use. The tool joints are heavy coupling elements CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–2



having coarse, tapered threads and sealing shoulders designed to sustain the weight and to transmit torque along the drillstring. The threads of the tool joints are specially designed to offer strength (axial and torsional), easy handling, fast connectio connections ns (number (number of turns to make the connection connection), ), and leak-proof leak-proof sealing sealing (metal to metal). Tool joints might be welded or screwed to the ends of the drill pipe body.

Figure Figure 3.2: 3.2: Typical ypical tool joint joint design designs. s. (A) Internal Internal upset upset DP with full–hol full–hole e shrink–grip TJ, (B) External upset DP with internal–flush shrink–grip TJ, (C) External upset DP with flash–weld unitized TJ, (D) External–internal upset DP with Hydrill™–pressure welded TJ. Two other common properties of drill pipes are capacity and displacement . Pipe Capacity: The capacity A p of a drill pipe is a measure of its internal area, expressed as volume/length, usually gal/ft or bbl/ft. 1 If Di is the inside diameter (ID) of a drill pipe in inches, then

A p =

π 2 Di2 Di2 Di [in2 ] = [gal/ [gal/ft] = [bbl/ [bbl/ft] . 4 24. 24.51 1029. 1029.41

Pipe Displacement: The displacement As of the drill pipe is the measure of its cross-sectio cross-section n area, expressed expressed as volume/l volume/lengt ength, h, normally normally bbl/ft. bbl/ft. If D o 1

Sometimes Sometimes the capacity capacity is expressed expressed as the reversal reversal of the area, usually in ft/bbl. The reader should be attended to the units.

CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–3



is the outside diameter (OD) of a drill pipe in inches, then

π As = Do2 4



Di2

−



Do2 Di2 Do2 Di2 [in ] = [gal/ [gal/ft] = [bbl/ [bbl/ft] . 24. 24.51 1029. 1029.41

−

2

−

Annulus Capacity: The annulus capacity A a is not a property of the pipe because it depends on the diameter of the hole opposite to the pipe. If D W is the diameter of the well, the annulus capacity A a in bbl/ft is given by:

π 2 Aa = DW 4



Do2

−



2 2 DW Do2 DW Do2 [in ] = [gal/ [gal/ft] = [bbl/ [bbl/ft] . 24. 24.51 1029. 1029.41

−

2

−

The capacity and displacement formulas above do not take into account the tool joints, and manufacturer tables must be consulted when more accurate values values are required. In particular, particular, the nominal nominal weight that specify a given given drill pipe represents neither the pipe body linear weight, nor the the average linear weight (body plus tool joint divided by its length). It is just a nominal value. For example, a typical 5in DP with 19.5 lb/ft has an internal diameter of 4.276in. The density of steel is 489.5 lb/ft 3 . Therefore, one foot op pipe body weights

π 2 5 4



2

− 4.276

1ft3 144in3

× ×

489. 489.5 lb/ lb/ft3 = 17. 17.93 lb/ lb/ft .

Considering a 30 ft long DP (Range II), the tool joints (pin and box) comprise about 21 /2 ft of its length. Outside and inside diameters of the tool joints are 6in and 3 1 /2 in respectively. Therefore, the linear weight of the tool joint is

π 2 6 4



2

− 3.5

×

1 ft3 144 in3

×

489. 489.5 lb/ lb/ft3 = 63. 63.41 lb/ lb/ft .

The weight of the drill pipe (body plus TJ) is

27. 27.5 ft

17.93 lb/ lb/ft + 2. 2.5 ft × 63. 63.41 lb/ lb/ft = 651. 651.6 lbm . × 17.

Consequently, the adjusted linear weight of the drill pipe is

651. 651.6 lb = 21. 21.72 lb/ lb/ft . 30 ft Drill pipes are subjected subjected to wear wear during operation operation.. In particular, particular, reduction reduction of tool joints OD and wall thickness reduce tensile and torsion capacity of the element. Used drill pipes are classified as Premium or Class I I if the minimum wall thickness is at least 80% of the wall thickness of a new pipe, and Class II when at least 70%. A new pipe that for the first time is connected to a drillstring is immediately re–classified to premium DP. Table A.1 presents dimensional data for new drill pipes. CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–4



Figure 3.3: A DP elevator and the links to the hook body.

3.1.1 3.1.1

Drill Dri ll Pipe Pipe Elevat Elevator or

Drill pipes are handled during tripping using a drill pipe elevator . (The (The swive swivell and and kell kelly y are are set set asid aside e in the rat rat hole hole.) .) It is connec connecte ted d by two links links to the hook body (See Figures Figures 3.3 and 2.4) . A hinge and latch allows opening opening and closing closing the bi–parted bi–parted collar around the drill pipe. pipe. The elevator elevator is operated operated by the roughnecks at the rotary table level, and by the derrick man at the monkey board. Drill pipes extend across almost the whole length of the drillstring and, although relatively light, they contribute with a significant part of the drillstring weight weight (50% or more). more). Howev However er,, drill pipes are, in general, general, used only under under tension. They should not be subjected to compression due to its low resistance to buckling. Therefore, they cannot be used to apply weight on the bit. 2

3.2 3. 2

Dril Dr illl Coll Collar ars s

Since drill pipes cannot be used to apply weight on bit, this role is played be the drill collars (and (and also by heavy heavy weight drill pipes as shown next). next). Drill col2

In horizontal wells, drill pipes can be put under compression if located in a suitably curved sect sectio ion n of the the hole hole;; in addi additi tion on,, comp compres ressi sion on servi service ce drill drill pipe pipes s (CSD (CSDP P, S-13 S-135 5 grad grade e DP with with 2 or 3 wear knots) are specially designed to work under compression to drill short radius horizontal wells.


Page 3–5



lars (DC) are thick walled steel pipes located normally right above the bit, and their purpose is to provide weight (axial force) to the bit. Drill collars are manufactured with carbon steel (AISI 4115), or some non-magnetic alloy (stainless steel, monel metal). The outside of drill collars may be slick (small diameters) or spiral grooved grooved (any (any size.). Figure Figure 3.4 shows shows a spiraled spiraled and a slick drill collars. lars. The purpose purpose of the grove groves s is to reduce reduce or avoid avoid the risk of differe differenti ntial al sticking opposite to permeable formations . The depth of the grooves is made larger than the average thickness of a flocculated mud cake (see Figure 3.5). Average length of drill collars is 34 ft, but re–threading normally makes them shorter.

Figure Figure 3.4: A spiraled and a slick drill collars. collars.

Figure Figure 3.5: 3.5: Spiral Spiraled ed DC cross– cross–sec sectio tion. n.

Figure Figure 3.6: 3.6: A DC eleva elevator tor..

The elevators for drill collars are very similar to the elevators for drill pipes. They differ in the shape of the internal hole that clamps on the pipe. Most drill collars collars are recessed recessed so as to be handled with the elevator elevator.. If the drill collar is not recessed (sometimes even if it is!), a special sub called lift sub is is used. Lift subs have the shape of the upper end of a drill pipe, and connects to the top of sections of drill collars during trips. Then the drill pipe elevator can be used to lift or lower the drillstring.

3.3 3. 3

Heav He avy y Wal alll Dr Drill ill Pi Pipe pes s

In addition to drill pipes and drill collars, there are special pipes called heavy wall drill pipes (HWDP). (HWDP). They are intermediary pipes between drill pipes and drill collars, being strong enough to be put under compression (they contribute CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–6



to the available weight to apply to the bit), and they are flexible enough to be used used in direct direction ional al drillin drilling g (less (less torqu torque e and drag drag than than drill drill collar collars.) s.) The use of HWDP also allow a gradual transition between the flexible drill pipes and the stiff drill collars (less stress concentration, and therefore, less mechanical fatigue on the threads.) HWDPs look very similar to regular drill pipes, being of the same length of Range II DP (27 to 30 ft), but with longer tool joints (to permit re–threading). HWDPs HWDPs have have a centra centrall extern external al upset as shown shown in Figure Figure 3.8. This This centra centrall upset provides an additional third point of contact, increasing the overall stiffness and protecting the pipe sections from excessive wearing in high inclination wells (normally the tool joints and central upset have a band of hard material to prevent/reduce prevent/reduce wear).

Figure 3.8: Heavy wall drill pipes.

3.4 3. 4

Spec Specia iall Tools ools

Several drilling equipment are used in the drillstring. The most important are: • stabilizer stabilizers, s, • reamers, reamers, • hole–open hole–openers. ers.

3.4.1 3.4.1 Stab Stabil ilize izers rs Stabilizers provide localized additional support points (localized larger diameter) ter) in one one or more more posi positio tions ns alon along g the the drill drillst strin ring. g. For vertic vertical al well wells, s, the the stabi stabililize zerr preven prevents ts low frequency frequency vibration vibration in the drillstring drillstring during rotation. The advanadvantages are: • reduce reduce wear (of both drillstring drillstring and casing), casing), • reduce mechanical mechanical fatigue, fatigue, • reduce mechanical mechanical instability of formation formation (caving), • reduce reduce tortuous tortuous hole. hole. CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–7



Stabilize Stabilizers rs are essential essential equipmen equipmentt for for directiona directionall drilling. drilling. Suitable Suitable choice of number and position of stabilizers in the drillstring lend to the bottom hole assembly (BHA, the lower part of the drillstring composed of drill collars, stabilizers and HWDP) special characteristics in terms of inclination control: • angle build– build–up up,, • angle drop–off drop–off,, • angle hold. hold. Stabilizer may be of the following types: • integral integral blade blade,, • interchangeable interchangeable blade, blade, • non–rotating blades, • replaceable blades, • clamp–on clamp–on • near-bit, near-bit, The diameter of stabilizers can be in gauge or under gauge . More More recentl recently y, remote adjustable blade stabilizers were introduced. Changing suitably the diameter of the stabilizers provide a level of control in the directional behavior of BHAs.

(a)

(b)

(c)

(d)

Figure 3.9: Some Stabilizers: (a) integral, (b) interchangeable, (c) non–rotating, (d) replaceable. CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–8


3.4. 3.4.2 2


Ream Re amer ers s

The purpose of the reamer is to keep the diameter of the open hole in gauge, that is, with the expected original diameter of the bit. Two reasons may cause a decrease in the original diameter: 1. formation swelling swelling (hydrated shales, shales, moving salt), 2. bit diameter reduction (hard (hard and abrasive abrasive formations). The reamer also functions as a stabilizer since the rollers touch the borehole wall. wall. Diffe Differen rentt types types of roller rollers s can be select selected ed to suit suit the formati formations ons being being reamed. (See Figure 3.10.)

Figu Figure re 3.10 3.10:: A roll roller er ream reamer er..

Figu Figure re 3.11 3.11:: A fixed fixed hole hole–o –ope pene nerr.

3.4.3 3.4.3 Hole– Hole–op open ener ers s The hole–opener is a tool designed to enlarge the diameter of a previously (or simultaneously drilled smaller borehole. Three situations (at least) are possible: (See Figure 3.11.) 1. to drill the borehole borehole section with a smaller smaller bit, and later to enlarge enlarge to the final diameter (a special tool called bull nose is connected in the place of the bit, to guide the hole opener along the pre-drilled hole), 2. to drill the borehole borehole section section with a smaller smaller bit and simultane simultaneously ously enlarge enlarge to the final diameter, CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–9



3. to enla enlarg rge e a sect sectio ion n belo below w a casi casing ng with with a diam diamet eter er larg larger er than than the the inte intern rnal al diamet diameter er of the casing. casing. In this case, case, a specia speciall hole hole opener opener (also caller caller underreamer) with hinged arms actuated hydraulically is used (the drilling fluid pressure actuates in rams that open the arms forcing the cutters against the borehole wall).

3.5 Connec Connectio tions ns Make–u Make–up p and Break– Break–out out To make–up and break–out the pipe connections (during normal operations and drillstring trips), big self–locking wrenches called manual tongs are used to grip the drillstring drillstring and apply torque. torque. The tongs are kept suspended suspended at suitable height above the rotary table (3 to 5ft) balanced by counter–weights. They work in pairs, one turns to left (counterclockwise) and the other to right (clockwise). They are mechanically or pneimatically actuated by the cathead (special rotating spools connected to the drawworks). (See Figure 3.12).

(a) Left tongue (break-out).

(b) Right tong (make–up).

Figure 3.12: Manual tongs. To make–up a connection, the left tong grips the upper tool joint joint (box) of the lower pipe, and the right tong grips the lower tool joint (pin) of the upper pipe. pipe. The left tong is connected by a steel rope to a fixed fixed point in the derrick, derrick, and the right tong is connected to the cathead (turns the upper pipe). To break a connection, the left tong grips the lower tool joint (pin) of the upper pipe, and the right tong grips the upper upper tool joint (box) of the lower pipe. pipe. The right tong is connected by a steel rope to a fixed point in the derrick, and the left tong is connected connected to the cathead (also turns the upper pipe). pipe). Figure Figure 3.13 shows the tongs ready to make–up a connection. CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–10



Figure 3.13: Tongs in position to make–up a connection. When set and ready to apply torque, the angle between the arms of the tongs should be either 90◦ or 180◦ (the ideal is zero degrees but operational difficulties difficulties make this position position not practical) practical).. It is important important to leave leave the rotary table unlocked, to avoid damage to the pipe caused by the slips. Comp Compre ress ssed ed air air tong tongs s or spinners (see (see Figu Figure re 3.14 3.14)) are are also also used used to spee speed– d– up the operation, but the torque to make–up or to break–out the connection is always done using the manual tongs.

3.5.1

Maximum Maximum Height Height of of Tool Tool Joint Shoulders Shoulders

To make–up or break–out a connection, the drillstring must be resting on the master bushing using the slips (DP and HWDP) or slips+safety clamp (DC and any other slick equipment.) The angle between the arms of the tongs are either 90◦ or 180 ◦ as shown in Figure 3.16. The maximum height of the tool joint shoulder with respect to the master bushing is given by the following formulas:

Case 1: 90◦

H max max [f t] = 0.053


Y min min LT S T mu mu Page 3–11



Figure 3.14: A spinner.

Figure 3.15: Tongs position during make–up. Figure 3.16: Case 1: 180◦

H max max [f t] = 0.038

Y min min LT S T mu mu

where: Y min min = minimum yield stress of the pipe [psi] LT = tong’s arm length [ ft ] 4 4 π(Do −Di ) S = I c = 32D = sect sectio ion n modu modulu lus s [in [in3 ] 32Do T mu [ft lbf] mu = make up torque


Page 3–12



Example 5: Calculate maximum maximum height of the tool joint shoulder for a 5 in OD DP, 19.5 lb/ft, X-95, with NC50 65/8 in – 312 in new tool joint, using tongs with 31 /2 ft positioned at 90 ◦ and 180 ◦ . Solution:

Y min 95, 000psi min = 95 kpsi = 95, LT = 3.5 ft π (54 4.2764 ) S = = 5.71 in 71 in 3 32 5

− ×

T mu 27, 076 ft lbf mu = 27, a) 90 ◦ :

H max max = 0.053

95, 95, 000 3.5 5.71 = 3.72 f 72 f t 27, 27, 076

H max max = 0.038

95, 95, 000 3.5 5.71 = 2.76 f 76 f t 27, 27, 076

b) 180 ◦ :

3.5.2 3.5.2

·

× ×

× ×

Make–up Make–up Torque orque

It is very important to apply the right torque during the make–up of a connection. Too little torque will not provide provide a suitable suitable seal between between the pin and box shoulders, and leakage might wash out the threads causing failure of the connection. Too much torque may cause mechanical failure of the threads, either in the box or in the pin. The API RP7G tables present the maximum (tb.9) and minimum (tb.10) torques for each standard connection.

3.6 Dr Dril illl Bi Bitt The bit is connected to the lower end of the drill collars. Bits are manufactured with a pin, so that to connect to the lower pin of the drillstring, a bit sub is used. The bit sub is a short sub ( 11 /2 to 2 ft) with two box connections. There There are a large large variety variety of bits. Each Each type is design designed ed to drill drill rocks of different hardness, composition, abrasiveness, etc, encountered during drilling operation operations. s. It is a duty of the drilling enginee engineerr to select select the most appropriate appropriate bit and the drilling parameters (nozzle sizes, weight-on-bit, rotation speed, and flow rate) to optimize the performance of the operation. A more detailed study of drill bits is covered in Chapter 10. CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–13



3.7 Other Other Dri Drills llstri tring ng Equipm Equipment ent In addition to the kelly–rotary table assembly, two other methods can be used to promote rotation to the bit: • Top drive, drive, • Bottom hole motors (positive (positive displacement motors and and turbines).

3.7.1 3.7.1 Top Dr Driv ive e The top drive , also called power swivel , takes the place of the kelly, and the torque is applied to the top of the drill pipe section by mean of hydraulic or electric electric motors. The assembly assembly slides along tracks (most models incorporate incorporate a swivel swivel in the design,) design,) and is suspended suspended by the hook. The reactive reactive torque is transmitted to the rig structure directly through the tracks or by a torque reaction beam.

Figure 3.17: An electrical top drive. A great advantage of using a top drive is the possibility to drill a full stand (3 or 4 drill drill pipes) pipes) without without interru interrupti ption, on, savin saving g time time in conne connectio ctions. ns. Anothe Anotherr advantage is the possibility of rotating the drillstring during the trips reducing the drag to pull–out or slack–off the drillstring in the hole for high inclination drilling. CHAPTER 3 Drillstring Tubulars and Equipment

Page 3–14



3.7.2 3.7.2 Bott Bottom om Hole Hole Moto Motors rs Bottom hole motors are special engines located above the bit to promote bit rotation. rotation. Bottom Bottom hole motors motors convert convert hydraulic hydraulic power power of the drilling drilling fluid ( P = q ∆ p ∆ p) into mechanical (rotational) power. Turbines use fluid momentum conversion on the blades of stator/rotor to gene genera rate te rota rotati tion on and and torq torque ue.. Turbi urbine nes s oper operat ates es in high high spee speed d and and has has a narr narrow ow range of operation. The torque decreases steadily from the maximum at 0 rpm (stalled) to zero at the maximum speed.

Figure 3.18: A bottom hole turbine. Positive displacement motors (PDM) use continuous displacement of constant volume compartments created between an elastomer stator and a steel rotor to generate rotation and torque. The rotation speed of a PDM is function of the flow rate, and the torque is directed related to the pressure differential across across the motor (easily monitored monitored from the surface). surface). The use of bottom bottom hole motors is essential for directional drilling. The use of bent sub or bent housing provides a good deal of control of the inclination and the azimuth, allowing to drill complex trajectories.

Figure 3.19: A bottom hole PDM.


Page 3–15




Page 3–16



Chapter 4 Introduction to Hydraulics By definition, a static fluid cannot sustain shear stresses, otherwise it will enter in motion motion (and (and will will not be static static anymore anymore). ). The consequ consequenc ence e of this is that the state of stress inside a fluid is such that the normal stresses are the same in any direction. direction. This state of stress stress is called called hydrostatic state of stress . The magnitude of the stress is called pressure .

4.1 Hydro Hydrosta static tic Pressu Pressure re The hydrostatic pressure inside a homogeneous fluid comes from the pressure at the surface and the weight of the fluid above the point in question. To calculate the pressure at any point inside a column of fluid of density ρ (gas or liquid), we consider an infinitesimal element of fluid with volume dV = dx dy dz as shown in Figure 4.1.

· ·

Figure 4.1: Stress state about a point in a fluid. CHAPTER 4 Introduction to Hydraulics

Page 4–1



To remain in equilibrium, the resulting forces acting in all 3 directions must be zero. Then we have:



 

F x = p x dy dz

dz = 0 → dpx = 0 − ( px + dpx) dy dz =

F y = p y dz dx

dx = 0 → dpy = 0 − ( py + dpy ) dz dx = F z = p y dz dx − ( py + dpy ) dz dx − ρ g dx dy dz = = 0 → dpz = −ρ g dz

The horizontal gradients ( dpx /dx and dpy /dy) are zero zero. Using Using p instead of pz , we have, for the hydrostatic gradient:

dpz = dz

−ρ g

Since in general we will be dealing with depth D , and since a point at depth D has coordinate z = D, the expression for the pressure differential in terms of depth is: dp = dp = ρ ρ g dD (4.1)

−

Note that ρ is mass per volume , or specific mass , or density . Therefore, ρ g is weight per volume or specific weight , usually noted as γ . For a gravitational system of units as the British System, the acceleration of gravity is equal to 1 G. Such that 1 lbm 1 G = 1 lbf . .

×

That is the magnitude of the force is numerically equal to the magnitude of the mass. mass. For For a scientific scientific or non–gra non–gravitati vitational onal system as the International International System 2 9 .80665m/ 80665m/s , such that (SI), the acceleration of gravity is equal to 9.

1 kg

80665m/s2 = 9.80665 N . × 9.80665m/

To integrate (4.1), we need to know how the density depends on the pressure. sure. All fluids fluids are are compressible , but for some applications, some fluids can be classified as incompressible . Liquids, in general are incompressible up to a considerably high pressures. Gases are, in general, incompressible.

4.1.1

Hydrostati Hydrostatic c Pressure Pressure for Incompressib Incompressible le Fluids Fluids

For incompressible fluids (liquids in general are in this class), the density ρ is constant and integrating the (4.1) yields:

p1 = p = p 0 + ρ g (D ( D1

− D0) ,

which allows us to calculate the pressure at a point with depth D1 if we know the pressure p0 at a point of depth D 0 . For pressure in psi, density in lbm/gal, and depth in feet we have:

p1 = p 0 + CHAPTER 4 Introduction to Hydraulics

12 ρ(D1 231

0 .0519ρ 0519ρ(D1 − D0 ) − D0) ≈ p0 + 0. Page 4–2



For pressure in psi, density in lbm/gal, and depth in meter we have:

p1 = p = p 0 +

12 D1 D0 ρ 231 0.3048

−

0 .1704ρ 1704ρ(D1 − D0 ) ≈ p0 + 0.

Example 6: Calculate Calculate the absolute absolute pressure pressure 1 at the depth of 3,000 ft in a borehole filled with a drilling fluid with 9.2 lb/gal, in a location where the atmospheric pressure is 13.5psi. Solution:

13.5 psi + 0. 0.0519 p = p = 13.

lbm/gal × (3, (3, 000 ft − 0 ft) = 1, 1, 446 psia × 9.2 lbm/

Example 7: Calculate Calculate the fluid density required required to drill a permeable permeable formation at 12,000 ft if the pore pressure of the formation is 8,200 psig, with an overbalance of 50 psi.

Solution:

8200 psig + 50 psid = 0. 0.0519

× ρ × 12000 ft

ρ = 13. 13.25 lbm/ lbm/gal

4.1.1.1 4.1.1.1

Complex Complex Fluid Fluid Colum Column n and Equival Equivalent ent Density Density

If the column of fluid is composed of several segregated fluids with different densities (complex fluid column), we still can use the expression for hydrostatic pressure above, considering that at the top of a homogeneous column of fluid acts a pressure resulting from the fluid above:

p = p = g g



ρi T i ,

where ρ i is the density of the i th layer of fluid and T i its thickness. For a complex fluid column, the equivalent fluid density at a given depth is the density of a homogeneous fluid that would cause the same hydrostatic pressure pressure at that depth. Note that the equivalen equivalentt density density depends depends on the depth in consideration. 1

The zero point of an absolute reference is the absence of all matter. There is no pressure at absolute zero. zero. On the other hand, a gauge pressure pressure measures pressure relative to the local atmosphere atmosphere.. Changes Changes in local atmospheric atmospheric pressure pressure occur due to weather weather,, altitude, altitude, and/or depth. Gauge pressure indications usually use a "g" after the unit as in "psig". Absolute pressure may have the letter "a" after the unit as in "psia". For pressure differential, it is common to use a "d" after the unit.

CHAPTER 4 Introduction to Hydraulics

Page 4–3



D = A general expression for the equivalent density at depth D = ρeq =



ρi T i . D



T i is

Example 8: A borehole borehole is filled with fluid fluid of density 8.8 lbm/gal lbm/gal from the the surface to the depth of 8,000 ft, and a fluid with 10.0 lbm/gal below 8,000 ft. Calculate culate the absolute absolute pressure pressure at 12,200 12,200 ft. What is the equivalen equivalentt density density of the fluid at 12,200 ft. Assume standard atmospheric pressure .

Solution:

p12200 = 14. 14.696 psi+0. psi+0.0519 8.8

×

lbm gal

ft+00.0519 × 10 lbm ×8000 ft+ ×(12200 ft−8000 ft) gal

P 12000 12000 = 5848 psia 5848 psia = 14. 14.696 psi + 0. 0.0519

× ρeq × 12200 ft

lbm/gal ρeq = 9.21 lbm/

4.1.2

Hydrostati Hydrostatic c Pressure Pressure for Compressib Compressible le Fluids Fluids

The density ρ of a compressible fluid (gases in general are compressible) is not constant constant and depends depends on the pressure pressure and temperat temperature. ure. In order to account account for this, we must consider the equation of state for real gases:

¯ T = z m R ¯T , pV = z n R M where z is the real gas deviation factor (see Figure 4.2) of the real gas at ρ = m/V m/V results in: pressure p and temperature T . Solving for ρ =

ρ =

M ¯ T p . z R

(4.2)

Considering the expression (4.1) for the pressure differential we have:

Mg dp = dp = ¯ p dD . z R T Since z depends depends on p , separating variables and integrating results in: p2

z Mg dp = dp = ¯ (D2 p R T



p1 2

2

− D1 ) .

And assuming that the temperature is constant in the gas column.


Page 4–4



Figure 4.2: Real gas deviation factor.


Page 4–5



The pressure integral can only be calculated if we know how z depends on p. This is normally very complicated. For short columns, z cam be considered constant and we can write: p2

z



p1

dp Mg = ¯ ( D2 p R T M g

− D1 ) ,

(D (D2 −D1 )

p2 = p 1 e z1 ¯R T

,

(4.3)

where z 1 is the compressibility factor at pressure p1 and T . A more more accurat accurate e approach is to use an average value for z given by:

z z¯ =

z 1 + z 2 , 2

where z 2 is the deviation factor for p2 calculated using the expression above. Using this new average value of the compressibility factor, a new pressure p2 is obtained from M g ( D2 −D1 ) p2 = p = p 1 e z¯ ¯R T (D , and compared compared with the previous previous one. The process process is repeated repeated until converconvergence is obtained. ¯ for various units are: Values for the universal gas constant R

¯ = 10. R 10.732

psi ft3 lb–mole◦ R

lbf ft lb–mole◦ R kPa m3 ¯ R = 8.3145 kg–mole K J ¯ = 8314. R 8314.5 kg–mole K ¯ = 1545. R 1545.4

It is important to note that pressure and temperature must be given in absolute scales , as required by the gas equation of state. The absolute absolute temperatures temperatures are normally the Rankine and the Kelvin scales and given by:

T [ T [◦ R] = t[ t[◦ F] + 459. 459.67 , 67 , = t[[◦ C] + 273. 273.15 . 15 . T [K] T [K] = t The figures above are normally approximated to 460 and 273 respectively. The Fahrenheit and Celsius temperature are converted using the following relations: 9 ◦ t[◦ F] = (t (t[ C] + 40) 40 , 40 , 5 5 ◦ t[◦ C] = (t (t[ F] + 40) 40 . 40 . 9

− −


Page 4–6


Example 9: gas)


What is is the density density of the air at 13 psia psia and 60 60 ◦F? (assume ideal

Solution: With sufficient accuracy, the molecular weight of the air is:

M air air = 22%

lbm 28.88 × 32 + 78% × 28 = 28. lb–mole

Using Equation (4.2) gives:

M ρ = ¯ p = z R T 1

×

lbm 28. 28.88 lb–mole lbf ft ◦ R 13 psi 1545. 1545.4 lb–mole (60 + 460) R ◦

×

  144 in2 1 ft2

= 0.0673 lbm/ lbm/ft3

Example 10: Consid Consider er a 10,000 10,000 ft deep deep borehole borehole with a drillstri drillstring ng and bit to the bottom bottom.. The The annula annularr is comple completel tely y filled filled with with methan methane e (CH 4 ), and the drillstring drillstring is filled with a 8.4 lbm/gal lbm/gal mud. After closing closing the BOP, the pressure pressure in the drill pipe at the surface is 640 psia. What is the expected pressure in the casing at the surface, assuming ideal gas and average temperature of 150 ◦ F. Solution:

The mass of one lb–mole of methane is 1x12+4x1 = 16 lbm. The drillstring– annular system form a U–tube system. The pressure at the bottom of the borehole can be calculated using the fluid inside the drill pipe and the surface pressure:

pbottom = 640 psi + 0. 0.0519

lbm/gal × 10000 ft = 5000 psi × 8.4 lbm/

This pressure is balanced by the pressure of the casing at the surface and the hydrostatic pressure of the gas. Using Equation (4.3) gives:

psur = 5000 e 1

×

16×1 (0 1545.4×(150+460)

−10000)

psur = 4220 psi

4.2 4. 2

Buo Buoyanc ancy

Archimedes principle of buoyancy states that the buoyant force exerted on a body fully or partially immersed in a fluid is equal in magnitude (and opposite in direction) to the weight of the volume of fluid which is displaced by that body. For homoge homogeneo neous us bodies bodies immers immersed ed in homog homogen eneou eous s fluids fluids,, the net or or buoyed weight of the body can be calculated from

W net net =


−  1

ρf W ρb Page 4–7



where W is the weight of the body (in the air), ρf is the density of the fluid,

−  ρ

and ρb is the density density of the body. body. The term term 1 ρf b is called buoyancy factor . This expressio expression n is valid valid only for homogeneous homogeneous bodies bodies fully immersed immersed in homogeneous homogeneous fluids. For homogeneous bodies, the geometric center of the body coincides with the center of mass. For non–homogeneous bodies, an equivalent density (total mass mass/v /vol olum ume) e) can can be used used,, but it is impo importa rtant nt to keep keep in mind mind that that the the geom geomet etri ric c center (where the buoyant force applies) may not coincide with the center of mass. In these cases, stable or instable equilibrium may exist.

Example 11: What is the weight weight of 0.4 0.4 ft3 of carbon steel? What is its buoyed weight when submerged in a 9.3 lbm/gal fluid? What is the equivalent density of the buoyed body in lbm/gal? Solution:

The average density of carbon steel is 490 lb/ft 3 = 65.5 lb/gal, so that the weight of the body is

W = ρ s g V b = 490

lbm ft3

× 1 G × 0.4 ft3 = 196 lbf

The buoyed weight is:

W buoyed buoyed =

 −   − × ρf W = ρs

1

1

9.3 65. 65.5

196 = 168. 168.2 lbf

The equivalent density is the density that would result in the same buoyed weight: ρf ρf 1 W b = 1 ρb g V b = ρ = ρ eq g V b ρb ρb

−  − 

ρeq = ρ b

65.5 lbm/ lbm/gal − 9.3 lbm/ lbm/gal = 56. 56.2 lbm/ lbm/gal − ρf = 65.

Exampl Example e 12: A 12,0 12,000 00 ft long long drills drillstri tring ng is comp compos osed ed of the the follo ollowi wing ng elem elemen ents ts (bottom up): 420 ft of 8 in OD–3 in ID DC, 840 ft of 7 in OD – 3 in ID DC, and 5 in E-75 19.5lb/ft drill pipe (22.28 lb/ft). The fluid density is 8.9 lb/gal. Calculate the expected hook load when the drillstring is hanging on the elevator (out of the bottom).

Solution: The linear weight of the DCs are π 1 ft2 w = (82 32 ) 8 in DC: w = 144 in2 4

×


−

  × ×

490 lbm/ lbm/ft3 = 147. 147.0 lbf /ft Page 4–8



Figure 4.3: Drillstring schematics for Example 12.


Page 4–9



π 1 ft2 w = (72 32 ) 7 in DC: w = 144 in2 4 The weight in the air of the drillstring is

×

W air air = 420

−

  × ×

490 lbm/ lbm/ft3 = 106. 106.9 lbf /ft

147.0 + 840 × 106. 106.9 + (12000 − 420 − 840) × 22. 22.28 = 390823 lbf × 147.

The buoyed weight of the drillstring is:

W hook hook = W buoyed buoyed =

 − × 1

8.9 65. 65.5

390823 = 337719 lbf

If the body is either not totally submerged, or submerged in a inhomogeneous neous fluid, the expression expression above above cannot be used. used. This may be complicated complicated for complex shape bodies. A more general way to calculate the buoyed weight, even for partially immersed bodies and for complex fluid column is to calculate the effect of the hydrostatic pressure on the body. This is demonstrated in the next example.

Example 13: Recalculate the hook hook load of Example 12 using the effect effect of the hydrostatic pressure on the drillstring.

Solution: Here we face a problem: the tool joints of the drill pipes causes a considerable increase increase in the average average linear weight weight os the drill pipe body. body. Although Although we can consider each tool joint of the drillstring, a most appropriate way is to consider an equivalent cross section of the drill pipe. Since the purpose is to determine the effect of the fluid in the weight, it is not important if we choose to change the inside or the outside diameter diameter (or both). both). In the present present example, example, we chose to change both, such that the average diameter remains the same. In this case, 5+4.276 = 4.638 in, and the area is given by the average diameter is D ave = 5+4. 2

As = π Dave t , where t is the thickne thickness ss of the equiva equivalen lentt drill drill pipe. pipe. The equiva equivalen lentt outsid outside e + t and the inside diameter is Dave t. Theref diameter is Dave + t Therefore ore,, for for a drill drill pipe with 22.28 lb/ft, we have:

−

π

× 4.638 in × t ×

lb 490 3 ft

 ×

1 ft2 144 in2



= 22. 22.28

lb ft

t = 0.449 in Do = 4.638 + 0. 0.449 = 5. 5.087 in Di = 4.638

− 0.449 = 4.4.189 in

Note that these figures are for buoyancy calculation only, and should never be used for strength calculations (torque, tensile, burst, collapse, etc). CHAPTER 4 Introduction to Hydraulics

Page 4–10



Now, the fluid pressure acts in the cross section areas at 10,740 ft, 11,580 ft, and 12,000 ft. In the first and second areas, the forces are downwards, and in the third case the force is upward. These forces add and subtract to the weight of the drillstring. drillstring. To calculate the force we need the pressures pressures at each depth and the area exposed to the fluid. The values are:

p10740 = 0.0519

× 8.9 × 10740 = 4961 psi p11580 = 0.0519 × 8.9 × 11580 = 5349 psi p12000 = 0.0519 × 8.9 × 12000 = 5543 psi A10740 =

π [72 4

[4.1892 − 32 ] = 24. 24.87 in2 − 5.0872] + [4. π 2 A11580 = 8 − 72 = 11. 11.78 in2 4 π 2 A12000 = 8 − 32 = 43. 43.20 in2 4



The hook load is:

W hook hook = 390823 + 4961

   



24.87 + 5349 × 11. 11.78 − 5543 × 43. 43.20 = 337756 lbf × 24.

This result should be compared with the previous example. The discrepancy is due rounding errors. This method should always be used when either the annular or the drillstring is filled with non–homogeneous fluids, partially filled, etc.


Page 4–11




Page 4–12



Chapter 5 Drillstring Drillstring Design The drillstring must be designed to support (with a safety margin) all the static and dynamic dynamic loadings loadings that occur during normal and special operations operations.. It also must must support support some some extre extreme me situat situation ions s like like pipe pipe stickin sticking g probl problems ems,, curved curved holes holes,, harsh environment, etc. In addition, it must be able to provide a suitable conduit for the drilling fluid flow without causing excessive frictional pressure drop both inside inside and outside outside the pipe. pipe. The mechani mechanical cal aspects aspects of drillin drilling g design design are covered in this chapter.

5.1 5. 1

Leng Length th of Drill Drill Colla Collarrs – Ne Neut utra rall Poin Pointt CalcuCalculation

One of the purposes of the drillstring is to apply weight on the bit, and as mentioned before, this is obtained by slacking part of the drillstring weight on the bit. When this is done, a portion of the lower end of the drillstring will be put in compression, and the upper portion will remain in tension. Since drill pipe can not be compressed, a sufficient length of drill collars and/or heavy weight drill pipes should be used so that the required weight on bit can be applied without compressing the drill pipes. The reason drill pipes can not be put in compression is that the moment of inertia1 of drill pipes are small compared to its length (a parameter called slenderness = length/radius of gyration). On the other hand, the slenderness of the drill collars are relatively low and compression is allowed. The slenderness is an important parameter to determine the mechanical buckling resistance of a column. The cause of buckling is the moment created by compressive forces. Therefore, a lot of discussion occurred to determine the right way to account the effects of hydrostatic pressure in the tendency to buckle a column. One of the considerations was that both the compressive force due to the weight on bit and the compressive force due to the hydrostatic pressure of the CHAPTER 5 Drillstring Design

Page 5–1



Figure 5.1: Assumption 1 – pressure contributes to buckling. fluid would contribute to buckle the drillstring, since the pressure acting in the lower area of the drillstring causes a substantial compressive force at the bit, even even with no weight weight applied against against the rock. Based on this assumption, assumption, the position of the neutral point of stress can be determined, and a sufficient length of drill collars must be used such that the neutral point lays in the drill collars (neutral (neutral point occurs where the axial stress is equal equal to zero.) zero.) Using this point of view, we can determine the position of the neutral point using the diagram in Figure 5.1. Consid Considerin ering g the elemen elementt of length length x in the lowe lowerr end end of the column column.. The forces acting on this element is shown in the diagram, and the equilibrium of forces results in the following equation:

F T T = W p A

− −

− F b

where F T force at acting acting at the top section of the element, element, W is the T is the axial force weight of the element of length x , p is the pressure at depth D, A is the cross section of the column, and F b is the reaction of the force applied to at the bit. The force due to the pressure p acting on the lower area A is given by

p A = ρ = ρ f g D A , and the weight of the element is given by

W = ρ s g A x . CHAPTER 5 Drillstring Design

Page 5–2



Substituting in the expression above we have for F T T :

F T Assumption1 T = ρ s g A x ρf Assumption1

pressurecontr ibutestobuckling.g .g D A−F b = g = g A (ρs x − ρf D ) −− pressurecontributestobuckling

−

The stress at x is calculated dividing F T T by the area A :

σ =

F T T = g (ρs x A

− ρf D) − F Ab .

σ = 0, that is: The neutral line is the position x n where σ = 0=

F T T = g (ρs xn A

Solving for x n we get:

xn =

− ρf D) − F Ab .

ρf F b D + ρs ρs g A

xn =

ρf F b D + . ρs wc

where wc = ρs g A is the linear weight weight of the drill collar. collar. Note that even even if the weight on bit is zero (drillstring hanging off the bottom), this expression tells us that the neutral point is quite above the bit, and that the lower end of the drillstring will be under compression.

Example 14: Calculate the position position of the neutral neutral point for for a column 10,000 10,000 ft long hanging off bottom and submerged in a borehole filled with a 9.3 ppg (poun (pound d per gallon gallon)) fluid. fluid. What What is the length length of drill drill collar collar with 147 lb/ft lb/ft is required to apply 100,000 lbf on the bit, assuming that, for safety, only 85% of the total length of DC is compressed. compressed. Calculate Calculate the suspended suspended weight weight of the drillstring drillstring using nominal nominal 19.5 lb/ft DP (actual 22.28 lb/ft). Use the assumption assumption that pressure contributes to buckle the drillstring. Solution:

Off bottom:

xn =

ρf F b 9.3 D + = ρs wc 65. 65.5

0 = 1420 ft × 10000 + 147

On bottom:

xn =

ρf F b 9.3 D + = ρs wc 65. 65.5

= 2100 ft × 10000 + 100000 147

Since this length should be 85% of the total, the minimum length of DC’s is

LDC = CHAPTER 5 Drillstring Design

2100 = 2471 f 2471 f t 85% Page 5–3



Figure 5.2: Assumption 2 – pressure does not contribute to buckling. The hook load (buoyed weight) of the drillstring is:

W h = [2471

22.28] × 147 + (10000 − 2471) × 22.

  × − 1

9.3 65. 65.5

= 455592 lbf 455592 lbf

The fact, however, is that the neutral point position calculated as above is calculated assuming that the forces created by hydrostatic pressure will contribute to the buckling of the column. But this is not true. If we recall the Archimedes’s effect, the resultant force due to hydrostatic pressures is equal in magnitude and opposite in direction to the weight of the displaced displaced fluid. fluid. But that is not all. In addition addition to that, the resulting resulting moment moment of the hydrostatic pressure must be ZERO, otherwise, the fluid would rotate about its center of mass. This means that the hydrostatic pressure acting on an immersed column can not contribute contribute to buckle buckle it. The easiest easiest (and correct) way to figure out the required length to apply a weight on bit is exactly doing this, that is, calculating the length of a column whose buoyed weight is equal to the required weight on bit. Considering the diagram in Figure5.2 we have:

W = ρ s g A xn B = ρ f g A xn F b = W

− − B = (ρs − ρf ) g A xn =

CHAPTER 5 Drillstring Design

−  1

ρf ρs

ρs g A xn

Page 5–4


xn =


F b

−  ρf ρs

1

wDC

where wDC is the linear linear weight weight of the drill drill collar collar.. Note Note that that the denomin denominato atorr corresponds to the buoyed linear weight of the drill collar. The length calculated with this expression determines the neutral point of buckling . Above this point the drillstring will not buckle. Bellow this point, which depends on the weight on bit, the drillstring might be buckled or not, and this will depend, among other factors, on the weight on bit and on the polar moment of inertia of the drill collars.

Example 15: Calculate the position position of the neutral neutral point for for a column 10,000 10,000 ft long hanging off bottom and submerged in a borehole filled with a 9.3 ppg (poun (pound d per gallon gallon)) fluid. fluid. What What is the length length of drill drill collar collar with 147 lb/ft lb/ft is required to apply 100,000 lbf on the bit, assuming that, for safety, only 85% of the total length of DC is compressed. compressed. Calculate Calculate the suspended suspended weight weight of the drillstring drillstring using nominal nominal 19.5 lb/ft DP (actual 22.28 lb/ft). Use the assumption assumption that pressure dos not contribute to buckle the drillstring. Solution:

Off bottom: The neutral point position is given by

xn =

F b

−  1

ρf ρs

. wDC

For the off bottom condition the weight on bit F b = 0 and the neutral point is at xn = 0 ft. For a weight on bit of 100,000 lbf we have

xn =

100000 = 793 f 793 f t 9.3 1 65. 147 65.5

 − ×

Since this length should be 85% of the total, the minimum length of DC’s is

LDC =

793 = 933 f 933 f t 85%

The hook load (buoyed weight) of the drillstring is:

W h = [933

22.28] × 147 + (10000 − 933) × 22.

  × − 1

9.3 65. 65.5

= 291008 lbf 291008 lbf

A more general expression for the neutral point position includes the possibility that the fluids inside and outside the pipe have different densities:

xn =


wDC

−

F b (ρo Ao

− ρi A i) . Page 5–5


5.2 5. 2


Design Desi gn for for Tensi ensile le Force Force,, Torqu orque, e, Burs Burst, t, and and Collapse

Once the length of the drill collars and the total weight of the drillstring had been determined, we must check if the drillstring will be able to resist the loadings it will be submitted submitted during during normal and special operation operations. s. The drill pipe section may be composed by one or several types of drill pipes (diameter, linear weight, weight, and steel grade). grade). We must check for maximum tensile tensile strength, maximum torque, torque, maximum burst, burst, and collapse pressure. pressure. In addition, addition, since these loadings very likely occur simultaneously (for example, tension and torque), the conjoined strength must be determined.

5.2.1 5.2.1

Maximu Maximum m Tensile ensile Force Force

During normal operation, the maximum tensile force occurs at the top of the drill pipe section during pick–ups. Since this point has (normally) the smallest section section area, it is also the point point of maximum maximum stress. In addition addition to static load (the buoyed weight of the drillstring), inertial effects (the force to accelerate the drillstring), friction effects between the drillstring and the borehole wall, and viscous effects must be considered. Also, in the event of stuck pipe 1 , the drillstring must be able to support the overpull applied during pipe freeing operations. Due to several uncertainties involved in the calculation of the various loadings, ings, relative relative large safety safety factors factors must be used. It is practice practice to use 125% of the static load as the design parameter (25% of overpull.) In addition, it is important to note that the API defines yield stress as the stress that will cause a certain permanent permanent (plastic) (plastic) deformation. deformation. Based on this, the minimum minimum yield strength of a pipe is defined. The minimum yield strength is the minimum axial tensile load that will cause the yield of the material. Normally we want to avoid any plastic deformatio deformation n of the drill pipes. pipes. Therefo Therefore, re, only a fraction fraction (normally 90%) of the minimum minimum yield strength strength is allowed allowed during drilling operations operations.. This figure is called tensile strength of the pipe. margin of overpu overpull ll (MOP The margin (MOP)) is defin defined ed as the the exces xcess s of the the tens tensilile e load load cacapacit pacity y of the the drill drillst strin ring g to the the norma normall tens tensilile e load load for norma normall oper operat atio ions ns.. Know Knowin ing g the MOP is important in case of stuck pipe. In practice, practice, the determined determined MOP must not be exceede exceeded d since the drillpipe would fail. Typical ypical values values of MOP requirements for drillpipe selections are in the range from 50,000 to 100,000 lbf.

Example 16: Calculate the minimum yield strength required required for the drill pipe of the previous example. 1

A stuck pipe is the situation in which the force required to move, or the torque required to rotate the drillstring is larger that its strength.


Page 5–6



Solution: The drillstring buoyed weight is 291000 lbf, and 125% of this value (25% (25% of overpu overpull) ll) is 1.25 1.25 x 29100 291000 0 lbf = 363750 363750 lbf. lbf. The minim minimum um tensile tensile strength required is 363750 F y = = 404167 lbf 0.9

If a new 5 in, 19.5 lb/ft (nominal) is used, the minimum yield stress is:

Y min min =

F y 404167 = π 2 = 76625 psi 2) As (5 4 . 276 4

−

501087 lbf ). Therefore, a X–95 grade pipe is required ( F y = 501087 ). The minimum yield strength in the example is obtained from drill pipe tables, or calculated from the pipe parameters:

F y =

π 4



OD 2

− I D2



Y s

where Y s is the yield strength of the material. For the example above we have:

F y =

π 4

×



52

− 4.2762 × 95000 = 501087 psi



This figure is for new pipe dimensions. For premium pipes (every new pipe is re–classified to premium in the first time it goes to operation), it should be considered that the wall thickness is reduced to 80% of the original wall thickness ness.. Norma Normalllly y, the wors worstt scen scenari ario o is used used and, and, in this this case case,, we assu assume me that that the the thickness thickness reduction reduction occurred in the outside diameter diameter (external wear). wear). ThereTherefore, the new dimensions are:

ID = 4. 4.276 in OD = ID + 80% (OD

− ID) = 4.4.276 + 80% × (5 − 4.276) = 4.4.8552 in

Therefore, for premium DP, the minimum yield strength is:

π F y = 4.85522 4



2

− 4.276

×

95000 = 394611 lbf . .

Note that if a premium DP is considered in the previous example, the X-95, a premium pipe can not be used in that operation. Situations like this suggest the use of tapered drill pipe section. In tapered drill pipe sections, the segment with the lowest capacity is placed right above the drill collar section (or HWDP when used), and the maximum length for that pipe is calculated. Then a higher capacity drill pipe is used and the maximum length for this drill pipe is calculated. The process is repeated until the expected total length length (the maximum depth depth for that drillstring) is reached. For For a multi multi diameter, multi weight, or multi grade drillstring, the topmost joint of each section must be checked for tension.


Page 5–7



Example 17: Calculate the drillstring drillstring for the previous previous example example data using premium drill pipes of 5 in–19.5 lb/ft and grades E-75, X-95, and G-105. Solution:

This problem is simplified by the fact that the total weight of the drillstring is the same, no matter the grade and length of the drillpipe. Using 25% of the buoyed weight, the MOP is

M OP = OP = 25%

72750 lbf . × 291000 = 72750 lbf

It is important to realize that any overpull applied to the top of the drillstring will manifest in every element of the drillstring (assuming that the stuck point is in the bit). The minimum yield strength and maximum tensile load for 5 in–19.5 lbf/ft (actual 22.28 lbf/ft) drill pipes for the various grades, and using a maximum of 90% as operational limits are: π (4. (4.85522 4.2762 ) 75000 90% = 280382 lbf 280382 lbf E–75: F y = 4 π (4. (4.85522 4.2762 ) 95000 90% = 355150 lbf 355150 lbf X–95: F y = 4 π (4. (4.85522 4.2762 ) 105000 90% = 392535 lbf 392535 lbf G–105: F y = 4 Starting with E–75 (on top of the drill collars), the maximum length for this grade is

− − −

(933

× × ×

22.28) × 147 + LE–75 × 22.

× ×

×

  × − 1

9.3 65. 65.5

= 280382

− 72750

LE–75 = 4705 ft The maximum length is 933 + 4705 = 5638 ft, and the partial weight is

933

22.28 = 241978 lbf . . × 147 + 4705 × 22.

Continuing with the next grade (X–95):

(241978 + L + LX–95

22.28) × 22.

  × − 1

9.3 65. 65.5

= 355150

− 72750

LX–95 = 3912 ft The maximum length is 5638 + 3912 = 9550 ft, and the partial weight is

241978 + 3912

22.28 = 329137 lbf 329137 lbf . × 22.

Continuing with the next grade (G–105):

(329137 + L + LG–105


22.28) × 22.

  × − 1

9.3 65. 65.5

= 392535

− 72750 Page 5–8



LG–105 = 1956 ft The maximum length is 9550 + 1956 = 11506 ft, enough to reach the depth of 10000 ft. Note that if different DP diameters or different linear weights are used, the final weight of the drillstring is not known before it has been designed, and the MOP cannot be imposed directly (and an iterative process will be needed). In this case, we can set the MOP absolutely, based, for example in the worst scenario (heaviest DP) or using field experience.

5.2.2 5.2.2

Maximu Maximum m Torque orque

The torque applied to the drillstring is the reaction due to the bit action on the bottom of the borehole, and due to friction forces between the drillstring eleme elements nts and the boreho borehole le wall. wall. In vertical vertical wells wells,, most most of the torque torque comes from the bottom hole assembly (bit, stabilizers, etc.). For directional wells, the torque is distributed along the borehole trajectory below the kick-off point (the point point of the traject trajectory ory where where the boreho borehole le leave leaves s the vertical vertical). ). In any case, case, the torque accumulates and the point of higher torque is always at the top of the drillstring. drillstring. The torsional torsional yield strength strength of a pipe is given given by the follow following ing expression: J T y = 0.577 Y s , r where T y is the torsional yield strength, J is is the polar moment of inertia of the circular section, and r is the outside diameter diameter of the section. section. The coefficient coefficient 1 √ comes 0.577 comes from the von Mises–Henc Mises–Hencky ky distortion distortion energy energy theory of fail3 ure of ductile materials, which determines the shear yield strength based on the tensile yield strength. The polar moment of inertia for circular pipes is given by: π J = OD 4 I D4 . 32 For T y in ft lbf , the expression for the torsional yield strength is given by:

≈



·



−

Y s OD 4 I D 4 T y = . 105. 105.92 OD

−

Example 18: Calculate Calculate the torsional torsional yield strength strength for a new and for a premium 5 in–19.5 lbf/ft, grade E–75 drill pipe. Solution:

75000 54 4.2764 = 41166 ft lbf a) new: T y = 105. 105.92 5 75000 4.85524 4.2764 = 32285 ft lbf b) premium: T y = 105. 105.92 5

−

·

−


·

Page 5–9



The figures in the example above can be found in Tables A.2 and A.4 (API RP7G). Note that the calculation of the torsional yield strength for a premium DP uses the worst worst scenario for the wall thickness thickness reduction. reduction. Although Although these values represent the maximum torque allowed in the body of the drill pipe, the maximum should not exceed the actual make–up torque of the connection.

5.2.3

Internal Internal (Burst) (Burst) and External External (Collapse) (Collapse) Pressures Pressures

The API criteria for internal pressure strength (burst resistance) assumes that drill pipes are thin–walled pipes. pipes. Since API accepts accepts pipes with a minimum minimum of 87.5% of the nominal wall thickness, the formula for the internal pressure for new pipes (and also for casing) is:

pint = 0.875





2 (OD I D) Y s . OD

−

If thick–wall and von Misses–Hencky theories are used, the formula is:

pint =

OD 2 I D4 Y s . 3 OD O D4 I D 4

√

−

−

Note that in this formula, the allowed wall reduction to 87.5% has not been considered. Data for internal pressure strength for new and premium drill pipes are shown in Tables A.3 and A.5 (API RP7G) respectively. In operations like drill stem testing, the drill pipe may be subjected to external pressure pressure higher than the internal pressure. pressure. The most critical point is the lower end of the drill pipe section. The net collapse pressure is determined by the depth of the fluid inside the pipe, the depth of the lower end of the drillstring section, and the density of the fluids in the annular and inside the drill pipe. The expression for the net external pressure is:

pext = 0.0519[ρ 0519[ρ0 D

− ρi(D − d)] ,

where pext is the external net pressure, D is the depth of the lower end of the drill pipe section, d is the depth of the fluid surface inside the pipe (make d = D if d > D), ρ o and ρ i are the densities of the fluid in the annular and inside the drill pipe. A safety factor of 1.125 is normally used for collapse pressure. The minimum collapse pressure is also in Tables A.3 and A.5. The collapse strength is calculated using API Bulletin 5C3. The discussion of the formulation is beyond the scope of this work, and will be covered in the Casing Design chapter in Advanced Drilling. Collapse pressure is detrimentally affected by tensile force in the drill pipe (the beneficial beneficial effect in burst resistance is neglected). neglected). the combined combined effect of CHAPTER 5 Drillstring Design

Page 5–10



tensile load to collapse resistance is obtained using the biaxial ( σr = 0) expression of the von Misses failure theory:

σz2

− σz σt + σt2 ≤ Y s2

The equality holds when for a given axial stress σz , an effective tangential tangential yield  stress Y s occurs. Then we have:

− Y s σt + Y 2s = Y s2 .

σz2

Solving for the appropriate Y s (tangential stress is compressive for collapse) results in: 2 Y s σz σz = 1 3 . Y s 2 Y s 2 Y s

   − −

API define 4 types of collapse: yield, plastic, transition, and elastic. All collapse types but the elastic depends on the yield strength (elastic collapse depends on the modulus of elasticity E ). ). API indicates that the collapse pressure s of drill pipes are reduced due to tensile load by the fraction Y above. Y s 

Example Example 19: Determi Determine ne the the collap collapse se resist resistanc ance e correc corrected ted for for tensio tension n loading for a premium 5 in–19.5 lbf/ft, grade E–75 drill pipe subjected to a tensile load of 50,000 lbf. Solution:

The outside and inside diameters of the DP are 4.8552 in and 4.276 in respectively (premium DP). Therefore the cross section area is:

A = and the axial stress is:

π 4.85522 4



σz =

− 4.2762



= 4.1538 in2 ,

50000 = 12037 psi . 4.1538

The ratio of the equivalent yield stress to the nominal yield stress is:

Y s = Y s

  − 1

12037 3 2 75000

×

2

−

12037 = 0.9100 . 2 75000

×

The minimum collapse resistance for the premium pipe is 7041 psi (see Table A.5), and the corrected collapse pressure is

7041


× 0.9100 = 6407 psi .

Page 5–11


5.2.4 5.2.4


Drillst Dri llstrin ring g Elonga Elongatio tion n

Due to the weight of the drillstring and the elastic characteristic of the steel (and many other metals and alloys), an elongation occurs when the drillstring is suspended inside the borehole. In addition, if an additional force acts at the bottom end (be it tensile or compressive) additional elongation (or shortening) occurs. The expression for the total elongation of a drillstring of length L is:

ρs g L2 F L ∆L = + , 2E A E where ρ s is the density density of the drillstring metal, E is the longitudinal modulus of rigidity (or modulus of elasticity or Young modulus), F is the force acting on the bottom, and A is the area area of the the right right sectio section n of the drills drillstri tring ng.. Note Note that this this expression is valid only for a homogeneous drillstring suspende pended d in the air. air. If the drillstrin drillstring g is submer submerged ged in a fluid fluid with density ρf , a hydrostatic pressure will act at the lower end (upward) such that the total elongation becomes: ρs g L2 F L ∆L = + 2E A E

−

∆L =

(ρs

ρf g L2 , E

− 2ρf ) g L2 + F L .

2 E A E For ρ in lbm/gal, E in psi, L and ∆L in ft, and A in in2 we have: (ρs 2ρf ) L2 F L ∆L = 0.0519 + . 2 E A E

−

This formulation formulation is not considering considering the effect of the Poisson’ son’s s rati ratio o due due to the the (vary (varyin ing) g) pres pressu sure re acti acting ng alon along g the the drill drill-string.

Example 20: Calculate Calculate the elongation elongation of 10,000 10,000 ft of drill pipe hanging hanging offbottom bottom inside a borehole borehole filled with 10 ppg drilling fluid. fluid. The modulus modulus of elas6 ticity of the steel is 30 10 psi. Solution:

×

∆L = 0.0519

(65. (65.5

10.0) × 100002 − 2 × 10. = 3.94 ft 2 × 30 × 106

For a tapered drillstring we calculate the elongation of each section, considering force acting at the bottom of each section (including pressure effects), and add all contributions.


Page 5–12



Example 21: Calculate Calculate the elongatio elongation n of a drillstring drillstring made up off 10,000ft 10,000ft of 5 in–19.5 in–19.5 lb/ft drill pipe and and 600 ft of 8 in OD, OD, 3 in ID drill collar (147lb/ft) (147lb/ft),, inside a borehole filled with 10 ppg drilling fluid and 20,000 lbf of WOB (weight on bit). Solution:

a) elongation of the drill collars:

ADC = F =

π 2 8 4

32 = 43. 43.197 in2

−

43.197 in = −257644 lbf −20000 lbf − 0.0519 × 10 ppg × 10600 ft × 43. ∆LDC = 0.0519

65. 65.5 6002 257644 600 + = 2 30 106 43. 43.197 30 106

× × ×

−

× × ×

−0.099 ft

b) elongation of the drill pipes:

ADC = F =

π 2 5 4



− 4.3762



= 5.2746 in2

(43.197 − 5.2746) in2 −257644 lbf + 147 lbf /ft × 600 ft + 0,0, 0519 × 10 ppg × 10000 ft × (43. F = F = 27373 lbf 65. 65.5 100002 27373 10000 ∆LDP = 0.0519 + = 7.40 ft 2 30 106 5.2746 30 106 ∆L = ∆DC + ∆DP = 0.099 + 7. 7.40 = 7. 7.30 ft

×

× ×

−

× × ×

Note that in this case (and in most cases), the elongation of the drill collars can be neglected, compared with elongation of the drill pipes.


Page 5–13




Page 5–14



Chapter 6 Drilling Drilling Hydraulics Hydraulics During most of the drilling operations, the drilling fluid is circulating through the circulating system and/or the drillstring is moving with respect to the fluid. These These dynami dynamic c compon component ents s cause cause pressu pressure re gradi gradient ents s and pressu pressure re losses losses that that must be determined, predicted, and controlled in order to perform the drilling operation operation safely safely and optimally optimally.. The most important parameters parameters to be determined are: • The pressure along the borehole borehole while circulating circulating (equival (equivalent ent circulating circulating density – ECD) • The pressure pressure along the borehole borehole while moving the drillstring (surge (surge and swab pressures) • The optimum optimum circulating circulating parameters parameters and bit nozzle sizes sizes • The cuttings cuttings transport transport capacity capacity of the fluid (hole cleaning) cleaning) • The pressure pressure along along the borehole borehole during well control operation operations s (kick removal) In order to proceed with these studies, we need to use basic physical principles like mass conservation and energy conservation. Also, the flow behavior of the fluid must be understood and determined (rheological properties of the fluid, laminar and turbulent flow, etc.)

6.1 6. 1

Mass Ma ss and and Ener Energy gy Ba Bala lanc nce e

Mass and energy (two different forms of the same thing), can be neither created nor destroyed and, for non–relativistic scenarios, one cannot be converted into the other. other. This results in the mass and energy conservation laws. CHAPTER 6 Drilling Hydraulics

Page 6–1



6.1.1 6.1.1 Ma Mass ss Cons Conserv ervati ation on “the net flow of mass into any volume is equal to the rate of increase of mass within the volume”

This statement is mathematically expressed by (see Figure 6.1):

Figure 6.1: Mass balance

dm =m ˙ = ρ = ρ i q i ρo q o . dt where m is the mass within the volume, ρ is density, and q is flow flow rate. rate. It is evident that a volume under study may have several influxes and several out fluxes so that a general expression can be written:

−

dm =m ˙ = dt



ρi q i

−



ρo q o .

Considering only steady-state conditions (the mass inside the volume under control does not change) this expression reduces to:

0=



ρi q i

 −

ρo q o .

In addit addition ion,, for for incomp incompres ressib sible le fluids fluids (most (most drillin drilling g fluids fluids are consid considere ered d incompressible, except those with gaseous components), this expression reduces duces further to: 0= q i q o q i = q o .

 −

→

The expression above is important when considering flows other than generated by the fluid pumps, like kick influx, fluid loss (to underground formations), mud treatment (material addition and fluid discard,) et cetera. For a closed, single way system (as the circulating system is considered most of the times) we have: q i = q o = q , that is, the flow rate at any point of the system is constant equal to q (the flow rate) rate).. There Therefo fore re,, it suffice suffices s to know know the flow area to calcul calculate ate fluid averag average e velocity at any point of the circulating system. CHAPTER 6 Drilling Hydraulics

Page 6–2



Example 22: A 11 lb/gal lb/gal fluid is being circulated at 280 gal/min. The diameter of the hole is 81 /2 in. The drill drill collars collars have have 8 in OD and 3 in ID, ID, and and the drill pipes have 5 in OD and 4.276 in ID. Calculate the average fluid velocities inside the DP’s and DC’s, and in the annular space opposite to the DP’s and DC’s. Solution:

A 11 lb/gal fluid can be consider incompressible. Therefore we can write for the flow rate at any point of the circuit:

→ ¯v = Aq

q = A = A ¯v

where q is the fluid flow rate, A is the area in the point of interest, and v¯ is the average velocity at the point. or v in ft/s, A in in2 , and q in in gal/min we have:

q [gal/ [gal/min] v¯[ft/ [ft/s] = A[in2 ]

 ×

231 in2 1 gal

v¯[ft/ [ft/s] = 0. 0.3208

× ×  1 min 60s

1 ft 12 in

[gal/min] q [gal/ A[in2 ]

The average velocities are:

vDP = 0.3208 vDC = 0.3208

6.1.2 6.1.2

×

280 gal/ gal/min = 12. 12.71ft/ 71ft/s π 2 3 4

vannDP = 0.3208 vannDC = 0.3208

280 gal/ gal/min = 6.26ft/ 26ft/s π 2 4 . 276 4

×

π 4

280 gal/ gal/min = 2.42ft/ 42ft/s (8. (8.52 52 )

×

π 4

−

280 gal/ gal/min = 13. 13.86ft/ 86ft/s (8. (8.52 82 )

×

−

Energy Energy Conserv Conservatio ation n

“the net flow of energy crossing the boundaries of a system is equal to the rate of increase of the internal energy of the system”

Energy crosses the boundary of a system as work and/or heat, and internal energy may be of several forms: • potential, potential, CHAPTER 6 Drilling Hydraulics

Page 6–3



• kineti kinetic, c, • chemical, chemical, • thermal thermal,, • etc. The best way to apply the energy conservation in drilling is using the Bernoulli’s equation:1 “at any point of an ideal steady state incompressible flow system the sum of the specific kinetic energy, the pressure, and the specific potential energy and pressure is constant”

or, in mathematical terms:

1 ρ ¯v 2 + p + p + ρ g h = cons = constan tantt , 2 where ρ is the density of the fluid (considered incompressible) and h a height with respect respect to any fixed fixed reference. reference. The value of the constant constant is not relevant, relevant, becau because se we can chose chose any any arbitr arbitrary ary refere reference nce to the height. height. Since Since we will will be more concerned to depth instead of height, the following expression for the Bernoulli equation is more appropriate:

1 ρ ¯v 2 + p 2

= constan tantt , − ρ g D = cons

where D is depth. If we take two points in a flow field (points 1 and 2), we can write: 1 1 ρ ¯v 2 + p ρ g D = ρ ¯v 2 + p ρ g D . 2 2 2 1



 

−



−

Note that this expression can be used to express hydrostatic pressure as presented before. In this case, the velocities are zero and we have:

p2

= p 1 − ρ g D1 , − ρ g D2 = p p2 = p = p 1 + ρ g (D2 − D1 ) .

The meaning of the Bernoulli equation is that the total energy at point 1 is equal equal to the total total energ energy y at point point 2. This This is true for ideal system system in which no energy loss occurs in the path between point 1 and point 2. For real systems, however, a loss of energy always occurs because of the internal viscous friction in the fluid. This energy loss reveals itself as a friction pressure drop. Therefore, if the fluid flows from point 1 to point 2 we have, for real flow:

 1

1 ρ ¯v 2 + p 2

−ρ g D

  =

2

1 ρ v¯2 + p 2

−ρg D

−

∆ pf .

1

Named after the Swiss physicist and mathematician Daniel Bernoulli.

CHAPTER 6 Drilling Hydraulics

Page 6–4



Figure 6.2: Schematic of a circulation system. In addition, if between points 1 and 2 we have a pump that increases the pressure by a value we end up with the general energy conservation equation:



1 ρ ¯v 2 + p 2

−ρ g D

  =

2

1 ρ v¯2 + p 2

−ρg D



+ ∆ p ∆ p p

1

− ∆ pf .

Most frequently we will be interested in calculating the pressure downstream. In these cases we can write:

p2 = p 1 + ρ g (D2 In field units we have:

p2 = p 1 + 0.0519 ρ 0519 ρ (D2

− D1) − 12 ρ



v¯22

− v¯12

− D1) − 8.073 × 10−4 ρ



v¯22



+ ∆ p ∆ p p

− v¯12



− ∆ pf

+ ∆ p ∆ p p

(6.1)

− ∆ pf , (6.2)

for pressure in psi, density in lb/gal, depth in ft, and velocity in ft/s.

Example 23: Determine Determine the pressure pressure at the lower lower end of the drillstring drillstring if the frictional pressure loss in the drillstring is 1650 psi, the flow rate is 340 gal/min, the mud density is 11 lb/gal, and the well depth is 11500 ft. The internal diameter of the drill collars at the lower end of the drillstring drillstring is 213 /16 in, and the pump increases the pressure by 3000 psi. Solution:

Consider Consider the simplified diagram diagram beside. beside. We can assume that the velocity velocity of the fluid at the tank (point 1) is very low so that v¯1 = 0. In addition we have CHAPTER 6 Drilling Hydraulics

Page 6–5



p1 = 05 psig, D 1 = 0 ft, D 2 = 11500 ft. The velocity at point 2 is calculated with the mass conservation formula: 340 gal/ gal/min vDC = 0.3208 π = 17. 17.56ft/ 56ft/s 2.81252 4

×

Using these values in the expression for the downstream pressure we obtain: p = 0+0. 0+0.0519 11 (11500 0) 8.073 10−4 11 17. 17.562 02 +3000 1650 2

× ×

− −

×

× ×





− − p2 = 0 + 6565 − 2.74 + 3000 − 1650 = 7912. 7912.26 psi ≈ 7912 psi

Note the small contribution of the kinetic energy to the pressure at point 2 (2.74 psi.) In most of the situations, we can neglect the kinetic energy term. A situation, however, where we can’t neglect the kinetic energy term is in the flow through bit nozzles.

6.2 6. 2

Flow Flow Thr Through ough Bit Nozzl Nozzles es

The drilling fluid exits the drillstring through nozzles at the bit (normally 3 or 4 nozzles). The nozzles accelerate the fluid forming a high-velocity jet below the nozzles, which are hurled against the recently cut formation. The fluid jets help to clean the bit cones and to remove the cuttings from beneath the bit (to avoid re–grinding them) and so improving the drilling efficiency.

Figure Figure 6.3: Longitudi Longitudinal nal cut of bit nozzles. nozzles. (Courtesy (Courtesy SPE) Bit nozzles are made of hard–erosion resistant materials such as cast tungsten carbide, sintered carbide or a ceramic material.

6.2.1 6.2.1

Pressur Pressure e Dro Drop p Across Across the Bit

As the fluid exits the drillstring through the nozzles, its velocity accelerates from v¯1 inside the bit to ¯v2 at the jet at the expense of a pressure drop across the bit. CHAPTER 6 Drilling Hydraulics

Page 6–6



Figure 6.4: Pressure drop across the bit. The pressure drop is determined determined using the mass conservation conservation and the energy conservatio conservation n laws. The length length of the nozzles nozzles is very short (about 11/2 in), so that so that the the potential potential energy energy terms can be neglected. neglected. In addition, addition, the friction loss is neglected (temporarily), and since there is no other source of pressure, the Equation (6.1) becomes:

p2 = p 1

− 12 ρ



v¯22

− v¯12



.

The velocity of the fluid inside the bit is much less than the velocity in the jet. Therefore we can write for the pressure drop across the bit:

∆ pb = p = p 1

− p2 = 21 ρ ¯v j2 ,

(6.3)

where ¯v j is the average velocity of the fluid in the jet. The pressure drop across the bit is an important parameter in drilling engineering. gineering. If we solve solve Equation (6.3) for the average average nozzle nozzle jet velocity velocity,, we obtain: 2 ∆ pb v¯ j = . ρ



This ideal exit velocity, however, is never obtained because of the frictionless assumption assumption made in the beginning beginning of the derivation. derivation. The friction pressure loss depends on parameters of the nozzle like the shape, material, and surface roughness. Based on experimental measurements comparing ideal to real nozzle jet velocities, the following expression was proposed:

v¯ j = C d




2 ∆ pb , ρ

(6.4)

Page 6–7



where the coefficient C d is called called discharge discharge coefficient. coefficient. Typical values values for C d are in the range 0.95 to 0.98. Using field units, we can write:

v¯ j = C d



∆ pb = 35. 35.2 C d 8.073 10−4 ρ

×



∆ pb . ρ

A drilling bit has several nozzles distributed in the bit body (normally equal to the number of cones.) cones.) If they have have different different areas, areas, the total flow q splits splits to the nozzles depending on their areas. The pressure drop is the same for each nozzle (equal to the pressure inside the bit less the pressure outside the bit), therefor therefore e the nozzle velocity velocity is must be the same for each nozzle. nozzle. Since the flow q is be equal to the sum of the flows in the nozzles we have (assuming 3 nozzles):

q = q = q 1 + q 2 + q 3 = v¯ j A1 + v¯ j A2 + v¯ j A3 = v¯ j (A1 + A2 + A3) = v¯ j At , q , At where At is the total total area of the bit nozzles nozzles.. Theref Therefore ore,, the pressu pressure re drop across the bit can be expressed by: v¯ j =

v¯ j =

q = C d At

∆ pb =



2 ∆ pb , ρ

ρ q 2 . 2 C d2 At2

(6.5)

In field units we have: 2 ρ q 2 − 5 ρ q ∆ pb = 8.310 × 10 = , C d2 At2 12034 C 12034 C d2 At2

(6.6)

∆ pb in psi, ρ in lbm/gal, q in for ∆ p in gpm, and A t in in2 .

6.2.2 6.2.2

Hydraul Hydraulic ic Pow Power er Acr Across oss the Bit

The power across the bit is obtained by multiplying the pressure drop across the bit by the flow rate: ρ q 3 P b = q ∆ p ∆ pb = . (6.7) 2 C d2 At2

6.2.3 6.2.3 Impa Impact ct Forc Force e of of the the Jets Jets The Jet of fluid exert a force at the bottom of the hole called jet impact force . It is due to the change in the jet momentum as it hits the bottom. An infinitesimal volume dV of fluid in the jet with velocity v¯ j has a linear momentum given by (see Figure 6.5): CHAPTER 6 Drilling Hydraulics

Page 6–8



Figure 6.5: Jet impact force.

dM = ρdV v¯ j . As this volume hits the surface, the momentum is totally transferred to the surface (the vertical velocity becomes zero), and the force due to this moment transfer is given by: dM dV F j = = ρ v¯ j = ρ q v¯ j . dt dt Using Equation (6.4) results in:

F j = ρ q C d



2 ∆ pb = C d ρ

F j = C d In field units we have:

F j =



2 ρ q 2 ∆ pb ,

 

2 ρ q 2 ∆ pb .

(6.8)

ρ q 2 ∆ pb .

(6.9)

C d 54. 54.85

Note that the impact force as derived is only true for nozzles close enough to the formation so that the viscous friction between the fluid in the jet and the fluid in the vicinity of the bit can be neglected.

1 /32nd of inch (hole diamThe sizes of the nozzles are usually measured in 1/ eter), and are reported in “thirty-seconds” of inch. Nozzles range from 6/32 to 32/32. The area of a #14 nozzle for example is: A(14)


π = 4

 14 32

2

= 0.1503 in2 .

Page 6–9



Example 24: A 12 ppg drilling fluid fluid is flowing through through a bit containing containing three three #13 #13 nozzle nozzles. s. The pump pump pressu pressure re increase increase is 3000 3000 psi at a flow of 400 gpm. gpm. Calculate (a) the pressure drop across the bit, (b) the hydraulic power delivered by the pump, (c) the power spent at the bit, and (d) the hydraulic impact force. (Use 0.95 for the discharge coefficient.) Solution:

The total area of the nozzles is:

At = 3 A 13 = 3

π 4

×

2

 13 32

= 0.3889 in2

The pressure drop across the bit is

ρ q 2 ∆ pb = = 1169 psi 12034 C 12034 C d2 At2 The hydraulic power delivered by the pump is [Equation (2.6)]

P H H =

3000 400 = 700 hp 1714. 1714.29

×

The power spent at the bit is [Equation (6.7)]

P b =

1169 400 = 272 hp 1714. 1714.29

×

The hydraulic impact force is [Equation (6.8)]

F j =

√

0.95 12 54. 54.85

× 4002 × 1169 = 821 lbf

6.3 Requir Required ed Hydrau Hydraulic lic Power ower Consider the expression for the pressure downstream from the fluid tank to the bottom of the borehole along the drillstring (see Figure 6.2). If the pressure in the tank is p 1 , the pressure at the bottom p 2 , after the bit is:

p2 = p 1 + ρ g (D2

− D1) − 12 ρ



v¯22

− v¯12



+ ∆ p ∆ p p

− (∆ pf )ds − ∆ pb ,

where ∆ pf ds is the frictional frictional pressure pressure drop along the drillstring. Consider Consider now the expression for the pressure downstream from the bottom of the borehole to the fluid tank along borehole annular.

p1 = p 2 + ρ g (D1 CHAPTER 6 Drilling Hydraulics

− D2) − 12 ρ



v¯12

− v¯22 − ∆ pfann ,



Page 6–10



where p fann is the frictional pressure drop along the annular. Adding these two equations we obtain: ∆ p p = ∆ pb + ∆ p ∆ pf , (6.10)

∆ pf = (∆ pf )ds + (∆ ( ∆ pf )ann is the pressure drop along the whole (closed) where ∆ p flow flow circui circuit. t. This This express xpression ion says says that that the pressure pressure increase increase in the pump is equal to the friction pressure drop (considered the whole circuit) plus the pressure drop across across the bit. Therefor Therefore, e, since the pump adds energy energy to the fluid ∆ pf , and part system, we clearly see that this energy is spent part in friction, ∆ p to accelerate the fluid in the nozzles, represented by the pressure drop across ∆ pb . the bit, ∆ p Multiplying the expression above by the flow rate q , and recalling that power P is equal to q ∆ p ∆ p we obtain the following relation:

P H H = P b + P f f where P H H is the hydraulic power delivered by the pump to the fluid, P b is the power spent at the bit to accelerate the fluid through the nozzles, and P f f is the power spent to overcome viscous friction.

6.4 Bit Hydrau Hydraulic lics s Optimiz Optimizati ation on The drilling of a well is made in several phases . A phase phase is determ determin ined ed by the bit diameter and starts when the first bit of that size is lowered into the borehole and the first interval of formation is cut, and ends after the casing or liner is landed and cemented cemented.. The common common practice is to drill a phase using a drilling fluid dense enough to keep the formation fluids from entering into the borehole, borehole, and light enough enough to preven preventt fracturing fracturing the formation formations. s. The fluid density, if correctly determined, is kept constant along the whole phase. When a new bit is lowered into the borehole it is expected that to drill a long interva intervall before before it become becomes s dull. dull. The nozzles nozzles of the bit will will play play an important important role in the performance of the bit, and, therefore, must be carefully selected. A wrong selection can be costly. The basic parameter for the bit nozzle selection is the flow rate. A minimum flow rate is required to keep the borehole clean from the cuttings generated by the bit. An unsatisfactory cuttings transport may put the operation at risk. It will be shown later that the frictional pressure drop (also called parasitic pressure loss) in the circulation system can be modeled quite accurately by an expression of the following form:

∆ pf = c q m ,

(6.11)

where c and m are constants that determine the behavior of the pressure drop of the whole circulation system. CHAPTER 6 Drilling Hydraulics

Page 6–11



The values of both c and m change as the borehole deepens, but not much. The appropriate way to determine the values of c and m (to be used for the next bit) is to run a simple test just before a dull bit is changed. The test consists in circulating the fluid at two different flow rates and measuring the pump pressures. Normally one of the flow rates is close to the minimum required to lift the cuttings and the other close to the expected flow rate to be used for the next bit. Using the nozzles sizes of the dull bit and the values of the pump pressures and flow rate, the friction pressure drop is calculated for the two flow rates. Then these two values are used to determine the values of c and m for the next bit run.

Example 25: In a drilling operation operation,, a flow test was was performed performed before before start a drillstring trip to change the bit. The following result was obtained:

q p p [gpm] gpm] [psi] psi] 475 3000 250 1000 The dull bit has one #13 nozzle and two #12 nozzles. nozzles. The fluid density density is 10 ppg. Determine Determine the constants constants c and m for for the next bit run. The discharge discharge coefficient is C d = 0.95. Solution:

The total nozzle area is:

π At = 4

 13 32

2

+2

×

π 4

 12 32

2

= 0.3505 in2

= 475 gpm we have (Equation [6.6]): For q =

ρ q 2 10 4752 ∆ pb = = = 1691 psi 12034 C 12034 C d2 At2 12034 0.952 0.35052

×

∆ pf = 3000 = 250 gpm we have: For q =

×

×

− 1691 = 1309 psi

10 2502 ∆ pb = = 468. 468.4 psi 12034 0.952 0.35052

×

×

∆ pf = 1000

×

468.4 = 531. 531.6 psi − 468.

Using these values in Equation (6.11) we have:

× 475m 531. 531.6 = c × 250m 1309 = c = c


Page 6–12



Dividing the first expression by the second expression we obtain:

1309 = 531. 531.6

  475 250

m

→

m = 1.404

Using this result in the first expression we find:

1309 = c = c

1.404

× 475

→

c = 0.2286

  psi gpm1.404

Therefore, for the next bit run, the frictional pressure drop should quite accurately be expressed by:

∆ pf = 0.2286 q 2286 q 1.404

6.4.1 6.4.1

Nozzle Nozzle Size Selectio Selection n Cri Criteri teria a

The two most common criteria used to select the nozzles are • Maximum Maximum hydraulic hydraulic power power at the bit, • Maximum Maximum jet impact impact force. force. 6.4.1.1 6.4.1.1

Maximum Maximum Hydra Hydraulic ulic Power Power at the the Bit Bit

The hydraulic power spent at the bit is given by:

P b = P H H

− P f f = ∆ p p q − ∆ pf q .

∆ pf (Equation [6.11])we have: Using the expressions for ∆ p P b = ∆ p p q

− c q m+1 .

Note that the power developed at the bit is function of the flow rate q . To obtain the flow rate of maximum power at the bit, we make

dP b = ∆ p p dq

1) c q m = ∆ p p − (m + 1) ∆ p ∆ pf = 0 . − (m + 1) c

∆ pf gives: Solving for ∆ p

∆ p p m = c q opt . m + 1 Therefore, the flow rate for maximum power at the bit is: ∆ pf =

q opt opt




∆ p p = c(m + 1)



1 m

.

(6.12)

Page 6–13



The pressure drop at the bit for maximum power is:

∆ pb = ∆ p p

− ∆ pf = ∆ p p − m∆ + p p1 ,

∆ pb =

6.4.1.2 6.4.1.2

m ∆ p p . m + 1

(6.13)

Maximum Maximum Jet Jet Impact Impact Force Force at at the Bit

The jet impact force at the bit is given by (Equation [6.8]):

F j = C d ∆ pb = ∆ p p Substituting ∆ p F j = C d



2 ρ



2 ρ q 2 ∆ pb .

− c q m we obtain: q 2

(∆ p (∆ p p

−c

q m)



2 ρ (∆ p p q 2

= C d

− c q m+2) .

Calculating the derivative of F j j with respect to q and and setting to zero, we find:

dF j 2 ∆ p p q (m + 2) c 2) c q m+1 = C d ρ = 0, dq 2 ρ (∆ p p q 2 c q m+2 ) which results in

2 ∆ p p q

−



−

2) c q m+1 = 2 ∆ p p q − (m + 2) ∆ pf q = = 0 . − (m + 2) c

∆ pf gives: Solving for ∆ p

2 ∆ p p = c q o ptm . m + 2 Therefore, the flow rate for maximum jet impact force is: ∆ pf =

q opt opt



2 ∆ p p = c(m + 2)



1 m

.

(6.14)

The pressure drop at the bit for maximum jet impact force is:

∆ pb = ∆ p p ∆ pb =

6.4.1.3 6.4.1.3

2 ∆ p p , −m + 2

m ∆ p p . m + 2

(6.15)

Total Bit Nozzle Nozzle Area

∆ pb (calculated using one of the For a given optimum pressure drop at the bit ∆ p criteria above), the total nozzle are is calculated using Equation (6.5): ρ q 2 ∆ pb = , 2 C d2 At2 CHAPTER 6 Drilling Hydraulics

Page 6–14


(At )opt


q opt opt = C d



ρ . 2 (∆ pb )opt

In field units we have:

(At )opt =

q opt opt 109. 109.7 C d



ρ . (∆ p (∆ pb )opt

∆ p p = 3000 psi, determine the Example 26: For For the data of of Example Example 25 and and ∆ p optimum flow rate, the pressure drop at the bit, and the total nozzle area for (a) maximum hydraulic power at the bit, and (b) maximum jet impact force. Solution:

The frictional pressure drop model for Example 25 is

∆ pf = 0.2286 q 2286 q 1.404 . Then we have (a) maximum hydraulic power at the bit



∆ p p q opt opt = c(m + 1) ∆ pb =

(At )otm

1

 

3000 = 0.2286 (1. (1.404 + 1)

m

×



1 1.404

= 459 gpm

m 1.404 ∆ p p = m + 1 1.404 + 1

× 3000 = 1752 psi

q opt opt = 109. 109.7 C d



ρ 459 = (∆ p (∆ pb )opt 109. 109.7 0.95

×



10 = 0.333 in2 1752

(b) maximum jet impact force



2∆ p p q opt opt = c(m + 2) ∆ pb =

1

  m

2 3000 = 0.2286 (1. (1.404 + 2)

× ×

1 1.404

= 587 gpm

m 1.404 ∆ p p = m + 2 1.404 + 2

× 3000 = 1237 psi

587 (At )otm = 109. 109.7 0.95

×






10 = 0.506 in2 1237

Page 6–15



Figure 6.6: Line of maximum hydraulic power.

6.4.2 6.4.2

Graphi Graphical cal Analys Analysis is

The hydraulic condition of a borehole is continuously changing as the borehole deepens. deepens. Therefo Therefore, re, it would would be necessary necessary to change continuously continuously the nozzle nozzle total area to attain optimum condition all the time. Since this is not economically possible, the best we can do is to keep the hydraulic parameters close to the optimum point.

(P H It is always always desirable to operate the pumps in their maximum power (P H )max . The relationship between hydraulic power, pressure and flow rate is given by P H H =

∆ p q . 1714. 1714.29

(2.6)

Taking the logarithm of this expression and solving for log∆ p p we obtain:

log∆ p p = log (1714 (1714..29 P 29 P H H )

− log q .

log ∆ p p versus logq reFor the maximum pump pressure (P H H )max , plotting log sults in a a straight line with slope -1, as shown in Figure 6.6 (other values for P H H plots as lines parallel to the maximum power line). The pump can theoretically operate in any condition below the maximum hydraul hydraulic ic power line. In addition, addition, three other constraints constraints limit the operation operational al area:

(∆ p p )max , 1. the maximum allowable allowable surface pressure (∆ p 2. the minimum minimum flow rate to carry the cuttings cuttings q min min , CHAPTER 6 Drilling Hydraulics

Page 6–16



Figure 6.7: Additional hydraulic constraints. 3. The maximum maximum flow rate for a given pump liner liner q max max . These three constraints are shown in the Figure 6.7. The shaded area represents represents the possible possible operation operational al condition conditions. s. The bold line corresponds to the maximum operational pressure for those constraints. However, in order to be able to operate on both the slanted bold line and on the horizontal bold line (maximum power or maximum pressure) it would be necessary to change the pump liner frequently as the borehole deepens. The common practice, however, is to select a liner to operate the pumps at the maximum allowed surface pressure at the maximum hydraulic power, and use this this liner liner,, if possib possible, le, in the entire entire well or, or, at least, least, for for the whole whole phase. phase. The maximum flow rate for the maximum surface pressure is given by

q max max =

1714. 1714.29 (P H H )max , ∆ pmax

and shown in the graph in Figure 6.8. Under these limitations, the pump should operate such that the surface operationa erationall paramete parameters rs remain on the bold line of the graph in Figure 6.8. That is, the pump will be operating at the maximum flow rate during the shallow portion of the well (the pressure is dictated by the resistance to flow), and at the maximum pressure during the lower portion of the well. The constants c and m , which determine the frictional pressure drop in the fluid circuit, circuit, depend depend on the depth depth (and other factors factors too). For For each depth, an optim optimum um nozzl nozzle e tota totall flow flow area area At exists, xists, for a given given optimi optimizat zation ion criterio criterion, n, which which depends on the constants c and m . Assuming the constant m will not change CHAPTER 6 Drilling Hydraulics

Page 6–17



Figure 6.8: Ideal surface operational parameters. ∆ p

too much as the well deepens, the optimum ratio ∆ pf p (for one of the criteria) remains quite constant. Therefore, as the pump operational conditions change during the drilling, the optimum condition is that for which the values of the frictional pressure drop ∆ pf follo follows ws the bold line line in Figure Figure 6.9. 6.9. This This line is called the path of optimum hydraulics . Consider now the expression for the pressure drop due to friction, ∆ pf = c q , where c and m depend on the depth. Taking the logarithm of this expression results in the following expression: m

log∆ pf = log c + m log q log ∆ pf Therefore, the graph of log as shown in Figure 6.10.

× log q plots plots as a straight line with slope m

The point where a frictional pressure drop line intercepts the path of optimum hydraulics defines the pair q opt These value values s of optim optimum um opt and (∆ pb )opt . These flow rate and optimum pressure drop at the bit are used to calculate the total bit nozzle area for the next bit run.

Example 27: Determine the proper proper pump operating conditions conditions and bit nozzle sizes for maximum hydraulic power for the next bit run. The bit currently in use has three #12 nozzles. The driller has recorded that when the 9.8 lbm/gal lbm/gal fluid is pumped at a rate of 485 gal/min, a pump pressure of 2900 psig was observed and when the pump was slowed to a rate of 260 gal/min, a pump pressure of 980 980 psig psig was was observ observed. ed. The nomina nominall powe powerr of the pump pump is 1250 1250 hp with an efficiency of 95%. The minimum flow rate to lift the cuttings is 230 gal/min. The CHAPTER 6 Drilling Hydraulics

Page 6–18



Figure 6.9: Path of optimum hydraulics. maximum allowable surface pressure is 3000 psig. The fluid density will remain unchange unchanged d in the next bit run. (Use C d = 0.95 Solution:

The hydraulic power delivered by the pump is

( pH )max = 1250

× 95% = 1188 hp

For a maximum surface pressure of 3000 psi, the maximum flow rate is

q max max =

1714. 1714.29 1188 = 679 gpm 3000

×

Now we need to calculate the frictional pressure drop for the two flowing test points. The total nozzle area is

At = 3

×

π 4

 12 32

2

= 0.3313n2

q = 485 gpm The The pres pressu sure re drop drop acro across ss the the bit bit and and the the fricti friction onal al pres pressu sure re drop drop for for q are: ρ q 2 9.8 4852 ∆ pb = = = 1934 psi 12034 C 12034 C d2 At2 12034 0.952 0.33132 ∆ pf = ∆ p p

×

×

×

− ∆ pb = 2900 − 1934 = 966 psi

q = 260 gpm The pressure drop across the bit and the friction pressure drop for q are: 9.8 2602 ∆ pb = = 556 psi 12034 0.952 0.33132 ∆ pf = 980 556 = 424 psi

×


×

−

×

Page 6–19



Figure 6.10: Frictional pressure drop lines. Therefore, we have for c and m :

× 485m 424 = c = c × 260m 966 = c = c

m = 1.321 c = 0.2741

  psi gpm1.321

For the maximum hydraulic power criterion, the optimum pressure drop due to friction and the optimum pressure drop across the bit are:

(∆ pf )opt = (∆ pb )opt =

∆ p p 3000 = = 1292 psi m + 1 1.321 + 1

m 1.321 ∆ p p = m + 1 1.321 + 1

× 3000 = 1708 psi

which defines the optimum hydraulics for the maximum hydraulic power at the bit. bit. A plot plot of the path path of optim optimum um hydraul hydraulics ics,, the line of frictio frictional nal pressur pressure e drop line (the straight line passing at the two test points), and the optimum operational point are shown in the graph in Figure 6.11. The frictional pressure drop line intercepts the path of optimum hydraulics at 1292 psi. The optimum flow rate is calculated from the frictional pressure drop model for the condition of the operation:

1292 = 0. 0.2741 q 2741 q 1.321 CHAPTER 6 Drilling Hydraulics

→

q = = 603 gpm Page 6–20



Figure 6.11: Graph for Example 27.

∆ pb = 1708‘psi at the flow rate of Therefore, the pressure drop across the bit is ∆ p 603 gpm. The total nozzle bit area is calculated calculated from: (At )otm

603 = 109. 109.7 0.95

×



9.8 = 0.4383 in2 1708

Sets of nozzles to use are

(13, (13, 14, 14, 14) (12, (12, 14, 14, 15) (11, (11, 14, 14, 15)


→ → →

At = 0.403 in2 At = 0.434 in2 At = 0.438 in2

Page 6–21




Page 6–22



Chapter 7 Introduction to Drilling Fluids The objective of a drilling operation is to drill, evaluate, and complete a well that will produce oil and/or gas efficiently. Drilling fluids perform numerous functions that help make this possible. possible. The responsibility responsibility for performing performing these functions is held jointly by the mud engineer and those who direct the drilling operation. The duty of those charged with drilling the hole – including the oil company representative, drilling contractor, and rig crew – is to make sure that the correct drillin drilling g proced procedure ures s are conduct conducted. ed. The chief chief duty duty of the mud mud engine engineer er is to assure that mud properties are correct for the specific drilling environment. The mud engineer should also recommend drilling practice changes that will help reach the drilling objectives.

7.1 Functi Functions ons of Dri Drillin lling g Fluids Fluids Drilling fluid (also called drilling mud) is a mixture of water, oil, clay, and various physical physical chemical chemical additives. additives. It performs various various functions functions in drilling drilling and contribute tributes s with with a large large portion portion to the total total well cost. In this way the drilling drilling fluid fluid system (or mud program) has to be carefully designed to ensure a successful drilling project. The drilling fluid serves many purposes, which may not all be achieved simultaneously for all parts of the well. In this way, an individual prioritization has to be followed. Below is a summary of some main drilling fluid functions: 1. Transport ransport the drilling cuttings cuttings from the bottom of the hole to the surface. surface. For this, a higher fluid circulation velocity and a higher fluid viscosity are favorable. 2. Create Create an overbalance overbalanced d drilling drilling condition condition to control control the formation formation pressure. Fluid density (or mud weight) is increased with additives like barite, iron oxide, etc, to maintain a hydrostatic pressure inside the well which is slightly slightly higher higher than the formation formation pressure. pressure. Normally Normally,, an overbalance overbalance of 100 to 200 psi has proven to be adequate to establish safe drilling. CHAPTER 7 Introduction to Drilling Fluids

Page 7–1



3. Reduce Reduce chemical and physical physical interactions interactions with formation. formation. This avoid avoid or reduce the occurrence of formation swelling, formation cave–in, and problems associated associated with cementing. cementing. The invasion invasion of fluid, filtrate, filtrate, or small solids into the formation can cause damage to the formation and result is less productivity 4. Produce an impermeable and thin seal (mud cake) along the walls of permeable formations to reduce fluid and filtrate loss to the formations, reducing reducing the damage potential potential of the fluid. This function function is obtained obtained by adding bentonite to the fluid, and by appropriate chemical treatment to enhances deflocculation and solids distribution. 5. Cool and lubricate lubricate the drilling bit and drillstring drillstring to minimize minimize wear. wear. Bentonite, oil, various emulsifying agents, graphite, and others favor the cooling and/or lubrication capacity of the fluid. 6. Allow Allow efficient solids control at the surface surface (separation (separation of cuttings, cuttings, formaformation gas, etc, from the fluid) by using appropriate solids control equipment. (To be recirculated, a maximum of 2% of sand contends are allowed to avoid early wear of mud pumps and drilling equipment in general.) 7. Hold drilling cuttings in suspension suspension when circulation is interrupted. Failure Failure of this function would allow cuttings to fall down the hole and settle around the drillstring possibly possibly causing causing pipe sticking sticking problems. problems. The thixotropic thixotropic 1 behavior of drilling fluids determines its capacity to keep solids in suspension (including weighting material). 8. Produce Produce buoyan buoyancy cy force to partly support the weight weight of the drillstring and casing string. An increase in mud weight, which increases buoyancy, results in smaller surface hoisting equipment requirements. 9. The drilling fluid should not interfere interfere with logging logging measurements. measurements. 10. Transmit ransmit he hydrauli hydraulic c power power to the bit and allow maximum penetration penetration rates. 11. Minimize Minimize the torque and drag of the drillstring drillstring to reduce reduce wear and failure of drillstring. 12. As a mean to transmit data between between bottom hole equipment equipment and surface surface equipment.

7.2 7. 2

Types ypes of Dr Drill illin ing g Flui Fluid d

The composition of a particular drilling fluid system depends on the actual requirement quirements s of the individual individual well or well section. Wells Wells are drilled drilled through different formations, which require different mud properties to achieve optimum 1

The ability of a fluid, such as cement or drilling mud, to develop gel strength over time when not subject to shearing, and then to liquefy when agitated.

CHAPTER 7 Introduction to Drilling Fluids

Page 7–2



penetrati penetrations ons and stable stable borehole borehole conditions. conditions. Economics, Economics, additives additives (cost and availability), temperature profile, contamination, and environment are some important factors that determine the design of the drilling fluid. The basic classification of drilling fluids is (sub–classifications exist): 1. Water–b Water–base ase fluids fluids (a) (b) (c) (d) (e) (f)

Clear Clear water and native native muds muds Inhibitive water–based water–based muds muds – calcium muds Dispersed muds muds – lignosulfonate lignosulfonate muds muds Non–dispersed muds muds – KCL/Polymer KCL/Polymer muds Flocculate Flocculated d muds Salt–saturated muds

2. Oil–base Oil–base fluids and Emulsio Emulsion n fluids (a) Oil–in–water Oil–in–water emulsion emulsion (b) Water–in–oil Water–in–oil emulsions 3. Synthetic Synthetic Fluids Fluids 4. Aerated Aerated fluids fluids (a) Gaseous Gaseous fluids (air, (air, nitrogen, nitrogen, CO 2 , natural gas) (b) Foams Foams (c) Mists

7.2.1 7.2.1

Water–Ba ater–Base se Fluids Fluids

Water–base fluids are any drilling fluid in which the continuous phase, where some materials are in suspension and others are dissolved, is water. Thus any water–base fluid system consists of a water phase, inert solids, reactive solids phase, phase, and chemical additives additives.. Each of these parts contribute contribute to the overal overalll fluid properties. The individual contributions are: water: create initial viscosity, inert solids: (low-gravity (low-gravity solids like sand and chert, and high-gravity high-gravity solids like barite and lead sulfides) produce required mud weight, reactive solids: ( low-gravity solids like bentonite and attapulgite clays) cause additional viscosity and gel, chemical additives: (thinners [phosphate, chrome, lignosulfonate lignosulfonate,, lignites, surfactants] and thickeners [lime, cement,polymers]) provide control to viscosity cosity,, yield yield point, point, gel gel streng strength, th, fluid fluid loss, loss, pH, filtrat filtration ion behav behavior ior,, etc. etc. Caustic soda (NaOH) is used to keep a high pH required to control corrosion, hydrogen embrittlement and the solubility of Ca 2 + and Mg2 +. CHAPTER 7 Introduction to Drilling Fluids

Page 7–3


7.2.1.1 7.2.1.1


Clear Water Water and Native Native Muds

To drill compact formations which are normally pressured 2, fresh water and salt–satur salt–saturated ated water water can be used as drilling drilling mud. Native Native muds are the result of mixing water water and and clays clays or shales shales from from the cutting cuttings s drilled drilled.. Here Here the clays clays or shales are dissolved dissolved by the water and returned to the surface. surface. Clear Clear water water and native muds are the cheapest mud systems since no additional material is needed to form the mud. They are also environmentally best accepted. 7.2.1.2 7.2.1.2

Inhibitive Inhibitive Water– Water–Base Base Muds – Calcium Calcium Muds Muds

When swelling and hydration of clays and shales are expected, inhibitive water– base base muds can be used. used. Calciu Calcium m muds are best suited suited to drill formati formations ons that contain gypsum and hydrite. hydrite. A subclassifica subclassification tion of inhibitiv inhibitive e water–bas water–base e muds distinguishes seawater muds, saturated saltwater muds, lime muds, and gypsum muds. 7.2.1.3 7.2.1.3

Dispersed Dispersed Muds – Lignosu Lignosulfo lfonate nate Muds

Dispersed muds are used when the following characteristics are required: • relative relative high mud weight weight (larger (larger than 14 ppg), • tolerance to moderately moderately high formation formation temperatures, • low filtratio filtration n loss • high tolerance for for contamination by by drilling solids. Some of the disadvantages of dispersed muds are sloughing of shales and formation damage due to dispersing of formation clays in the presence of lignosulfonate. Dispersed mud systems consist of: • fresh or salty salty water, water, • bentonite bentonite,, • lignosulfonate, lignosulfonate, • caustic caustic soda soda • colloidal polymers (carboxymethylcellulose (carboxymethylcellulose or stabilized stabilized starch). In general, these mud systems exhibit better control of viscosity, higher solids tolerance, and better control of filtration than non–dispersed muds. 2

Formation with pore pressure gradient equal to brine density


Page 7–4


7.2.1.4 7.2.1.4


Non–Disp Non–Disperse ersed d Muds Muds – KCL/P KCL/Poly olymer mer Muds Muds

To drill sloughing shales and water sensitive rocks such as productive sands, which are prone to formation damage, fresh water non–dispersed muds are used. used. Commonly Commonly,, non–dispe non–dispersed rsed muds are associated associated with low mud weights weights and low solid concentrations. Non–dispersed mud systems consist of: • fresh water water or brine, brine, • potassium potassium chloride chloride (KCl), (KCl), • inhibiting inhibiting polymer polymer,, • viscosifier viscosifier,, • stabilized starch or carboxymethyl carboxymethyl cellulose, • caustic caustic soda, soda, • lubricants. lubricants. Low–solids polymer mud systems are widespread in the industry since they offer advantages like increased penetration rate, hole stability, shear thinning ability, hole cleaning with maximum hydraulics, and lower equivalent circulation density over conventional deflocculated muds. Besides these advantages, they also have disadvantages like instability at temperatures above 250 ◦ F, irreversible absorption of the polymer on clay, higher dilution, the requirement of adequate solids removal equipment, and the fact that they are more corrosive. 7.2.1.5 7.2.1.5

Flocculate Flocculated d Muds

Flocculated muds generally causes a dynamic increase in filtration, viscosity, and gel strength. strength. Flocculati Flocculation on refers to a thicken thickening ing of the mud due to edge– edge– to–edge and edge–to–face association of clay particles. The flocculation is commonly caused by high active solids concentration, high electrolyte concentration, and high temperature. To reduce the flocculating fl occulating tendency of the mud, chemical additives, also called deflocculants or thinners, are used. Thinners like phosphates, tannin, lignin, and lignosulfonate are used to lower the yield point and gel strength. When deflocculants are added, the pH is controlled by NaOH. 7.2.1.6 7.2.1.6

Salt–Satu Salt–Saturated rated Muds

Sever Several al mud systems have been included included in this classificatio classification. n. Saturate Saturated d salt systems have a chloride concentration near 190,000 mg/l (saturated) and are CHAPTER 7 Introduction to Drilling Fluids

Page 7–5



used used to drill drill salt salt format formation ions. s. Saltw Saltwate aterr systems systems have have a chlorid chloride e conten contentt of 10,000 10,000 to 190,000 190,000 mg/l. The lower lower levels levels are usually usually referred referred to as brackish brackish or seawater systems. Salt–saturated muds are used to drill through salt domes and salt sections. These mud systems are saturated with sodium chloride (NaCl), which prevents severe severe hole enlargements enlargements due to washouts washouts of the salt formations formations.. Swelling Swelling of bentonitic shales is controlled by adding of polymer. Various specialty products, such as attapulgite, CMC, starch, and others, are used to increase viscosity for hole–cleaning properties and to reduce fluid loss.

7.2.2 7.2.2 Oil–B Oil–Base ase Muds Muds In oil–base mud systems, crude or diesel oil forms the continuous phase in the water–in–oil emulsion. In this way oil–base mud can have as little as 3% to 5% or as much as 20% to 40% (invert emulsions) of water content. Oil–base mud systems are used when: 1. Drilling sensitive production zones zones or troublesome troublesome shales, 2. Directiona Directionall drilling and slim hole drilling, drilling, 3. Drilling Drilling of depleted depleted reservoirs, reservoirs, 4. Drilling salt sections and formations that contain contain hydrogen hydrogen sulfide, 5. High risk of pipe sticking sticking problems, problems, 6. Drilling formations with higher temperatures. temperatures. Low–gravity solids content has to be monitored closely when drilling with oil– base base muds. muds. The reason reason is becaus because e there there is no hydrat hydration ion of solids solids (clays), (clays), which which freq freque uent ntly ly caus causes es the the cont conten ents ts of low– low–gr grav avit ity y soli solids ds to excee xceed d acce accept ptab able le levels levels.. This results in reduction reduction of penetrati penetration on rate, increase risk of formation formation damage, and increase risk of differential sticking. Since oil–base muds contain substantially less colloidal particles, they exhibit an increased spurt fluid loss 3 . Due to the higher filtration rates, the monitoring of high–pressure/high–temperature filtration and the drilling conditions are important to ensure that excessive filtration or filter cake buildup does not lead to drilling problems. 3

The instantaneous volume (spurt) of liquid that passes through a filter medium prior to deposition of a competent and controlling filter cake.


Page 7–6



7.2.3 7.2.3 Synt Synthe heti tic c Flui Fluids ds Synthetic fluids are designed to mirror oil–based mud performance without the environmental hazards. Primary synthetic fluids are esters, ethers, poly alpha– olefin and isomerized isomerized alpha–olefin alpha–olefin.. They They are environment environmentally ally friendly, friendly, can be discharged offshore, and are non–sheening and biodegradable.

7.2.4 7.2.4 Ae Aerat rated ed Flui Fluids ds Four basic operations are included in this specialized category. These include: 1. Dry air drilling, drilling, which involves involves injecting injecting dry air or gas into the wellbore wellbore at rates capable of achieving annular velocities that will remove cuttings; 2. Mist drilling, which involve involves s injecting injecting a foaming foaming agent agent into the air stream that mixes with produced water and coats the cuttings to prevent mud rings, allowing drill solids to be removed; 3. Foam uses surfactants and possibly clays or polymers to form a high carrying-capacity foam; 4. Aerated fluids rely on mud with injected air (which reduces hydrostatic head) to remove drilled solids from the wellbore. These fluids are discussed in details in advanced drilling courses (air drilling, underblanced drilling, etc).

7.3 7. 3

Labo Labora rato tory ry Tes ests ts

The API has recommended standard methods of conducting field and laboratory tests for drilling fluids and detailed procedures may be found in the API publicati publication, on, “Recommen “Recommended ded Practice: Practice: Standard Standard Procedure Procedure for Field Testin Testing g Water–Based (Oil–Based) Drilling Fluids,” API RP13B–1, 13B–2 and supplements (also see 13I for Laboratory Testing Drilling Fluids, 13J for Testing Heavy Brines and supplements).

7.3.1 7.3.1

Water–Ba ater–Base se Mud Tests

7.3.1.1 7.3.1.1

Density Density of Fluid (Mud Weight: Weight:))

The density or weight of the mud may be determined by the use of any instrument of sufficient accuracy to permit measurement to within 0.1 lb/gal (0.5 lb/ft 3 or 5 psi/1000 ft of depth). CHAPTER 7 Introduction to Drilling Fluids

Page 7–7


Figure 7.1: A mud balance.


Figure 7.2: .2: A Marsh funnel.

For all practical purposes, density means weight per unit volume and making a density density test means means weigh weighing ing the mud. The weight weight of mud mud may may be expressed as a hydrostatic pressure gradient in psi/1000 ft of vertical depth, as a density in lb/gal, lb/ft 3, or specific gravity.

7.3.1. 7.3 .1.2 2

Marsh Marsh Viscos Viscosity ity::

The Marsh funnel is used for routine field measurement of the viscosity of drillin drilling g mud. mud. It is a simple simple device device for indicati indicating ng viscosity viscosity on a routin routine e basis. basis. When used with a measuring cup, the funnel gives an empirical value for the consistency of a fluid.

7.3.1.3 7.3.1.3

Dynamic Dynamic Viscosity Viscosity,, Gel:

The rotary viscometer is used to supplement the information obtained from the Marsh funnel, particularly with respect to the gel characteristics of the mud. It is capable of giving the apparent viscosity, plastic viscosity, yield point, and gel strengths strengths (initial and timed). Drilling Drilling fluid contained contained in the annular annular space between two concentric cylinders is sheared by the rotation of the outer cylinder (rotor (rotor sleeve) sleeve) at a constant RPM (rotational (rotational velocity). velocity). The torque torque at the rotor sleeves sleeves is transmitted transmitted by the fluid to the inner cylinder cylinder (bob). A torsion spring restrains the movement of the bob, and a dial indicates displacement of the bob. Instrument constants are adjusted so that plastic viscosity and yield point are obtained by using readings from rotor sleeve speeds of 600 and 300 RPM. A six-speed model (600, 300, 200, 100, 6, and 3 rpm) can be used to fully characterize a fluid.

7.3.1. 7.3 .1.4 4

API Fluid Fluid Loss: Loss:

Determines the filtration properties of drilling muds and cement slurries. Generally consists of a mud reservoir mounted in a frame, a pressure source, a filtering medium, and a graduated cylinder for receiving and measuring filtrate. CHAPTER 7 Introduction to Drilling Fluids

Page 7–8



Figure 7.3: A rotational viscometer (rheometer). Working pressure is 100 psi and the filtering area is 7.1 in 2, as specified by the American Petroleum Institute (API RP13B–1 and RP13B–2).

7.3.1. 7.3 .1.5 5

HTHP HTHP Filtra Filtratio tion: n:

Similar to API filter press (with half filtration area), but can be used in pressures up to to 1200 psi and temperatures up to 500 ◦ F.

Figu Figure re 7.4: 7.4: A API API filte filterr pres press. s.


Figu Figure re 7.5: 7.5: A HTHP HTHP filte filterr pres press. s.

Page 7–9


Figure 7.6: Sand content sieve. 7.3.1.6 7.3.1.6


Figure 7.7: Retort. rt.

Filter–Cak Filter–Cake e Compress Compressibilit ibility: y:

This is not yet an API test. The purpose is to determine some mechanical properties of the mud cake like penetration resistance, adherence, compressibility, etc.

7.3.1. 7.3 .1.7 7

Solid Solid Conten Contents: ts:

A retort is used to determine solid contents including dissolved ions.

7.3.1. 7.3 .1.8 8

Sand Sand Conten Content: t:

Determines sand content by sieve analysis. The volume of sand, including void spaces between grains, is usually measured and expressed as a percentage by volume of the drilling fluid.

7.3.1.9 7.3.1.9

Methylen Methylene e Blue Capacity: Capacity:

Determines the capacity of a clay to absorb cations from a solution, and thereby pred predic ictt how how the the cla clay will will reac reactt in its its inte intend nded ed use use. The The cla clay may may be comp compon onen entt of a drilling fluid, a binder in foundry sand, or a clay used for some other purpose. The Methylene Blue Test is based on the property of clays known as base exchange capacity, that is, clays can exchange some of their ions for the ions of certain certain other other chemic chemicals als.. The number number of ions ions avail availab able le for for this this excha exchange nge varies with different types of clay.

Flocculant Efficiency Test: CHAPTER 7 Introduction to Drilling Fluids

Page 7–10



Figure 7.8: Methyl blue capacity test kit.

Figure 7.9: A pH meter. meter.

7.3. 7.3.1. 1.10 10 pH: pH:

Indicator sticks and/or pH meter are used to determine the hydrogen ion concentration in the mud and in the filtrate. 7.3.1.11 7.3.1.11 Chemical Chemical Analys Analysis is of Water Water–Bas –Base e Fluids Fluids (Titration) (Titration)

• Alkalinity Alkalinity (Pf, (Pf, Mf, Pm and Lime Content) Content) • Garrett Gas Train Train (GGT) Test Test for Carbonates • Chloride Chloride (Cl) • Calcium Calcium Qualitativ Qualitative e • Total Hardness Hardness • Hardness Hardness in Dark Filtrate Filtrates s • Sulfate Sulfate • Potassium otassium (K+) (K+) • Nitrate Nitrate Ion Concentra Concentration tion • PHPA PHPA Polymer Concentration Concentration

7.3.1.12 7.3.1.12 Chemical Chemical Anal Analysis ysis Rela Relating ting to Corros Corrosion ion

• SULF-X SULF-X (ZnO) and Basic Zinc Carbonate Carbonate (ZnCO (ZnCO 3 Zn(OH)2 ) • Iron Sulfide Sulfide Scale (Qualitativ (Qualitative) e) • Hydrogen Hydrogen Sulfide Sulfide (H2 S) • Phosphate Phosphate CHAPTER 7 Introduction to Drilling Fluids

Page 7–11



Figure Figure 7.10: A titration kit.

Figure 7.11: A permeameter kit. • Oxygen Oxygen Scaven Scavenger: ger: SO 32- Content • Resistivity Resistivity • Resistivity of the mud, filtrate and filter cake is sometimes sometimes needed in order to help evaluate the electric logs.

7.3.1.13 7.3.1.13 Permea Permeability bility Plugging Plugging Test

The Permeability Plugging Apparatus (PPA) is designed to provide accurate simulation and measurement of downhole static filtration. The PPA is very useful in predicting how a drilling fluid can form a permeable filter cake to seal off depleted depleted/unde /underr pressure pressure intervals. intervals. The PPA PPA utilizes utilizes a conven conventiona tionall HTHP Heating Jacket to simulate reservoir temperature.

7.3.2 7.3.2

Oil-Ba Oil-Base se Mud Testing esting

7.3.2.1 7.3.2.1

Aniline Aniline Point Point Determina Determination tion

The test indicates if an oil is likely to damage elastomers that come in contact with the oil. It is the lowest temperatur temperature e (°F or °C) °C) at which equal equal volumes volumes of aniline (C6 H5 NH2 ) and and the oil form form a single single phase phase.. A low AP is indica indicativ tive e of higher aromatics, while a high AP is indicative of lower aromatics content. CHAPTER 7 Introduction to Drilling Fluids

Page 7–12


Figure Figure 7.12: 7.12: An anilin aniline e poin pointt kit. kit. 7.3.2.2 7.3.2.2


Figure Figure 7.13: 7.13: Elect Electrica ricall stab stabili ility ty tester tester..

Electrical Electrical Stability: Stability:

Indicates Indicates the emulsion and oil–wettin oil–wetting g qualities of the sample. sample. The test is per ◦ formed by inserting the ES probe into a cup of 120 F (48.9 ◦ C) mud and pushing a test button. button. The ES meter automaticall automatically y applies an increasing increasing voltage voltage (from 0 to 2000 volts) across an electrode electrode gap in the probe. probe. Maximum Maximum voltage voltage that the mud will sustain across the gap before conducting current is displayed as the ES voltage. 7.3.2.3 7.3.2.3

Chemical Chemical Anal Analysis ysis of of Oil-Base Oil-Base Drilling Drilling Muds: Muds:

• Alkalinity Alkalinity (Pom) (Pom) (VSAAPI) (VSAAPI) • Salinity Salinity Chloride Chlorides s • Whole Mud Mud Calcium Calculat Calculation ion • Sulfide Sulfides s

7.4 Fluid Fluid Densit Density y and and Viscos Viscosity ity Cal Calcul culati ations ons Additives are added to the drilling fluid in order to bring the fluid parameters to the required required values. values. Density Density and viscosity are the two most basic paramete parameters rs to control. The drilling fluid technician or engineer should carry some calculations, and laboratory measurements and tests to determine the correct additive and the correc correctt amount amount to be mixed mixed to the fluid fluid system. system. Fluid Fluid volum volumes es are normall normally y measured in barrels. Useful conversion factors are: 1 bbl = 42 gal 1 gal = 231 in 3 1 ft3 = 1728 in 3 1 in3 = 2.543 cm3 CHAPTER 7 Introduction to Drilling Fluids

Page 7–13



1 lbm = 453.59 g Powder owder and dry additive additives s are normally measured in pounds, pounds, and liquid liquid additives are normally measured in gallons or barrels. Pilot tests are are laboratory (small scale) tests that aim to determine the amount of additive required to bring some fluid parameters parameters to determinate determinate values. values. Small Small scale tests are fast fast and cheap to perform. A handy conversion for is that of lbm/bbl to g/cm3:

lbm 1 bbl

 ×

453. 453.59g 1lbm

× × × 1bbl 42gal

1gal 231in3

1in3 2.543 cm3



=

1g 350cm3

Therefore, in a pilot test with 350 ml of fluid, 1 gram of added additive correspo responds nds to the addition addition of 1 lbm of dry additiv additive e to 1 barre barrell of fluid. A simila similarr conversion shows that 25 ml of liquid additive in 350 ml or fluid corresponds to 3 gallons of additive per barrel of fluid.

7.4.1 7.4.1

Density Density Cal Calcul culatio ations ns

It is frequently necessary to compute the density of a mixture from the amount of the substanc substances es in the mixture mixture.. It is also also important important to be able able to calcul calculate ate the amount to be added of a given substance in order to increase or decrease the density density of the mixture. mixture. The density density and specific gravity gravity of some common common substances used in drilling fluids are shown in the following table: Subs Substa tanc nce e water Diesel oil bentonite attapulgite sand barite drilling cuttings cement CaCl2 NaCl

Spec Specifi ific c Grav Gravit ity y 1 0.86 2.6 2.89 2.63 4.2 2.6 3.14 1.96 2.16

Dens De nsity ity (lbm/ (lbm/ga gal) l) 8.33 7.2 21.7 24.1 21.9 35 21.7 26.2 13.3 18.0

The assumption that the mixture is ideal, that is, that the volume of the mixture is equal to the volume of the components (not valid for highly soluble substances like NaCl in water) facilitates the volume–density calculations. The relations are: V mix V i , mix =



where V mix mix is the volume of the mixture and V i is the volume of the component i of the mixture, and

M mix mix = ρ mix V mix mix = CHAPTER 7 Introduction to Drilling Fluids

  M i =

ρi V i , Page 7–14



where M is mass and ρ is density. In general, the final density of a mixture of substances (assuming ideal mixture) is:

ρmix =



ρi V i . V i

Example 28: Calculate the volume volume and density of a fluid composed of 25 lbm of bentonite, 60 lbm of barite, and 1 bbl of fresh water. Solution:

The volume and the mass of the mixture are:

V mix mix = 1bbl

M mix mix = 1bbl

  ×   × × 42gal 1bbl

42gal 1bbl

ρmix =

7.4.2 7.4.2

+

25lbm 60lbm + = 44. 44.87gal 21. 21.7lbm/ 7lbm/gal 35lbm/ 35lbm/gal 8.33

lbm + 25lbm + 60lbm = 434. 434.9lbm gal

434. 434.9lbm = 9.69 lbm/ lbm/gal 44. 44.87gal

Density Density Treatmen reatmentt

The density control of a drilling fluid is obtained usually with the use of barium sulfate sulfate (BaSO4) (BaSO4) commonly commonly called barite. The specific gravity gravity of pure barite is 4.5, and the average specific gravity of API barite is 4.2 or 35 lbm/gal. To keep the barite in suspension, a minimum gel strength of 3 lbf/100ft2 is required and normally obtained with water–bentonite fluid. The bentonite itself adds viscosity and density to the water. The most common weight treatment is that of increasing the density of the fluid by adding barite. Four basic different procedures exist and must be understood. They are: 1. Increase Increase fluid density density by adding barite barite (no volume limit). limit). In this case it is not necessary to discard part of the original fluid. 2. Increa Increase se fluid density density by adding adding barite barite (volum (volume e limit) limit).. In this case it is necessary to discard part of the original fluid before adding barite to the system. We assume that the final volume is equal to the original volume. 3. Increase Increase fluid density by adding adding barite and water (no volume volume limit). In this case, hydration water is added to avoid increasing the viscosity. CHAPTER 7 Introduction to Drilling Fluids

Page 7–15



4. Increa Increase se fluid fluid densi density ty by adding adding barite and water water (volu (volume me limit). limit). In this case it is necessary to discard part of the original fluid before adding barite and hydration water. 5. Reduce fluid density by adding water water and bentonite (no volume volume limit). 6. Reduce fluid density by adding water water and bentonite (volume (volume limit). 7.4.2.1 7.4.2.1

Increase Increase Density Density – No No Volum Volume e Limit – No Hydrati Hydration on Water Water

Let V 1 and ρ1 be respectively the volume and the density of the fluid before the treatme treatment. nt. The require required d densit density y of the fluid fluid after after treatment treatment is ρ2 . Since ince there is no limit for the final volume V 2 , no discard of fluid is needed before the treatment. Using volume and mass continuity we have:

V 1 + V B = V 2 , ρ1 V 1 + ρB V B = ρ 2 V 2 , where V B and ρ B are respectively the volume and density of the barite. Solving for V B in terms of V 1 we get:

V B =

ρ2 ρB

− ρ1 V . − ρ2 1

(7.1)

The barite is added to the fluid in mass quantity (usually in lbm), and since mB = ρB V B we can write: mB = ρ B V B . (7.2) This is the mass of barite needed to add to obtain a fluid with density ρ 2 . The final volume V 2 is given by: V 2 = V = V 1 + V B . (7.3) 7.4.2.2 7.4.2.2

Increase Increase Density Density – Volume Volume Limit Limit – No Hydration Hydration Water Water

In this case, since the volume is limited (and assumed to be the initial volume V1), it is necessary to discard discard a volume volume V d before increasing the density. Using volume and mass continuity we have:

V 1

− V d + V B = V 1 → V d = V B , ρ1 V 1 − ρ1 V d + ρB V B = ρ 2 V 1 .

Note Note that that the the discar discarded ded volum volume e will will be equal equal to the volum volume e of of barite barite added added to increase the density. density. Substituting V d for V B in the second equation and factoring out V d we obtain: ρ2 ρ1 V d = V = V B = V 1 . (7.4) ρB ρ1 With V B , the mass of barite to be added is given by

− −

mB = ρ B V B . CHAPTER 7 Introduction to Drilling Fluids

(7.5) Page 7–16


7.4.2.3 7.4.2.3


Increase Increase Density Density – No No Volum Volume e Limit – Hydration Hydration Water Water

If barite is added to the mud, a part of the free water in the mud will be adsorbed by the particles of barite (particle sizes in the silt and fine sand range - 2 µ m to 100 µ m). The water will come from the fluid and, as consequence, an increase of viscosity will occur. To avoid this increase in viscosity, viscosity, a volume of water is added along with the barite to serve as hydration water, and thus keeping the viscosity viscosity under control. control. The amount amount of water water added depends depends on the amount amount of barite added. added. Normally Normally,, 2 to 3 gallons gallons of water are required required for 100 lbm of barite. Let vh be volume of hydration water per mass of barite, and V w and ρ w the volume and density of the hydration water. Then we have:

V 1 + V B + V w = V 2 , ρ1 V 1 + ρB V B + ρw V w = ρ 2 V 2 . The volume of water and the volume of barite are related by the specific volume vh: V w = m B vh = ρB V B vh . Substituting this value in the system above we get:

V 1 + (1 + ρ + ρB vh ) V B = V 2 , ρ1 V 1 + (1 + ρ + ρw vh ) ρB V B = ρ 2 V 2 . Solving for V B in terms of V1 we get:

V B =

ρ2 (1 + ρw vh) ρB

− ρ1 − (1 + ρB vh) ρ2 V 1 .

(7.6)

With V B , the mass of barite, volume of water, and the final volume are given by:

mB = ρ B V B ,

(7.7)

V w = m B vh ,

(7.8)

V 2 = V 1 + V B + V w . 7.4.2.4 7.4.2.4

(7.9)

Increase Increase Density Density – Volume Volume Limit Limit – Hydration Hydration Water Water

The final volume is V 1 , and a volume V d must be discarded before the treatment. The continuity equations are:

V 1

− V d + V B + V w = V 1 → V d = V B + V w , ρ1 V 1 − ρ1 V d + ρB V B + ρw V w = ρ 2 V 1 .

Using V w = ρ B V B vh and the first equation above gives:

V d = (1 + ρB vh ) V B . CHAPTER 7 Introduction to Drilling Fluids

Page 7–17



Substituting these two results in the second equation and solving for V B results in: ρ2 ρ1 V B = V 1 . (7.10) (1 + ρw vh) ρB (1 + ρB vh ) ρ1

− −

With V B , the volume to discard, the mass of barite, and the volume of hydration water are: V d = (1 + ρB vh ) V B , (7.11)

mB = ρ B V B ,

(7.12)

V w = m B vh .

(7.13)

Example 29: It is required required to increase the the density density of 300 bbl of a drilling drilling mud from from 9.3 ppg to 10.5 10.5 ppg. Calcul Calculate ate the amount amount of barite barite and the suitabl suitable e procedure procedure (discard, (discard, hydrati hydration on water, water, etc) for the four four cases above. above. (Assume (Assume ρB = 35 ppg, v w = 2.4 gal/ gal/100 lbm of barite, and ρ w = 8.34 ppg.) Solution:

Case 1: No volume limit, no hydration water. Using Equations 7.1 to 7.3:

V B =

10. 10.5 9.3 35 10. 10.5

− −

lbm mB = 35 gal

14.69 bbl × 300 bbl = 14.

14.69 bbl × 14.

  × 42 gal 1 bbl

V 2 = 300 + 14. 14.69 = 314. 314.69 bbl

= 21, 21, 600 lbm

≈ 315 bbl

Case 2: Volume limit, no hydration water. Using Equations 7.4 and 7.5:

V d = V B = mB = 35

10. 10.5 9.3 35 9.3

− −

14.01 bbl × 300 = 14.

14.01 × 42 = 20, 20, 591 lbm × 14.

The final volume is equal to the original volume, i.e., 300 bbl. Case 3: No volume limit, hydration water. Using Equations 7.6, to 7.9:

V B =



1 + 8. 8 .34

×

2.4 100

10. 10.5 35

9.3 1 + 35

 × −−

mB = 35

15.87 bbl × 300 = 15. 2.4 10.5 × 100 × 10.



15.87 × (42) = 23, 23, 328 lbm × 15. 2.4 V w = 23, 23, 328 × = 560 gal = 13. 13.33 bbl 100 V 2 = 300 + 15. 15.87 + 13. 13.33 = 329. 329.2 bbl ≈ 329 bll

Note that the volume of barite is larger than in case 1 because its density is reduced by the hydration water. CHAPTER 7 Introduction to Drilling Fluids

Page 7–18



Case 4: Volume limit, hydration water. Using Equations 7.10 to 7.13:

V B =



1 + 8. 8 .34

10. 10.5 35

2.4 × 100

V d = 1 + 35 mB = 35

2.4 100

2.4 100

9.3

14.46 bbl × 300 = 14.

14. 14.46 = 26. 26.6 bbl

14.46 × (42) = 21, 21, 258 lbm × 14.

V w = 21, 21, 258

7.4.3 7.4.3

9.3 1 + 35

 × −−  ×  ×  × ×

2.4 = 510 gal = 12. 12.15 bbl × 100

Initial Initial Viscosi Viscosity ty Treatmen reatmentt

Bentonite is a clay mineral that is composed principally of three–layer clays, such as montmorillonite, and widely used as a mud additive for viscosity and filtration filtration control. Commercia Commerciall bentonite bentonite ores vary widely widely in amount amount and quality of the swelling swelling clay, clay, sodium sodium montmorillon montmorillonite. ite. Ores of lower lower quality quality,, those with more calcium–type montmorillonite, are treated during grinding by adding one or more of the following following:: sodium sodium carbonate carbonate,, long–cha long–chain in synthetic polymers, polymers, carboxymet carboxymethyl hylcellu cellulose lose (CMC), starch, or polyphosphate polyphosphate.. These These help make the final product meet quality specifications. Clays, in general when ground to colloidal size (less than 2 µm in diameter) and, mixed with water, form a colloidal suspension that causes the increase in viscosity and gel strength. strength. Differen Differentt clays clays produce produce different different levels of viscosity ity. The The term term yield is used to specify the quality of a clay according to the number of barrels of 15 cP viscosity mud (measured at 600 RPM in a API standard rotational viscometer) that one ton of the clay would produce (sometimes it is used used 30 cP). Clays Clays can be classifi classified ed as high–, high–, medium– medium– or low– low–yie yield. ld. A bentonite bentonite produces produces at least 85 bbl of mud per ton. High–yield High–yield clays clays produced produced 30 to 50 bbl/to bbl/ton, n, and low–yi low–yield eld clays clays produc produced ed 15 to 30 bbl/t bbl/ton. on. The graph graph in Figure 7.14 shows typical performances performances for different different types of clay clay. We can clearly see that bentonite can produce large viscosities with low densities. With the right selection of clay and other additives (like barite) a wide range of viscosity cosity/de /densi nsity ty can be obtai obtained ned.. It is importan importantt to stress that viscosity viscosity is also also affected by other additives, and pilot tests are always required to obtain the correct properties.

Example 30: It is required to produce produce 350 bbl of a water–base water–base drilling drilling fluid with 15 cP and 9.4 ppg using premium clay and barite. Specify the amount of material needed. Assume the same material properties of the Example 29. Solution: CHAPTER 7 Introduction to Drilling Fluids

Page 7–19



From the graph in Figure 7.14, we see that premium clay produces 15 cP mud at the density of 8.94 ppg. We have 3 continuity equations and 1 relationship to solve: Final mud volume and mass equations:

V w + V c + V h + V B = V m ρw V w + ρc V c + ρw V h + ρB V B = ρ m V m Mass of the initial mud (water and premium clay):

ρi (V w + V c ) = ρ w V w + ρc V c Hydration water requirement:

V h = ρ = ρ B V B vh With numbers (watch out for the units):

V w + V c + V h + V B = 350 8.34

2 1.7 × Vc + 8. 8 .34 × Vh + 35 3 5 × VB = 9.4 × 350 × Vw + 21

Mass of the initial mud (water and premium clay):

8.94

21.7 × V c × (V w + V c) = 8.34 × V w + 21.

Hydration water requirement:

V h = 35

2.4 × V B × 100

Solving the system we find:

V w = 323. 323.21 bbl V c = 15. 15.2 bbl V h = 5.29 bbl V B = 6.3 bbl Therefore, the total water volume, and the mass of clay and barite are:

V t = 323. 323.21 + 5. 5.29 = 328. 328.5 bbl mc = 15. 15.2

21.7 × (42) = 13, 13, 851 lbm × 21. mB = 6.3 × 35 × (42) = 9, 9, 261 lbm


Page 7–20



Figure 7.14: Clay performance for viscosity.


Page 7–21




Page 7–22



Chapter 8 Rheology and Rheometry Rheology is the science that study the flow behavior of fluids. Rheometry is the process or processes to measure the various rheological parameters of a fluid. Both are important to the study of drilling fluids.

8.1 Rheolo Rheologic gical al Cla Classi ssifica ficatio tion n of Fluids Fluids All fluids encountered in drilling and production operations can be characterized as either Newtonian fluids fluids or non–Newtonian fluids. fluids. Newtonian fluids, like water, gases, and thin oils (high API gravity) show a direct proportional relation˙ , assuming pressure and ship between the shear stress τ and the shear rate γ temperature are kept constant. They are mathematically defined by:

τ = µ γ˙ ,

(8.1)

˙ is where τ is shear stress, γ is shear rate. The proportionality coefficient µ is the (dynamic) (dynamic) viscosity viscosity of the fluid. A dimensional dimensional analysis analysis shows that dynamic dynamic − − − 2 1 1 [ F L T ] T ] or [M [ M L T ]. Typical units are Nm2s = viscosity viscosity has the dimension dimension [F s P a s, P (poise in regard to Poiseuille), cP (centipoise), and lbf Conversion sion f t2 . Conver factors are: g 1P=1 , cm s Ns 1 2 = 10 P = 1000 cP , m 1 lbf s 1 cP = . 47880. 47880.259 ft2 ˙ produces A plot of τ versus γ produces a straight line that passes through the origin and has a slop µ. Viscosity Viscosity usually changes changes with pressure and temperature temperature.. (See Figure 8.1). Most fluids encountered at drilling operations like drilling mud, cement slurries, ries, heavy heavy oil, oil, and and gelle gelled d fractu fracturing ring fluids do not not show show this this direct direct relati relationonship ship betwe between en shear shear stress stress and shear shear rate. rate. They They are charact characteriz erized ed as non– non– Newtonia Newtonian n fluids. fluids. To describe describe the behavior behavior of non-New non-Newtonia tonian n fluids, fluids, various various CHAPTER 8 Rheology and Rheometry

Page 8–1



Figure 8.1: Typical graph of Newtonian fluids.

models like the Bingham plastic and the Power Law fluid models (two parameter models), the Herschel–Bulkley and the Robertson–Stiff fluid models, which are time–independent models. There exist time–dependent fluid models, which present change of viscosity and other parameters based in time and shear history history.. Time–dep Time–depende endent nt fluids model model are sub-classified sub-classified as thixotropic (time– (time– thinning) and rheopetic (time-thickening). (time-thickening). It shall be understood that all the models mentioned above are based on different assumptions that are hardly valid for all drilling operations, thus they are valid valid to a certain certain exten extend d only only. Most Most common commonly ly drillin drilling g fluids fluids are treated treated behaving either aa a Bingham plastic or a power–law fluid. These two models can describe relatively well most of the common drilling fluid for all ordinary drilling operations.

8.1.0.1 8.1.0.1

Bingham–P Bingham–Plasti lastic c Fluid Model: Model:

The Bingham plastic fluid model is a linear model (although not proportional), and is expressed mathematically as follows.

 

τ = τ y + µ p γ ˙ τ = τ y + µ p γ ˙ ˙ =0 γ

−

γ˙ > 0 γ˙ < 0 τ τ y

(8.2)

−τ y ≤ ≤ ≤

The constant τ y is called the yield point usually denoted “YP” and µy is called plastic viscosity , usually denoted “PV”. The typical graph of a Bingham plastic fluid is shown in Figure 8.2. CHAPTER 8 Rheology and Rheometry

Page 8–2



Figure 8.2: Typical graph of Bingham-plastic fluids. 8.1.0.2 8.1.0.2

Power Power–La –Law w Fluid Model: Model:

The power law fluid model is non–linear and do not present a yield stress, and can be expressed as follows.



τ = K (γ ˙ )n γ ˙ 0 n τ = K ( γ ˙ ) γ˙ < 0

−

−

≥ ≥

(8.3)

The constant K is called the consistency index and n is called behavior index .The .The typical graph of a power–law fluid is shown in Figure 8.3.

Figure 8.3: Typical graphs of power–law fluids. When the plot is done on a log–log log–log scale it results in a straight straight line. line. Here the slope determines the flow behavior index n and the intercept with the vertical the value of the consistency index ( log K ). ). The flow behavior index, which ranges from 0 to 1.0, declares the degree of non–Newtonian behavior, where n = 1.0 indicates indicates a Newtonia Newtonian n fluid. Mathemati Mathematically cally n can be greater than 1.0, but drilling fluids do not present present this characteristi characteristic. c. Another Another characteristic characteristic (erroneously shown in the graph) is that the power–law presents an infinite slope ˙ = 0. This, and other experimental results lead to the proposal of other fluid at γ models. models. The consistency consistency index K on on the other hand gives the thickness fluid, CHAPTER 8 Rheology and Rheometry

Page 8–3



where larger K represen representt thick thicker er (more (more viscous viscous)) fluids fluids.. It should should be under under-stood, however, that K has not the dimension dimension of viscosity. viscosity. Consistenc Consistency y index index K is is usually expressed in equivalent viscosity units.

8.2 Rheo Rheom metry etry To determine the rheological properties of a particular fluid a rotational viscometer cometer or rheomete rheometerr is commonly commonly used. Rheomete Rheometers rs may have two, two, six, or continuous (variable) speeds. The typical rotational rheometer (see Figure 7.3 has an arrangement of two concentric cylinders in which the outer cylinder rotates and the inner cylinder actuate a torsion spring with a dial, as depicted in Figure 8.4.

Figure 8.4: Arrangement of a rotational viscometer.

The rotation of the rotor (external cylinder) shears the fluid between the rotor and the bob (internal cylinder), which transmits torque from the rotor to the bob. The torque causes a deflection of the dial (proportional to the torque), against the resista resistance nce of the torsion torsion spring. spring. The The wide wide use of rheome rheometer ters s in the field field leaded to the design of special dimensions that made easy the measurement of fluid parameters. CHAPTER 8 Rheology and Rheometry

Page 8–4


8.2.1 8.2.1


Viscosi Viscosity ty of Newton Newtonian ian Fluids Fluids

The dimensions of the rotor and bob, and the coefficient of the torsion spring are determined such that the viscosity of a Newtonian fluid in centipoise (cP) is obtained directly by reading the deflection of the dial in degrees, when the rotor rotates at 300 rpm.

8.2.2

Parameter Parameters s of Bingham–Plas Bingham–Plastic tic Model Fluids Fluids

Two measurements at different rotation speeds are required. The parameters are obtained with the following formulas (please note the units):

µ p =

300 (θN 2 N 2 N 1

−

τ y = θ N 1

− θN 1 N 1 )

[cP]

[lbf /100 ft2]

N 1 µ p − 300

Note that any two rotation speeds can be selected, but 300 rpm and 600 rpm lead to the simplification of the calculation, in addition to cover a range of shear rates rates that typically occurs in drilling drilling operation operations. s. The simplified expression expressions s for N 1 = 300 rpm and N 2 = 600 rpm are:

µ p = θ600

− θ300 τ y = θ 300 − µ p

[cP] (8.4) 2

[lbf /100 ft ]

The The gene genera rall formu formula las s are are usef useful ul to dete determi rmine ne the the para parame mete ters rs in othe otherr rang ranges es of interest (too small or too large).

8.2.3

Parameter Parameters s of of Power–Law ower–Law Model Fluids Fluids

The parameters are obtained with the following general formulas:

n =

log(θ log(θN 2 N 2 /θN 1 N 1 ) log(N log(N 2 /N 1 )

[1]

510 θ 510 θ N (1. (1.703 N 703 N ))n

[eq. [eq.cP]

K =

Again, any two rotation speeds can be selected. For 300 rpm and 600 rpm the expressions reduce to:

n =

log(θ log(θ600 /θ300 ) log 2

K =

CHAPTER 8 Rheology and Rheometry

510 θ 510 θ 300 511n

[1] (8.5)

[eq. [eq.cP]

Page 8–5



Example 31: A fluid rheometry rheometry test results results in dial reading readings s of θ300 = 13 and θ600 = 22. Calculate Calculate the parameters parameters using power–la power–law w and Bingham–p Bingham–plastic lastic models Solution:

(a) Power–law

log log (22/ (22/13) = 0.759 log 2 510 13 K = = 58. 58.3 eq. eq.cP 5110.759

n =

×

(b) Bingham

µ p = 22

− 13 = 9 cP τ y = 13 − 9 = 4 lbf /100 ft2 8.2.4 8.2.4 Gel Gel Stren Strengt gth h Gel strength is the shear stress measured at low shear rate after a mud has set quiescently for a period of time (10 seconds and 10 minutes in the standard API procedure, although measurements after 30 minutes or 16 hours may also be made). To measure the gel strength a rheometer with low rotary speed (3 rpm) is needed. needed. The sample sample is sheared at 600 rpm for a period (1 to 5 minutes) minutes) and set to rest for the determined time (10 seconds or 10 minutes). Then the shear is applied at 3 r pm and the maximum deflection deflection of the dial is read. read. The value value 2 indicates the gel strength at 10s or 10m in lbf/100ft .

CHAPTER 8 Rheology and Rheometry

Page 8–6



Chapter 9 Flow in Pipes and Annuli This chapter discuss the flow of fluids, Newtonian and non–Newtonian, both in pipes and concentric concentric annuli, for the laminar and turbulen turbulentt regimes. regimes. It is assumed sumed also that the flow is steady (not time dependent). dependent). We will be primarily dp interested in calculating the frictional pressure drop gradient dLf as function of the flow rate q or vice-versa. The formulas to be presented are valid for horizontal conduits but since the flow potential gradient is equal to the frictional pressure drop gradient they can also be used for inclined and vertical conduit. To force a fluid to flow inside a horizontal conduit, it is required a pressure differen differential tial between between upstream upstream and downstre downstream am points along along the conduit. Or, Or, conversely, if a fluid flows on a horizontal conduit, a pressure differential occurs between between an upstream upstream and a downstrea downstream m points. points. This pressure pressure differen differential tial is viewed as a friction pressure drop.

9.1 Lamina Laminarr Flow Flow in Pipes Pipes and Annuli Annuli For For this, we need to understan understand d the concept of flow potential. potential. Considerin Considering g the Bernoulli equation for real flow, we have:

p2 = p 1 + ρ g (D2

− D1) − 12 ρ



v¯22

− v¯12



+ ∆ p ∆ p p

− ∆ pf .

(6.1)

If there is no pump between points 1 and 2, and if the conduit has constant cross section, we can write for the friction pressure drop:

∆ pf = p 1

(∆ p − ρ g ∆D ) . − p2 + ρ g (D2 − D1) = − (∆ p

For incompressible fluids (constant density) the expression can be written as:

∆ pf =

−∆ ( p − ρ g D) .

The term ∆ ( p ρ g D) is called flow potential, which is responsible for the flow. In a conduit, it is balanced by the frictional pressure drop as shown in the

−

−

CHAPTER 9 Flow in Pipes and Annuli

Page 9–1



equation. If the flow is horizontal, the gravitational term is zero and the friction pressure drop between two points along the flow path is equal to the pressure differential between the upstream and downstream points. What characterizes a fluid is its ability to continuously deform when submitted to a shear stress τ (some restrictions apply for some non–Newtonian ˙ and there should fluids). fluids). The continuo continuous us deformation deformation is called shear rate γ ˙, exist a functional relationship between τ and γ

τ = f ( ( γ ˙) , which characterizes the fluid behavior and is called the rheological model or constitutive constitutive equation of of the fluid. Several rheological models were presented in Chapter 8. Independently of the rheological model, however, there exist a relationship involving the forces acting on the flowing fluid, associated directly with the geometry of the conduit. This is the equilibrium equation of of the flow (also called field equation equation ) and and we will will be inte intere rest sted ed in the the field field equa equati tion on for pipe pipes s and and annu annulili and open slots.

9.1.1 9.1.1

Equilib Equilibriu rium m Equati Equations ons

The following equilibrium equations can be easily obtained from the balance of forces acting in an infinitesimal element of fluid. The only forces considered are due to the pressure between the fluid and its surroundings and shear stresses between the fluid and conduit. Body forces due to mass are irrelevant for steady state situations. 9.1.1.1 9.1.1.1

Equilibriu Equilibrium m Equation Equation for Pipes Pipes

The equilibrium equation is

1 dpf r, (9.1) 2 dL where τ is the shear stress at any radial distance r from the axis of the conduit. For constant area conduits and incompressible fluids, it is assumed that dpf /dL is constant. Figure 9.1 (left) shows the velocity profile of laminar flow in pipes. The dash–dot line corresponds to the axis of the pipe. τ =

9.1.1.2 9.1.1.2

Equilibriu Equilibrium m Equat Equation ion for Annuli Annuli

The equilibrium equation is

τ =

1 dpf c r + , 2 dL r

(9.2)

where τ is the shear stress at any radial distance r from the axis of the conduit, and c is a constant to be determined determined.. Note that the equilibrium equilibrium equation equation for CHAPTER 9 Flow in Pipes and Annuli

Page 9–2



pipes pipes can be obtained obtained from the equilibrium equilibrium equation equation for annuli, in which, which, since the radius can be zero, the value of c must be zero. Figure 9.1 (right) shows the velocity profile of laminar flow in annuli. The dash–dot line corresponds to the axis of the annulus.

Figure 9.1: Velocity profiles of laminar flow.

9.1.1.3 9.1.1.3

Equilibriu Equilibrium m Equatio Equation n for for (open) (open) Slots

Although an open slot is not a typical cross section for drilling fluid conduits, in some situations, situations, it is advisable advisable to approximate approximate an annulus annulus by an slot. This assumption simplifies the expressions for frictional pressure drop gradient for non-New non-Newtonia tonian n fluids. fluids. It is usually usually accepted accepted that slot approximati approximation on are accu30%. By “open” rate enough for annuli in which Di /Do “open” slots slots we mean that that only the upper and lower surfaces of the slot are subjected to shear stresses. The lateral surfaces are free of shear stress. The equilibrium equation is

≥

τ =

dpf y. dL

(9.3)

where τ is the shear stress at any distance y from from the axis of the slot. slot. FigFigure 9.2 shows shows the velocity profile profile of laminar flow in a slot. The dash–dot dash–dot line corresponds to the plane in the median of the slot.

9.1.2 9.1.2

Contin Continuit uity y Equatio Equations ns

In was seen in Chapter 6 that the average velocity of the fluid in a conduit is given by q v¯ = , A and this expresses the assumption that every “particle” of fluid in a given section tion of the flow has the same same veloc velocity ity.. This This avera average ge velocity velocity is a very very good good representation of the flow velocity in a given section if all we want is to calculate CHAPTER 9 Flow in Pipes and Annuli

Page 9–3



Figure 9.2: Velocity profile of laminar flow in a slot. the kinetic kinetic energy of the fluid at that point of the flow path. path. Since the velocity velocity of any interior “particle” of fluid in a cross section is the same, no shear exists in the fluid. In Figures 9.1 and 9.2, the dotted lines correspond to the average velocity profile. However, the real fluid velocity a section varies with the position of the in the section. section. In special, the fluid velocity velocity adjacent adjacent to the wall of the pipe is zero, zero, otherwise the shear rate at that point would be infinite. If at the contour of the section the velocity is zero, and in order to have a non–zero average velocity (the fluid is flowing), then the flow should present a non–zero continuous velocity profile inside the cross section. It the velocity profile is known, then the flow rate can be calculated using the expression

q =



v dA

A

where v is the velocity profile of the flow and dA is an infinitesimal area of the flow cross section. Due to the axial symmetry of pipes and annuli, we can write dA = 2π r dr. r = 0 to r = r = R R : For pipes we integrate from r = R

q = = 2π



vrdr.

0

v (R) = 0.) If this equation is integrated by parts gives:(Note that v( R

q = 2π



R

v r dr = dr =

0

 −   

R πvr πv r2 0

dv dr

r2

π

R

dr = dr =

0

−π

   r2

dv dr

dr . (9.4)

0

r = R R i to r = r = R R o For annuli we integrate from r = Ro

q = 2π



v r dr dr .

Ri


Page 9–4



v (Ro) = 0.] Integratin Integrating g by parts gives: [v (Ri ) = v( Ro

q =

−π

   dv dr

r2

dr .

(9.5)

Ri

D A = w = w dy and we integrate from For slots of width w and thickness t , DA t/ 2: to t/2

t/2 −t/2

t/2 t/2

q = w = w



vdy.

−t/2 t/2

t/2) = v( v (t/2) t/2) = 0.] Integratin Integrating g by parts gives: [v ( t/2)

−

t/2 t/2

q = =

−w

   y

dv dy

dy .

(9.6)

t/2 −t/2

The shear rate is in general defined for pipes annuli as

γ ˙ =

. − dv dr

γ ˙ =

. − dv dy

For slots is is defined as

The minus sign is an arbitrary choice, and its purpose is to produce positive shear shear rates in pipes and outer flow regions regions of annuli annuli and slots. On the other hand, the inner flow regions for annuli and slots will have negative shear rates. All expressions so far are completely general and independent of the fluid. In the following, constitutive equations of fluids are coupled with equilibrium equations of conduits to produce various flow models. The whole process will be presented for Newtonian fluids. Only the final expressions will be presented for non–Newtonian fluids, but the steps are basically the same.

9.1.3

Newtonian Newtonian Flow in Pipes Pipes – Poiseuille’ oiseuille’s s Equation Equation

A Newtonian fluid is expressed by Equation (8.1)

τ = µ γ ˙ =

. −µ dv dr

Considering the equilibrium equation for flow in pipes [Equation (9.1)] we obtain: dv 1 dpf τ = µ = r, dr 2 dL dv 1 dpf = r. dr 2µ dL

−

−


Page 9–5



To obtain the flow rate (and ultimately the relationship with the pressure drop gradient) we use the expression for dv and Equation (9.4): dr

π dpf q = 2µ dL

R



r 3 dr = dr =

π dpf 4 R , 8µ dL

0

A = π π R2 , and we can write: The area of the pipe is A = q = = Solving for

dpf we dL

A dpf 2 R . 8µ dL

obtain the Poiseuille’s equation:

dpf 8µq = , dL A R2 or, in terms of average velocity

dpf 8µv¯ 32µ 32µv¯ = 2 = . dL R D2

(9.7)

In field units this expression is written as:

dpf µv¯ = , dL 1500 D 1500 D 2 for

dpf in dL

9.1.4 9.1.4

(9.8)

psi, µ in cP, v¯ in ft/s, and D in inches.1

Newtonia Newto nian n Flow Flow in Concent Concentric ric Annuli Annuli – Lamb’ Lamb’s EquaEquation

Considering the equilibrium equation for flow in annuli [Equation (9.2)] we obtain: 1 dpf dv c τ = µ = r + , dr 2 dL r dv 1 dpf C = r + . dr 2µ dL r

−

 −



Separating variables and integrating, and considering that for pipes the velocity is zero for r = Ri , we obtain: v (r)



dv =

0

v (r ) = 1

r

  −



1 dpf C r + 2µ dL r

Ri

− 41 µ

dpf 2 r dL



dr ,

− Ri2 − C ln Rr



. i

A more accurate figure for the conversion factor is 1496.26.


Page 9–6



v (R0 ) = 0, we can determine the value of the constant C : Using the fact that v( v (Ro) = 0 =

− 41 µ

C =

−

dpf 2 Ro dL

− Ri2 − C ln RRo ,





i

1 dpf Ro2 Ri2 4µ dL ln RRoi

−

(9.9)

To obta obtain in the the flow flow rate rate we agai again n use use the the expre xpress ssio ion n for Ro

q = π

 

1 dpf 3 r +C r 2µ dL

Ri

q = π



Ro2

−

Ri2





dv and Equati Equation on (9.5): (9.5): dr

dr ,



 

−



1 dpf C Ro2 + Ri2 + 8µ dL 2

Since π (Ro2 Ri2 ) is the area of the annulus, and using the definition of the average velocity for annuli and the value of the constant C [Equation (9.9)] yields: 1 dpf Ro2 Ri2 2 2 v¯ = Ro + Ri 8µ dL ln RRoi

−



Solving for

dpf ,results dL

v¯ =

−

in the Lamb’s equation:

1 dpf 8µ dL



dpf 8µv¯ = dL Ro2 + Ri2

−

Ro2 + Ri2

−

Ro2 −R2i o ln R R

=

Ro2

−

Ri2

ln RRoi



32µ 32µv¯ Do2 +

Di2

−

i

(9.10)

Do2 −Di2 o ln D D i


dpf = dL

for

dpf in dL

µv¯ 1500



Do2 + Di2

−

Do2 −Di2 o ln D D i



,

(9.11)

psi, µ in cP, v¯ in ft/s, and D o and D i in inches.

Example 32: A 9 lbm/gal Newton Newtonian ian fluid having having a viscosity viscosity of 15 cp is being being circulated in a 10,000 ft well containing a 7 in ID casing and a 5 in OD, 19.5 lb/ft drill pipe at a rate of 250 gal/min. Compute the pump pressure and the equivalent circulation density (ECD) at the bottom of the well. Assume laminar flow regime, and slick drill pipes (no effect of tool joints. Solution: CHAPTER 9 Flow in Pipes and Annuli

Page 9–7



The The pump pump pres pressu sure re in a typi typica call circu circula lati tion on syste system m is gove governe rned d by Equa Equati tion on (6.1 (6.10) 0),,

∆ p p = ∆ pb + ∆ p ∆ pf ,

(6.10)

(∆ pf )ds + (∆ pf )ann. There is no bit ( ∆ pb = 0), and we need the where ∆ pf = (∆ p frictional frictional pressure drop along the drillstring drillstring and along along the annulus annulus.. Then we have: Pressure drop gradient in the drillstring: [Equation (9.8)]

250 gal/ gal/min q v¯ = = π A (4. (4.276 in)2 4

×

  dpf dL

 ×

= ds

231 in3 1 gal

× ×  1 min 60 s

1 ft 12 in

= 5.585 ft/ ft/s

15 5.585 = 0.00305477 psi/ psi/ft 1500 4.2762

× ×

The pressure drop across the drillstring is

(∆ p (∆ pf )ds = 10, 10, 000 ft

psi/ft = 30. 30.55 psi × 0.00305477 psi/

Pressure drop gradient in the annulus: [Equation (9.11)]

v¯ =

  dpf dL

π 4

×

−

231 60 12

×

× 4.255 72 + 52 − 7ln−5



15

= 1500

ds

 ×

250 (72 52 )



2

7 5

The pressure drop across the annulus is

(∆ pf )ann = 10, 10, 000 ft

2



= 4.255 ft/s 255 ft/s

= 0.015927psi/ 015927psi/ft

psi/ft = 159. 159.27 psi × 0.015927 psi/

The total frictional pressure drop is

∆ pf = 30. 30.55 + 159. 159.27 = 189. 189.82 psi

≈ 190 psi

The ECD is the density of a fictitious fluid whose hydrostatic pressure is equal to the circulat circulating ing pressur pressure e at a given given depth. depth. The circulat circulating ing pressur pressure e at the bottom is obtained using the Bernoulli’s Equation:

p2 = p 1 + 0.0519 ρ 0519 ρ (D2 0 = p 1 + 0. 0 .0519

− D1) − 8.073 × 10−4 ρ

v¯22



− v¯12

10, 000) − 159. 159.27 → × 9 × (0 − 10,



+ ∆ p ∆ p p

− ∆ pf , (6.2)

p1 = 4830 psi

The ECD is given by:

4830 = 0. 0.0519


10, 000 → × EC D × 10,

EC D = 9.31 ppg

Page 9–8


9.1.5


Slot Approxi Approximation mation for Newtonian Newtonian Fluids Fluids

Lamb’s equation is an explicit exact formulation for the flow of Newtonian fluids in concentric annulus. The purpose of this section is to obtain the slot approximation, imation, and compare it with the exact formulatio formulation. n. The reason is to establish establish the procedure to be applied for non–Newtonian fluids since, in general, models for flow in annulus for non-Newtonian fluids result in implicit formulation. A slot approximation of an annulus is such that the thickness t of the slot is equal equal to the radial radial clearance clearance of the annulus, annulus, and the areas are equal. This is shown shown in Figure Figure 9.3. Therefo Therefore, re, the thickness thickness and width of an approximat approximating ing slot are given by: t = (Ro Ri ) ,

−

w = π = π (Ro + Ri ) . Considering the equilibrium equation for flow in annuli [Equation (9.3)] we obtain: dv dpf τ = µ = y , dy dL dv 1 dpf = y. dy µ dL

−

−

To obtain the flow rate we use the expression for

dv and dy

Equation (9.6):

t/2 t/2

w dpf q = = µ dL



y2 dy = dy =

w dpf 3 t , 12µ 12µ dL

t/2 −t/2

Solving for

dpf and dL

using average velocity we obtain:

12µv¯ dpf 12µ = 2 , dL t t = R R o Using t =

− Ri gives dpf 12µ 12µv¯ 48µ 48µv¯ = 2 = . dL Ro Ri2 (Do Di )2

−

−

(9.12)

Figure 9.3: Slot approximation of an annulus. CHAPTER 9 Flow in Pipes and Annuli

Page 9–9




dpf µv¯ = , dL 1000(D 1000(Do Di )2

−

for

dpf in dL

(9.13)

psi, µ in cP, v¯ in ft/s, and D o and D i in inches.2

Example 33: Calculate Calculate the frictional frictional pressure pressure drop in the annulus of Example 32 using the slot approximation model and compare the results. Solution: Using Using the Equation (9.13) we have:

dpf 15 = dL 1000

× 4.225 = 0.015957 psi/ft 015957 psi/ft × (7 − 5)2

The pressure drop for 10,000 ft is

(∆ p )ann = 10, 10, 000 ft

psi/ft = 159. 159.56 psi × 0.015957 psi/

Comparing with the result using exact approximation (159.27 psi) we see that the difference is of less than 0.2%.

9.1.6

Pressure Pressure Drop Drop Gradient Gradient for for Non–Newto Non–Newtonian nian Fluids Fluids

The derivation of the expressions for the pressure drop gradient in pipes and annu annulili (slot (slot appro approxim ximati ation) on) for for both both powe power–l r–law aw and and Bingha Bingham–p m–plas lastic tic fluids fluids is beyond the scope of this work. The derivations, however follow basically the same steps presented before (for pipes and slot approximation), with some peculiarities, which result in the expressions presented next. One peculiarity (shared by all fluid models that present yield point) is the fact that for Bingham–plastic fluid model there exist a region of the cross section where the fluid moves as a plug (no shear between the adjacent layers). 9.1.6.1 9.1.6.1

Pressure Pressure Drop Drop Gradient Gradient for for Power– Power–Law Law Fluids Fluids in Pipes Pipes

The expression in field units is:

dpf K v¯n = dL 143, 143, 640 D 640 D n+1



3n + 1 24 n



n

(9.14)

dp

for dLf in psi, K in equivalent centipoises, ¯v in ft/s, and D in inches (n is dimensionless). 2

A more accurate figure for the conversion factor is 997.51.


Page 9–10



9.1.6.2 Pressure Drop Gradient for for Power–L Power–Law aw Fluids Fluids in Annuli (Slot Approximation)


dpf K v¯n = dL 143, 143, 640 (D (Do Di )n+1

−

for

dpf dL in

9.1.6.3 9.1.6.3



2n + 1 48 n



n

(9.15)

psi, K in equivalent centipoises, ¯v in ft/s, and D o and D i in inches. Pressure Pressure Drop Drop Gradient Gradient for for Bingham–P Bingham–Plasti lastic c Fluids Fluids in Pipes


dpf µ p v¯ τ y = + dL 1500 D 1500 D 2 225 D 225 D for

dpf in dL

9.1. 9.1.6. 6.4 4

(9.16)

psi, µ p in cP, τ y in lbf/100 ft2 , ¯v in ft/s, and D in inches. Pres Pressu sure re Dr Drop op Grad Gradie ient nt for for Bing Bingha ham– m–Pl Plas asti tic c Flui Fluids ds in Annu Annuli li (Slo (Slott Approximation)


dpf µ p v¯ τ y = + dL (Do 1000 (D (Do Di )2 200 (D

−

for

dpf in dL

− Di)

(9.17)

psi, µ p in cP, τ y in lbf/100 ft2 , ¯v in ft/s, and D o and D i in inches.

Example 34: Calculate the pressure loss across a 10,000 ft long long annulus with 8.5 in OD, 5 in ID. The fluid follows a power–law model with behavior index of 0.3, and consistency index of 850 eq.cP. The flow rate is 250 gpm (assume laminar flow.) Solution:

The average velocity is:

v¯ =



231 60 12

×



250

dpf 850 2.1610.3 = dL 143, 143, 640 (8. (8.5 5)0.3+1

× ×

−

∆ pf = 10, 10, 000


ft/s − 52) = 2.161 ft/ 0.3 2 × 0.3 + 1 48 = 0.00772 psi/ psi/ft

π (8. (8 .52 4



0.3



77.22 psi × 0.00772 = 77.

Page 9–11



9.2 Turbule urbulent nt Flo Flow w in Pipes Pipes and and Ann Annuli uli All formulas for fluid flow in the previous section were developed assuming that the flow is laminar. In a laminar flow, the “particles” of the fluid move in orderly layers layers (laminae) (laminae),, although although the velocities velocities may change change between between those layers. layers. The particle velocities along the conduit change with regularity, and there is no eddies or streamline crosses. If a dye is carefully poured in the flow it flows as a streamline. streamline. On the other other hand, hand, in a turbulen turbulentt flow, flow, particles follows follows irregular irregular paths, with large change in velocity and flow direction compared with other near particles. particles. Eddies Eddies and vortices may occur, occur, and if a dye is poured in the flow, flow, it prom prompt ptly ly loos looses es it stre stream amliline ne shap shape e and and get get mixe mixed d with with the the flowi flowing ng fluid fluid.. Facto actors rs that govern the laminar flow are: 1. the velocity velocity of the the fluid, 2. its density density, and 3. the size, shape, shape, and surface roughness roughness of the conduit. conduit. There There is a transition between between laminar laminar and turbulent turbulent flow. flow. In these transition transition periods or regions, the flow is normally turbulent in the central portion of the flow section but may remain laminar close to the boundaries. Flow equations for laminar flow exist for various conduit shapes and fluid types. However, analytical mathematical models for turbulent flow have not yet been suitably developed. Studies of turbulent flow of Newtonian fluids in pipe have been extensively done. Most results are empirical and the subject is still not exhausted, although the results are quite accurately. Treatment for other geometries (annulus, open channels, slots, etc), and also for non-Newtonian fluids is normally made by some analogy analogy with Newtonia Newtonian n fluid in pipes. pipes. Therefo Therefore, re, it is important important to study the latter case in detail, and then to extend the results for other cases (in particular to those related to drilling engineering).

9.2.1

Turbulent urbulent Flow Flow of Newtonian Newtonian Fluids Fluids in Pipes

9.2.1.1 9.2.1.1

Reynold Rey nolds s Number Number

An important step is to determine or at least to evaluate the boundary between laminar and turbulent flow. In a series of classical experiments in 1883, Reynolds 3 showed that the flow would remain laminar (continuous dye streak) if a dimensionless group 3

Osborne Reynolds (1842-1912), scientist, engineer.


Page 9–12



now called Reynolds number (Re) is less than 2000, and that the flow will be turbulent if Re greater than 3000. The Reynolds number is given by

Re = Re =

ρ ¯v D , µ

(9.18)

where D is the diameter of the pipe in cm (glass pipe in Reynolds experiments), v¯ is the average velocity of the fluid in the conduit in cm/s, ρ is the density in g/cm3 , and µ is the dynamic dynamic viscosit viscosity y in poises. poises. A dimens dimension ional al analysis analysis will confirm that the expression is dimensionless: In field units the expression for the Reynolds number is:

Re = Re = 928

ρ ¯v D , µ

(9.19)

for D in inches, v¯ in ft/s, ρ in lbm/gal, and µ in cP. Another common notation for Reynolds number is N RE RE . The value of Re for a particular situation serves as an indicative of the flow regime. In most scenarios the ranges [0 to 2000] for laminar, [2000 to 3000] for transitional, and [3000 to infinite) are reasonably correct.

Example 35: Determi Determine ne the flow pattern pattern for a brine brine of 8.9 ppg and and 1.1 cP flowing inside a 5 in drillpipe (4.276 in ID) at 480 gpm. Solution:

The average velocity is:

v¯ =



231 60 12

Re = Re = 928

×

8.9



π 4

480 = 10. 10.73 ft/ ft/s 4.2762

×

10.73 × 4.276 × 10. = 252, 252, 630

1.5 Note that the Reynolds number is far above 3000, which indicates that the flow inside the drillpipe is clearly turbulent.

9.2.1.2 Friction Factor Factor,, Fanning Fanning Equation, and Colebrook Equation

When a fluid flows in a pipe, the resistance to the flow, measured by the pressure drop gradient, is caused by the effect of the boundary of the pipe (its internal wall) upon the fluid, more specifically, by the shearing stress occurring in the the fluid in contact with the pipe wall. A dimensionless group that relates shear stress τ at the surface between a length of fluid in a circular conduit and the specific kinetic energy ek (energy per CHAPTER 9 Flow in Pipes and Annuli

Page 9–13



Figure Figure 9.4: Fluid Fluid particle flowing flowing in a pipe. unit of volume) is the friction factor f (because it relates the resistance to flow to the velocity of flow), which is defined by the following relation:

f =

τ ek

(9.20)

∆ L flowing with average velocity Figure 9.4 shows a fluid element of length ∆L v¯ = q = q//A p . The force equilibrium in the flow direction results in the following expression relating the shear stress and the pressure drop gradient:

0=



F = P p p

 − − − P

−

dpf ∆L dL

p

τ Ac ,

where A p = π4 D2 is the cross–sectional area of the pipe and Ac = π D ∆L is the lateral area of the fluid element. Simplifying we obtain:

τ =

1 dpf D. 4 dL

The specific kinetic energy e k of the fluid element is:

E k ek = = V

1 2

m v¯2 1 = ρ v¯2 . V 2

where ρ is the density of the fluid. Substituting these two expressions into the expression for the friction factor and solving for the pressure drop gradient we obtain: 1 dpf D τ 1 dpf D f = = 41 dL 2 = , 2 ek 2 dL ρ ¯ v ¯ ρ v 2

dpf 2 ρ v¯2 = f. dL D This is called the Fanning equation . In field units we have dpf ρ v¯2 = f, dL 25. 25.8 D CHAPTER 9 Flow in Pipes and Annuli

(9.21)

(9.22) Page 9–14



Figure 9.5: Stanton chart. for

dpf in dL

psi/ft, ρ in lbm/gal, ¯v in ft/s, and D in inches.

If we comp compar are e (and (and equa equate te)) the the Fanni anning ng equa equati tion on with with Equa Equati tion on (9.7 (9.7)) (Poi (Poise seui uille lle’’s equation) we obtain: dpf 32 µ 32 µ v¯ 2 ρ v¯2 = = f, dL D2 D 16 µ 16 µ 16 f = = (9.23) ρ v¯ D Re This expression for the friction factor f is valid only if the flow is laminar . For turbulent flow, however, the friction factor must be determined experimentally imentally.. Extensive Extensive series of experimen experiments ts performed performed by Nikuradse Nikuradse in 1933 resulted in an empirical correlation obtained by Colebrook 4 for the friction factor f , which depends on the Reynolds number Re and on the pipe relative roughness D , where  is the absolute roughness (the average depth of the micros microscop copic ic irregul irregularit arities ies of the internal internal surfa surface ce of the pipe). pipe). The empirica empiricall correlation called modified Colebrook equation , is an implicit formula given by:

1 = f

√

−4 log10



 √

/D 1.255 1 + 3.7 Re f

.

(9.24)

This implicit expression requires an iterative solution. A log–log plot of this expression for various values of relative roughness, called Stanton chart , is shown in Figure 9.5. The solution of the implicit expression, however results in much more accurate values for the friction factor and should. The Stanton chart also include the curve for the friction factor for laminar flow. 4

Cyril Frank Colebrook, “Turbulent flow in pipes”


Page 9–15



R e = 125, 500 Example 36: Calculate Calculate the friction friction factor factor with four figures figures for Re ( /D)) = 0.0037. Compare the result with the Stanton chart. and (/D Solution:

Substituting the value for the Reynolds number and relative roughness into the modified Colebrook equation, and simplifying we obtain:

1 = f

√

−4 log10



 √

0.0037 1.255 1 + 3.7 125, 125, 500 f

Any value can be used to start the iterative process. A reasonable guess is

f 0 =

16 16 = = 0.00012749 Re 125500

Starting and proceeding the iterations we obtain:

f 0.000 0.000127 1275 5 0.008 0.008419 4196 6 0.007 0.007157 1572 2 0.007 0.007174 1746 6 0.007 0.007174 1744 4

→

f 0.0084 0.0084196 196 0.0071 0.0071572 572 0.0071 0.0071746 746 0.0071 0.0071744 744 0.0071 0.0071744 744

→

f = 0.0071744 Comparing this value with the Stanton chart we notice that a much higher degree of accuracy is obtained with the iterative procedure. In most drilling situations, the relative roughness is less than 0.0004 and the Reynolds number is less than 100,000. For these conditions we can consider the flow as occurring in a hydraulically smooth pipe, pipe, and the simplified equation is 1 1.255 1 = 4 log10 . (9.25) Re f f

√



−

 √

Attention: It is common in the industry another definition of friction factor whose value is four times the value obtained with the modified Colebrook equation. The log-log plot of this friction factor is called the Moody chart. If this chart is used to obtain the friction factor, it must be divided by four before it is is used in the pressure drop gradient formula obtained previously. 9.2.1.3 9.2.1.3

Alternative Alternative Expre Expression ssions s for the the Friction Friction Factor Factor

Blasius Formula– An approximation approximation for the the friction factor for for 2100 < R e < 100000 and hydraulic smooth pipes was presented by Blasius, and is given by:

f = CHAPTER 9 Flow in Pipes and Annuli

0.0791 . Re0.25

(9.26) Page 9–16



This expression is shown in the Stanton chart (Figure 9.5), where the curves for the various relative roughnesses converge to the Blasius approximation in the range mentioned. The range of applicability of Blasius’s approximation covers quite well most condit condition ions s in drillin drilling. g. If this this expre expressio ssion n is used used in Equati Equation on (9.21) (9.21) (Fann (Fanning ing equation) gives:

dpf 2 ρ v¯2 0.0791 2 ρ ¯ ρ ¯v 2 = = dL D Re0 .25 D

0.0791

  ρ v¯ D µ

0.25

µ0.25 ρ0.75 v¯1.75 = 0.1582 . D1.75

Expressing ¯v in terms of flow rate we obtain:

dpf µ0.25 ρ0.75 q 1.75 = 0.2414 . dL D4.75 In field units we have:

dpf µ0.25 ρ0.75 q 1.75 = . dL 8624 D 8624 D 4.75

(9.27)

(9.28)

dp

for dLf in psi/ft, µ in cP, ρ in ppg, q in in gpm, and D in inches. This expression for the frictional pressure drop gradient for turbulent flow in pipes was the motivation of the general expression

∆ pf = c q m used to determine the total frictional pressure drop in our study of nozzle optimization. timization. The reason reason for using "m" in place of the “theoretical” “theoretical” value value 1.75 is for the various different flow regime (turbulent, laminar, transition) that occurs in the circulation system.

Swamee–Jain Swamee–Jain Formula– Another explicit explicit expression for the friction factor is  − 6 000 < Re < 108, 108, 000 which given by Swamee–Jain for 10 < D < 10−2 and 5, 000 < diff differ ers s less less than than 1% from from the the valu value e obta obtain ined ed with with the the modi modifie fied d Cole Colebr broo ook k impl implic icit it formula: 1 f = (9.29) 2 . /D 5.74 4 log10 3.7 + Re0.9







Example 37: Calculate Calculate the the pressure pressure loss loss across across a 5,000 5,000 ft of 5 in OD, OD, 4.276 4.276 in ID drillstring. The fluid (assumed Newtonian) has density 9.3 ppg and viscosity 20 cP. cP. The flow rate is 370 370 gpm. gpm. Calcu Calculat late e the friction friction factor factor using (a) the modifie modified d Colebr Colebroo ook k formul formula, a, (b) the Blasiu Blasius s appro approxim ximati ation, on, and (c) the Swame Swamee e and Jain explicit formula. Solution: CHAPTER 9 Flow in Pipes and Annuli

Page 9–17



The first step is to confirm if the flow is turbulent. The average velocity inside the pipe is: 231 370 v¯ = = 8.266 ft/ ft/s 60 12 π4 4.2762





×

Re = Re = 928

9.3

×

× 8.266 × 4.276 = 15, 15, 253

20 This in an indication that the flow is turbulent.

(a) Assuming smooth pipe, we determine by iterative calculation the value for the Colebrook friction factor:

1 = f

√ Starting with f 0 =

16 15253



−4 log10

 √

1.255 1 15, 15, 253 f

= 0.0010490, and proceeding iteratively we obtain: f 0.001 0.001049 0490 0 0.009 0.009280 2807 7 0.006 0.006638 6380 0 0.006 0.006964 9644 4 0.006 0.006916 9162 2 0.006 0.006923 9231 1 0.006 0.006922 9221 1 0.006 0.006922 9223 3 0.006 0.006922 9222 2

f 0.0092 0.0092807 807 0.0066 0.0066380 380 0.0069 0.0069644 644 0.0069 0.0069162 162 0.0069 0.0069231 231 0.0069 0.0069221 221 0.0069 0.0069223 223 0.0069 0.0069222 222 0.0069 0.0069222 222

→

→

f = 0.0069222 In this this case case (Col (Coleb ebro rook ok)) the the fricti friction onal al pres pressu sure re drop drop grad gradie ient nt usin using g Equa Equati tion on (9.2 (9.22) 2) (Fanning equation) is

dpf ρ ¯v 2 9.3 8.2662 = f = dL 25. 25.8 D 25. 25.8 4.276

× ×

psi/ft , × 0.0069222 = 0.0.03987 psi/

and the pressure drop for 5,000 ft is:

∆ pf = 5, 000

× 0.03987 = 199 psi

(b) Using the formula resulting from Blasius approximation [Equation (9.27)] we have: dpf µ0.25 ρ0.75 q 1.75 200.25 9.30.75 3701.75 = = = 0.04100 dL 8624 D 8624 D 4.75 8624 4.2764.75

×


∆ pf = 5, 000 CHAPTER 9 Flow in Pipes and Annuli

×

×

× 0.04100 = 205 psi Page 9–18



Figure 9.6: Selection of the correct pressure drop value. (c) Using Swamee and Jain approximation [Equation (9.29)] we have:

f =

1



4 log10



0 3.7

+

5.74 152530.9



2

= 0.0069163

In this case the frictional pressure drop gradient using Equation (9.22) is

9.3 8.2662 dpf = dL 25. 25.8 4.276

× ×

psi/ft , × 0.0069163 = 0.0.03984 psi/


∆ pf = 5, 000

9.2.2

× 0.03984 = 199 psi

Criterion Criterion for for Laminar Laminar – Transiti Transition on – Turb Turbulent ulent Flow Flow

One of the questions usually raises is the criterion for laminar, transition, and turbulent flow for pressure drop calculation. The best procedure is to calculate the friction factor using the laminar relation, and using one of the turbulent friction factors (Colebrook, Blasius, Swamee and Jain, etc) and use the larger to obtain the frictional pressure drop gradient using the Fanning formula. This is depicted in the graph in Figure 9.6. In fact, it suffices to select the larger of the friction factors and use Fanning equation. CHAPTER 9 Flow in Pipes and Annuli

Page 9–19



It is important to stress that this approach does not determine the correct flow regime. regime. In some situations, situations, as for example example in cementing cementing operations operations,, it is desirable that during the cement displacement the flow regime of the flush 5 in the annulus is turbulent.

9.2.3 9.2.3

Other Other Geometri Geometries es – Turbule urbulent nt Flow Flow in Annuli Annuli (Newto(Newtonian)

A large amount of experimental data has been obtained for pipes, but relatively few few have have been obtained obtained for other geometries geometries,, in special for annuli. annuli. There There are several empirical procedures to apply the results for pipes to other geometries. All these empirical procedures try to determine an equivalent pipe diameter Deq of the flow cross section geometry, and use this value to determine the equivalent Reynolds number and in the Fanning equation. Note, however, that the average velocity must be calculated using the real area of the cross section. Three criteria are used to determine the equivalent diameter D eq : 1. The hydraulic hydraulic radius radius 2. Exact annulus annulus to pipe analogy analogy 3. Slot annulus annulus to pipe analogy analogy

9.2.3. 9.2 .3.1 1

Hydrau Hydraulic lic Radius Radius

One approach to calculate pressure drop for other conduit geometries is using the hydraulic radius concept. The hydraulic radius r H is defined as the ratio of the area of flow to the wetted perimeter. For a circular pipe, the hydraulic radius is: π R2 R D (rH ) pipe = = = . 2 π R 2 4 For a concentric annulus, the hydraulic radius is:

(rH )annulus

π (Ro2 Ri2 ) Ro R i Do Di = = = . 2 π (Ro + Ri ) 2 4

−

−

−

The concept states that “sections with the same hydraulic radius are equivalent alent for turbulent turbulent flow calculation calculation purposes. purposes.”” Therefo Therefore, re, equivale equivalent nt diameter diameter is Deq = D o Di . (9.30)

−

5

The flush is a special fluid displaced in front of the cement to help remove mud and mud cake in the annulus to improve the bonding between the cement sheath and the formation.


Page 9–20


9.2.3.2 9.2.3.2


Exact Exact Annu Annulus lus to Pipe Analogy Analogy

An equivalent diameter is obtained if we compare (and equate) Equation (9.7) for pressure drop in pipes (Poiseuille’s equation) with Equation (9.10) for pressure drop in annuli (Lamb’s equation). The purpose is to determine the diameter of the pipe that will produce the same pressure drop gradient for the same average velocity (for laminar flow).

dpf 32µ 32µv¯ 32µ 32µv¯ = = . Do2 −Di2 2 2 dL D2 Do + Di ln Do

−

Di

Therefore, the equivalent diameter is:

Deq =

9.2.3.3 9.2.3.3



Do2 +

Di2

−

Do2 Di2 . o ln D Di

−

(9.31)

Slot Annulus Annulus to Pipe Analogy Analogy

In this criterion (useful for non–Newtonian fluids) we compare (and equate) Equation (9.7) for pressure drop in pipes (Poiseuille’s equation) with Equation (9.12) for slot approximation of the pressure drop in annuli.

dpf 32µ 32µv¯ 48µ 48µv¯ = = . dL D2 ( Do D i ) 2

−

Therefore, the equivalent diameter is:

Deq =



2 (D ( Do 3

(Do − Di ) . − Di) = 0.8165 (D

(9.32)

Example 38: Calculate Calculate the frictional frictional pressure drop gradien gradientt using the three criteria above for the following flow condition: 9.2 ppg brine with 8 cP viscosity, circulating in an annulus of 8 1 /2 in OD and 5 in ID, at a flow rate of 280 gpm. Use the Blasius formula for the friction factor. (Compare with the friction factor for laminar flow.) Solution:

The average velocity in the annulus is:

v¯ =



231 60 12

×



π 4

×

280 (8. (8.52

ft/s − 52) = 2.421 ft/

a) Hydraulic radius criterion:

Deq = D o CHAPTER 9 Flow in Pipes and Annuli

− Di = 8, 5 − 5 = 3.5 in Page 9–21



9.2

Re = Re = 928 f =

× 2.421 × 3.5 = 9043 8

0.0791 0.0791 = = 0.008111 Re0.25 90430.25

Comparing with laminar, f =

16 9043

= 0.001769, we see that turbulent dominates.

9.2 2.4212 dpf ρ ¯v 2 = f = dL 25. 25.8 D 25. 25.8 3.5

×

×

psi/ft × 0.008111 = 0.0.004844 psi/

b) Exact annulus to pipe analogy:

Deq =



Do2 +

Di2

−

Do2

−

ln

= 928 Re = Re f =

Di2 Do Di

9.2

=



8.52

+ 52

−

8.52 52 = 2.864 in ln 85.5

−

× 2.421 × 2.864 = 7400 8

0.0791 16 = 0 008529 > 008529 = 0.002162 . > 74000.25 7400

dpf 9.2 2.4212 = dL 25. 25.8 2.864

× ×

psi/ft × 0.008529 = 0.0.006224 psi/

c) Slot annulus to pipe analogy:

Deq =



2 (D (Do 3

Re = Re = 928 f =

− Di) = 9.2



2 (8. (8 .5 3

− 5) = 2.2.858 in

× 2.421 × 2.858 = 7384 8

0.0791 16 = 0 . 008533 > 008533 > = 0.002167 73840.25 7384

dpf 9.2 2.4212 = dL 25. 25.8 2.858

× ×

9.2.4

psi/ft × 0.008533 = 0.0.006240 psi/

Turbulent urbulent Flow Flow for for Non–Ne Non–Newton wtonian ian Fluids Fluids

Turbulent flow for non–Newtonian fluids is treated by analogy with the results for Newtonia Newtonian n fluids. The approach approach is similar similar for Bingham–p Bingham–plastic lastic fluids and power–law power–law fluids. The key problem is to determine the Reynolds number and the criterion for turbulence. CHAPTER 9 Flow in Pipes and Annuli

Page 9–22


9.2.4.1 9.2.4.1


Pressure Pressure Drop Drop Gradient Gradient for Bingh Bingham am Fluids Fluids

To calculate the Reynolds number we need to determinate an apparent viscosity for for the the fluid fluid at a give given n cond conditi ition on.. One One appr approa oach ch is comp compari aring ng the the expre xpress ssio ions ns for the pressure drop gradient for Newtonian and Bingham fluids and determine the apparent Newtonian viscosity µ a that would cause the same pressure drop gradien gradientt (for (for laminar flow). The strategy strategy is to determine the viscosity of a Newtonian fluid that would cause the same pressure drop as the Bingham fluid under the same flow conditions. Bingh Bingham am Fluids Fluids in Pipes: Pipes: Compa Comparing ring and equati equating ng Equati Equation on (9.8) (9.8) (Pois (Poiseui euille lle’’s equation in field units) with Equation (9.16) (pressure drop gradient in pipes for Bingham fluids) gives:

µv¯ µ p v¯ τ y = + 1500 D 1500 D 2 1500 D 1500 D 2 225 D 225 D Solving for the dynamic viscosity and renaming for apparent viscosity µa we obtain: τ y D µa = µ = µ p + 6. 6 .66 (9.33) v¯ The apparent viscosity is used to calculate the Reynolds number and then the friction factor using the same procedure for Newtonian fluids, and then the pressure drop gradient using Fanning equation. Bingh Bingham am Fluids Fluids in Annulu Annulus: s: C Com ompari paring ng and and equati equating ng Equat Equation ion (9.13) (9.13) (pres(pressure drop gradient in annuli for Newtonian fluids – slot approximation), with Equation (9.17) (pressure drop gradient in annuli for Bingham fluids – slot approximation) proximation) gives:

µv¯ µ p v¯ τ y = + (Do 1000(D 1000(Do Di )2 1000 (D (Do Di )2 200 (D

−

−

− Di)

Note that here we opted in using the slot approximation formula for Newtonian fluid for consistency, since there is no explicit exact formula for Bingham fluid in annuli. Solving for the dynamic viscosity and renaming for apparent viscosity µa we obtain: τ y (Do Di ) µa = µ p + 5 (9.34) v¯

−

For consistency also, we should choose to use the equivalent diameter for annulus based on the slot approximation [Equation (9.32)]. The apparent viscosity and the equivalent diameters are used to calculate the Reynolds number, and then the friction factor using the same procedure for Newtonian fluids, and then the pressure drop gradient using Fanning equation.


Page 9–23



Example 39: Calculate the frictional pressure drop gradient gradient for for the following following lbf flow condition: 10 ppg Bingham fluid with µ p = 40 cP and τ y = 15 100 ft2 , circulating in an annulus of 6 3 /4 in OD by 4 1 /2 in ID, at a flow rate of 600 gpm. Use the Colebrook formula for the friction factor (assume smooth pipe). Solution:




v¯ =

231 60 12

×



π 4

×

600 (6. (6.752

ft/s − 4.52) = 9.683 ft/

The apparent viscosity for the fluid at this flow conditions is:

µa = µ p + 5

τ y (Do v¯

(6.75 − 4.5) − Di) = 40 + 5 × 15 × (6. = 57. 57.43 cP 9.683

The equivalent diameter is:

Deq =



2 (Do 3

− Di ) =



2 (6. (6.75 3

− 4.5) = 1.1.837 in

The Reynolds number is:

Re = Re = 928

ρ ¯v Deq = 928 µa

683 × 1.837 = 2874 × 10 × 9.57. 57.43

16 = 0.0056672. This is used to start the iterThe laminar friction factor is f = 2874 ative ative process process and to compare compare with the converg converging ing value. value. Using the Colebrook Colebrook equation for smooth pipe [Equation (9.25)] we have:

1 = f

√

−4 log10



 √

1.255 1 2874 f

Proceeding iteratively gives:

f 0.005 0.005667 6672 2 0.012 0.012494 4949 9 0.010 0.010776 7768 8 0.011 0.011070 0702 2 0.011 0.011016 0160 0 0.011 0.011025 0259 9 0.011 0.011024 0241 1 0.011 0.011024 0244 4

→

f 0.0124 0.0124949 949 0.0107 0.0107768 768 0.0110 0.0110702 702 0.0110 0.0110160 160 0.0110 0.0110259 259 0.0110 0.0110241 241 0.0110 0.0110244 244 0.0110 0.0110244 244

→

f = 0.0110244 The pressure drop gradient is:

dpf ρ ¯v 2 10 9.6832 = f = dL 25. 25.8 D eq 25. 25.8 1.837

× ×


psi/ft × 0.0110244 = 0.0.2181 psi/

Page 9–24


9.2.4.2 9.2.4.2


Pressure Pressure Drop Drop Gradien Gradientt for Powe Power–L r–Law aw Fluids Fluids

Here a procedure similar to Bingham fluids is conducted. The only difference is that the Colebrook formula (and consequently Blasius and Swamee–Jain) does not provide an accurate calculation of the friction factor. An special formulation is Needed. The friction factor must be calculated with the following implicit formula:

√ 1f = n04.75

n

log10 Re f 1− 2



−

0.395 n1.2

(9.35)

This This impl implic icit it formul ormula a was was deve develo lope ped d by Dodg Dodge e and and Metzn Metzner er for for hydra ydraul ulic ic smoo smooth th pipe, and is acceptable for most drilling situations. Power–Law Fluids in Pipes: We calculate calculate the apparent apparent viscosity viscosity equating equating Equation (9.8) (Poiseuille’s equation in field units) with Equation (9.14) (pressure drop gradient in pipes for power–law fluids):

µv¯ K v¯n = 1500 D 1500 D 2 143, 143, 640 D 640 D n+1



3n + 1 24 n



n

Solving for the dynamic viscosity and renaming for apparent viscosity µa we obtain: n K v¯n−1 3n + 1 µa = 24 (9.36) 95. 95.9 D n−1 n





The apparent viscosity is used to calculate the Reynolds number and then the friction factor using Dodge and Metzner formula, and then the pressure drop gradient using Fanning equation. Power–Law Fluids in Annulus: We calculate calculate the apparent apparent viscosity viscosity equating Equation (9.13) (pressure drop gradient in annuli for Newtonian fluids – slot approximation) with Equation (9.15) (pressure drop gradient in annuli for power–law fluids – slot approximation):

µv¯ K v¯n = 1000(D 1000(Do Di )2 143, 143, 640 (D (Do Di )n+1

−

−



2n + 1 48 n



n

Again we need to use the slot approximation formula for Newtonian fluid for cons consis iste tenc ncy y, sinc since e ther there e is no expli xplici citt exact xact formul ormula a for for powe power– r–la law w fluid fluid in annu annulili.. Solving for the dynamic viscosity and renaming for apparent viscosity µa we obtain: K v¯n−1 2n + 1 n µa = 48 (9.37) n 143. 143.9 (Do Di )n−1

−





Again we should choose to use the equivalent diameter for annulus based on the slot approximation [Equation (9.32)]. CHAPTER 9 Flow in Pipes and Annuli

Page 9–25



The apparent viscosity and the equivalent diameters are used to calculate the Reynolds number, and then the friction factor using Dodge and Metzner formula, and then the pressure drop gradient using Fanning equation.

Example 40: Calculate the frictional pressure drop gradient gradient for for the following following = 205eq cP and n = K = n = 0.7885, circulating in an flow condition: 10 ppg fluid with K 3 1 annulus of 6 /4 in OD by 4 /2 in ID, at a flow rate of 600 gpm. (The fluid has the same θ 600 and θ300 of Example 39.) Solution:




v¯ =

231 60 12

×



π 4

×

600 (6. (6.752

ft/s − 4.52) = 9.683 ft/

The apparent viscosity for the fluid at this flow conditions is:

K v¯n−1 µa = 143. 143.9 (Do Di )n−1



−

×

n

 

2n + 1 48 n 205 9.6830.7885−1 = 143. 143.9 (6. (6.75 4.5)0.7885−1

−

=

× 0.7885 + 1 48 × 0.7885 2



0.785

= 56. 56.34 cP

The equivalent diameter is:

Deq =



2 (Do 3

− Di ) =



2 (6. (6.75 3

− 4.5) = 1.1.837 in

The Reynolds number is:

Re = Re = 928

ρ ¯v Deq = 928 µa

683 × 1.837 = 2930 × 10 × 9.56. 56.34

16 = 0.0054608. This The laminar friction factor is f = 2874 This is used used to start start the iterative process and to compare with the converging value. Using the (Dodge and Metzner) equation [Equation (9.35)] we have:

1 4 = log10 2930 f 0.78850.75 Proceeding iteratively gives:



√

f 0.005 0.005460 4608 8 0.011 0.011074 0749 9 0.009 0.009260 2606 6 0.009 0.009675 6750 0 0.009 0.009571 5711 1 0.009 0.009596 5966 6 0.009 0.009590 5903 3 0.009 0.009591 5918 8 0.009 0.009591 5915 5

→


1− 0.7885 2

× f

−

0.395 0.78851.2

f 0.0110 0.0110749 749 0.0092 0.0092606 606 0.0096 0.0096750 750 0.0095 0.0095711 711 0.0095 0.0095966 966 0.0095 0.0095903 903 0.0095 0.0095918 918 0.0095 0.0095915 915 0.0095 0.0095915 915

→

Page 9–26



f = 0.0095915 The pressure drop gradient is:

dpf ρ ¯v 2 10 9.6832 = f = dL 25. 25.8 D eq 25. 25.8 1.837

× ×


psi/ft × 0.0095915 = 0.0.1898 psi/

Page 9–27




Page 9–28



Chapter 10 Drilling Bits The bit is the active drilling drilling tool. Differen Differentt from any other drilling drilling equipment, equipment, it is designed to perform (drill a length of formation in an economical time), and eventua eventually lly it wears wears out and is discarded. discarded. The average average life of bits (dependin (depending g on the diameter and drilling conditions) can be as short as hours and they may cost tens of thousand Dollars. The drill bit performance is a very important issue in drilling design and operation operation.. There There is an extremely extremely large variety variety of bit types and models made currently by the bit industry, each one suitable to one or more different kind of rocks and drilling conditions. Formation rocks can be roughly classified as 1. Soft (low compressive compressive strength, high drillability). 2. Medium soft/hard (low compressive strength, medium medium drillability). 3. Hard (high compressive strength. strength. low drillability). 4. Abrasive (high (high compressive strength, low drillability). 5. Plastic (low compressive compressive strength, low low drillability). Drill bit selection is in general a complicated process but, when performed properly, has a major impact on the total well cost. First in this chapter, the different types of drill bits are discussed. discussed. Then, Then, applying applying their classification classification and wear wear considera considerations tions,, a drill bit selection selection is presented presented.. Finally Finally,, various various parameters parameters that influence the rate of penetration are discussed.

10.1 10 .1

Dril Dr illl Bi Bitt Types ypes

Different bits perform differently in different rocks. Bits can be generally classified in two major groups: roller cone bits, and fixed cutter bits (drag bits). CHAPTER 10 Drilling Bits

Page 10–1



Figure 10.1: Typical roller cone bits.

10.1. 10. 1.1 1

Rolle Rollerr Cone Cone Bi Bitt

Roller cone bits are designed to crush rock efficiently while incurring a minimal amou amount nt of wear wear on the the cutt cuttin ing g surf surfac aces es.. Inve Invent nted ed by Howa Howard rd Hugh Hughes es,, the the roll roller er– – cone bit has conical cutters or cones that have spiked spiked teeth around them. them. As the drillstring is rotated, the bit cones roll along the bottom of the hole in a circle. As they roll, new teeth come in contact with the bottom of the hole, crushing the rock rock immed immediat iately ely below below and around around the bit tooth. tooth. As the cone cone rolls, rolls, the tooth then lifts off the bottom of the hole and a high-velocity fluid jet strikes the crushed rock chips to remove them from the bottom of the hole and up the annulus. annulus. As this occurs, another tooth tooth makes contact with the bottom of the hole and creates creates new rock chips. Thus, Thus, the process process of chipping chipping the rock and removing removing the small rock chips with the fluid jets is continuo continuous. us. The teeth intermesh on the cones (but with no interaction), which helps clean the cones and enables enables larger teeth to be used. There There are two main types of roller-cone roller-cone bits, steel milled-tooth bits and tungsten carbide insert bits. Typical roller cone bits are shown in Figure 10.1. Figure 10.2 shows a cut view it a roller cone bit. The basic elements in a roller cone bit are the bit body, the bearings, and the cutting cones.

10.1. 10 .1.1.1 1.1 Bit Body Body

The bit body is composed of the following parts: • Connectio Connection n pin: API thread thread of various sizes dependin depending g on the size of the bit. CHAPTER 10 Drilling Bits

Page 10–2



Figure 10.2: Cut view of a roller cone bits. • Bit legs: 3 elements elements welded welded together together to form the bit body. body. Each leg has a pin in the shirttail with a bearing where the cone is mounted. • Fluid channels: channels: Flow Flow ways for the drilling drilling fluid. They They end in the nozzles. nozzles. Most bits have 3 nozzles (each between two cones.) Some bits have 4 nozzles. In this case, the fourth nozzle is centered with the bit, located above the cones and, in most of the cases, it is not interchangeable (fixed nozzle area). The bit body may also contain a lubricant reservoir (only in sealed bearings). 10.1. 10 .1.1.2 1.2 Be Beari arings ngs

There are basically three types of bearings for tricone bits: 1. non–seale non–sealed d roller bearing, bearing, 2. sealed sealed roller bearing, bearing, 3. sealed sealed journal bearing. bearing. The type of bearing and sealing will determine the range of weight on bit, rotary speed speed,, life, life, and also the price of the bit. bit. All three three types have have a ball ball bearing bearing whose function is to support axial load and keep the cone connected to the bearing pin. Roller bearing bits normally operate in medium load and medium to high speed ranges. CHAPTER 10 Drilling Bits

Page 10–3



Figure 10.3: Cut view of a non–sealed bearing bit. Non–sealed Roller Bearing Bits Non–sealed roller bearing bits are the the simplest plest and cheap cheapest est of all bits. bits. A channe channell conne connects cts the outsid outside e of the bit leg to inside the bearing. bearing. The drilling drilling fluid penetrates penetrates in the roller compartment, compartment, which helps to cool down the parts, but also causes wear in the parts due to the solids solids in suspen suspensio sion n in the fluid. fluid. The fluid also “lubrica “lubricates tes”” the bearbearing parts. Non–sealed roller bearing bits are manufactured normally with steel teeth. Figure 10.3 shows a cut view of a non–sealed bearing bit.

Sealed Roller Bearing Bits Sealed Sealed roller bearing bearing bits have have a seal between between the internal side of the cones (where the bearings are located) and the outside. Drilling Drilling fluid cannot cannot penetrate penetrate in the bearing. bearing. A lubrication lubrication system keeps the bearing lubricated. The lubrication system has a reservoir holding a supply of lubricant lubricant located located in the upper flank of each bit leg. A passage in the bit body extends from the reservoir to the bearing to allow flow of lubricant to the bearing. A diaphragm at the reservoir provides pressure compensation between the lubricant and the drilling fluid in the annulus between the bit and the wellbore. The lubrication lubrication system increases increases significantl significantly y the life of the bearings. bearings. Sealed Sealed bearing bits are in the intermediate range of price. The life expectancy expectancy however, however, in most cases, compensates the higher price compared to a similar non–sealed bearing bearing bit. Figure Figure 10.5 shows a cut of a sealed roller bearing cone. cone.

In roller bearings (both sealed and non-sealed) the rollers are responsible to support the radial force of the cones, and consequently the axial force applied to the bit. The ball bearing prevents the cone from falling. As the rollers wear, howev however er,, part of the radial radial load is transfer transferred red to the ball baring. baring. Ball bearing bearing CHAPTER 10 Drilling Bits

Page 10–4



Figure Figure 10.4: 10.4: A sealed sealed bearing bearing bit.Figure bit.Figure 10.5: 10.5: Cut view view of a roller roller bearing bearing cone. cone. is not designed to work under radial forces, and if the balls wear out, the cone may may fall fall out from from the bit. Theref Therefore ore,, it is very very importa important nt to keep keep the drilling drilling parameters within the working range and monitor the bit performance so that a worn bit is changed before extreme wear causes the fall of the cones. Sealed bearing type roller cutter bits further have a lubrication system includin cluding g a reserv reservoir oir holding holding a supply supply of lubrica lubricant. nt. A passag passage e in the bit body body extends from the reservoir to the bearing to allow flow of lubricant to the bearing. A seal is disposed disposed between between the roller cutter and the bearing bearing journal that holds holds lubricant lubricant in the bit. A diaphrag diaphragm m at the reservoir reservoir provides pressure pressure compensation between the lubricant and the drilling fluid in the annulus between the bit and the wellbore.

Journal Bearings Bits In bits with Journal Journal bearings, bearings, the internal internal surface surface of the cones cones maintai maintains ns direct direct contact contact with the bearing bearing pin of the bit legs. legs. An extremely rigorous tolerance and special metallurgical treatment are required in this type of bearing, in addition to permanent lubrication. In higher priced bits, a silver alloy is deposited in the internal surface of the cone, increasing the radial capacity of the bearing. Journal bearing bits are expensive but have longer life compared to roller bearing bits. Since Journal bearings are more compact than roller bearings, more room exists to increase the cone wall thickness and pin diameter so that the bit can be designed with more robust specifications (larger weight on bit rating). Figure 10.6 shows a cut of a sealed roller bearing cone.

10.1. 10 .1.1.3 1.3 Cuttin Cutting g Cones Cones

The cutting cones are of fundamental importance in the performance of the bits. bits. In general, general, two things determine determine the type of cone: The cutting cutting structure CHAPTER 10 Drilling Bits

Page 10–5



Figure Figure 10.6: Cut view of a journal bearing cone. cone. (steel teeth or tungsten carbide inserts,) and the cone offset. Steel (Milled) Teeth Bits In steel teeth teeth cones, cones, the teeth are machine machined d in the cone and normally receive receive heat and surface surface treatment. treatment. These These bits are very robust and tolerate severe drilling conditions but wear out relatively quickly. From this reason they are not well suited for deeper wells where tripping constitutes a larg large e time time facto actorr. A typi typica call mill milled ed teet teeth h bit bit is sho shown in the the left left side side of Figu Figure re 10.1 10.1..

Large Large teeth are designed designed for soft to medium medium hard formations. formations. Steel Steel teeth bits for hard formations have smaller and more robust teeth (shorter, and wider base). The teeth are distributed circularly in the cones in an intermeshing fashion but without interference. This promotes a better teeth cleaning action. Tungsten Carbide Inserts Bits Tungsten ungsten carbide, carbide, WC or W2C, W2C, is a chemical compound compound containing containing tungsten tungsten and carbon. carbon. Tungsten ungsten carbide insert is a powder metallurgy technology in fine powder of tungstan carbide cristal are cemented together using tough cobalt as agglutinant in a process called sintering.

Tungsten carbide insert bits designed to drill soft formations have large and chisel shaped chips, and those designed to drill hard formations have small and round shaped inserts. Inserts are mounted in the steel cones by pressing the compacts in precisely bored holes on the cone surface. These bits do not tolerate shock loadings but they can drill long sections befo before re being worn worn out. out. In genera general, l, insert insert bits bits of the same same bit size are more expensive than milled tooth bits. Coating the inserts with a thin layer (about 5 µm ) of titanium carbide made by chem chemic ical al vapo vapour ur depo deposit sitio ion n (CVD (CVD)) proc proces ess s may may impr improv oved ed the the lifet lifetim ime e of tool tools s CHAPTER 10 Drilling Bits

Page 10–6



Figure 10.7: Geometry of bit cones. by a factor of 2 to 5. 10.1. 10 .1.1.4 1.4 Cone Cone Geomet Geometry ry

The cone geometry geometry affects affects the selectivity selectivity and action of the bit. Three parameparameters are important in the cone geometry: 1. cone angle angle (or angles), angles), 2. offset angle, angle, 3. offset distan distance. ce. The shape of the cone is determined by one or more angles (normally two). The cone angles and the pin angle will define the crown profile of the bit. The pin angle (relative to hole bottom) is reduced for softer bits and increased creased for harder harder bits. This alters the cone profile which which in turn affects affects tooth action on the hole bottom and gage cutter action on the wall of the hole. Softer bits have more highly profiled cones than harder bits. This increases the scraping action of both bottomhole and gage teeth. The scraping action is beneficial for drilling soft formations but it will result in accelerated tooth and gage wear if the formation formation is abrasiv abrasive. e. Scraping Scraping action is minimized minimized on hard formation formation bits where where strength and abrasion abrasion resistance resistance are emphasize emphasized d in the design. design. (See Figure 10.7.) The bit profile also affects the direction tendency of the bit (to cut straight for straight drilling, or for direction changes in directional drilling). Concave crown profiles tend to keep the drilling straight, and more flatten crown profiles favor to direction change (some other important drilling parameters like bit tilt and lateral force define the actual tendency of the bit). During operatio operation, n, the cones cones are subjected subjected to two rotations. rotations. One about its own axis, and one due to the rotation of the cone axis itself about the axis of CHAPTER 10 Drilling Bits

Page 10–7



Figure 10.8: Cone offsets. the bit, as the bit rotates. The cone offset determines basically the action of the teeth teeth on the formati formations ons.. The offset offset is a measur measure e of the distan distance ce of the cone apexes to the center of the bit.(See Figure 10.8.) If the offset is zero, the cones roll as the bit rotates and the teeth crushes the formation. This is suitable to hard formations. If an offset exists, in addition to roll, the cones tend to scrape the formation. This is suitable to soft to medium hard formation formations. s. The offset distance distance is the distance distance of the cone centerline centerline to the bit centerline. This distance also increase the scrape action of the teeth.

10.1. 10. 1.2 2

Airr Dr Ai Dril illi ling ng Bi Bits ts

Bits used for air drilling have special special design. In particular, particular, air drilling is approappropriated for hard to very hard formations. Therefore, these bits have zero cone offset, offset, small and round tungsten tungsten carbide inserts and non–sealed non–sealed bearings. bearings. A system of channels conducts part of the drilling gas (air, N 2 , or natural gas), throug through h the bearing bearings. s. In addition addition to cool cool the moving moving parts, parts, the flowing flowing gas keep keeps s the bearing bearings s clean clean from small small abras abrasiv ive e particle particles. s. A screen screen in the entrance of the channels filters the gas from large particles, which could damage the bearings. Figure 10.9 shows a cut of an air drilling bit.

10.1.3 10.1.3

Fixed Fixed Cutter Cutter Bits Bits (Drag (Drag Bits) Bits)

Drag bits have have an integral integral cutting element element and no moving moving parts. In the beginbeginning of the drilling industry, the bits were all of the drag type, with cutters (or blades) blades) made out of hardened hardened steel. steel. Drag bits cut the formations formations plowing plowing the rock by the blades blades under the action action of axial force and rotation. rotation. It is the same mechanism used in drill bits to bore holes in metal and masonry. The fact that drag bits have no moving parts reduces the possibility of leaving junks in the borehole. CHAPTER 10 Drilling Bits

Page 10–8



(a)

(b)

Figure Figure 10.9: Air drilling bits. 10.1. 10 .1.3.1 3.1 Steel Steel Bla Blade de Bits Bits

Steel drag bits works well in soft formations, but wear rate increases rapidly in medium and hard formations, making them inappropriate for deep drilling. With the advent of the roller cone bits early last century, the steel blade drag bits becomes obsolete and drag bits, in general, were banned from the drilling industry (except diamond drag bits for hard and abrasive formations and for coring operations) until recently with the introduction of PDC bits. (See Figure 10.10). 10.1. 10 .1.3.2 3.2 Dia Diamo mond nd Bits Bits

Until early 70’s, diamond bits were used exclusively in hard and abrasive formations and for coring. They take advantage of the properties of the (natural) diamonds like extreme hardness, compressive strength, and thermal conductivity. tivity. Nowadays, Nowadays, however however,, with the advance of the manufacturing manufacturing technology, technology, diamond bits have been designed and made to drill also medium hard formations. The body of diamond bits is made from sintering powdered tungsten carbide using copper copper or cobalt cobalt as agglutinating agglutinating agent. agent. A graphite graphite cast with diamonds diamonds suitably distributed on the internal surface is filled with powdered tungsten carbide. A hollow rod is positioned centered with the cast creating the way for the drilling fluid. The cast is put in a high temperature (1050 ◦ C to 1170 ◦ C) and the molten molten agglutinan agglutinantt infiltrates infiltrates under pressure in the pores of the powder. powder. After the brazing process, a larger rod is welded to the previous one and the pin is lathed. When properly operated (right formation and right drilling parameters), only CHAPTER 10 Drilling Bits

Page 10–9



Figure 10.10: Steel blade drag bits.

Figure 10.11: A diamond bit.

CHAPTER 10 Drilling Bits

Page 10–10



Figure 10.12: Schematic and nomenclature of diamond bit. the diamonds contact the formation, letting a narrow space between the body of the bit and the formation. formation. Drilling Drilling fluid flows from the central central hole of the bit and across groves molded in the body of the bit, and then across these narrow spaces, spaces, cooling cooling the diamonds diamonds and carrying the small rock fragments. fragments. The number, size, and distribution of the diamonds in the bit determine its characteristics. acteristics. Bits designed designed for medium medium hard formations formations have fewer fewer and larger stones (0.75 to 2 carats) than bits designed for hard formations (0.07 to 0.125 carats.) The distribution of the diamonds is also important (manufacturers have their own design), and the shape of the cutting face (crown profile) basically determines the drilling characteristics (concave for straight holes, long taper to build build and drop angle in directiona directionall drilling, etc). Figure Figure 10.12 presents presents a typical schematics and the nomenclature of a diamond bit. Note that some grooves serve to conduct the fluids from inside the bit to the borehole, and others collect and drive the ground rock from below the bit. Since diamond bits have no nozzle, the hydraulic calculation is quite different from that for bits with nozzles. Experimental results from different manufacCHAPTER 10 Drilling Bits

Page 10–11



turers indicate that the ideal hydraulic parameters should be 2.0 to 2.5 hp/in 2 and pressure drop between 500 to 1000 psi across the face of the bit to properly cool and clean the diamond diamonds. s. Since Since the pressu pressure re drop depend depends s on the area open to flow, which in turns depends of the groves distribution and the distance of the bit body to the formation (and therefore on the weight on the bit and formation hardness), the pressure drop is determined by the difference between the pressure when the bit is off the bottom and when weight is applied to the bit. The hydraulic pressure is determined using Equation (10.1)

∆ pb q , 1714. 1714.29

P b =

(10.1)

∆ pb is the pressure increase when the bit is forced against the formation. where ∆ p Manufacturers provide tables with the estimate required flow rate and pressure drop for different bit models and formations.

Example 41: A 8 1 /2 in diamond bit is used to drill a hard formation at 9000 ft. The flow rate required to suitably carrying the cuttings is 400 gpm. How should the driller apply the weight on the bit. Solution:

Using the criterion of 2 to 2.5 hp/in2, the power developed at the bit must be in the following range:

P min [hp/in2 ] min = 2 [hp/

113.4 hp × π4 × 8.52 = 113.

[hp/in2 ] P max max = 2.5 [hp/

141.9 hp × π4 × 8.52 = 141.

The pressure drop across the bit for these power rates are:

∆ pmin = ∆ pmax =

1714. 1714.29 113. 113.4 hp = 486 psi 400 gpm

×

1714. 1714.29 1141. 1141.9 hp = 608 psi 400 gpm

×

Assuming that a minimum of 500 psi is required to properly cool and clean the diamonds, the driller should apply weight so that the pump pressure increase remains between 500 psi and 608 psi.

10.1 10 .1.3 .3.3 .3 PD PDC C Bits Bits

Since late 70’s, bits using artificial diamonds, called Polycrystalline Diamond Compact Compact (PDC), have have been used with increasing increasing success in the drilling industry. CHAPTER 10 Drilling Bits

Page 10–12



Figure 10.13: PDC bits.

A PDC cutter is made of a thin layer of synthetic diamond particles, bonded in a sintered tungsten carbide chip (in a process very similar to the construction of a diamond diamond bit, but in a small scale). scale). Important Important properties of the PDC chips are abras abrasion ion resist resistanc ance e and impact impact resista resistance nce.. These These properti properties es are influen influenced ced primarily primarily by the size distribution distribution of the synthetic diamonds diamonds particles, with large sized chips being more resistant to impact forces, and small sized chips more resistant to abrasive wear. In steel base PDC bits, the chips are inserted under pressure in holes bore in the bit body, and in tungsten carbide bits the chips are welded in steel supports in the bit body because the temperature of the sintering process is too high and may cause destabilization of the diamonds layer layer.. Figure Figure 10.14 presents a typical typical schematics schematics and the nomenclat nomenclature ure of a PDC bit. Some PDC bits (all steel base and most carbide base) have interchangeable able nozzles. The hydraulic hydraulic optimizatio optimization n criteria are similar to those for roller cone bits, when provided with nozzles. However, some carbide base PDC bits have structure of groves similar to those in natural diamond bits and the same criteria (hp/in²) and pressure drop apply (manufacturers also provide tables of parameters.) Other important aspects of the PDC bit performance is the number, size, position (back rake and side rake angles), and shape of the chips, as well as the shape of the bit body itself. More aggressive back angle is used for soft to medium formation, and less aggressive back angle is used for hard formations. These angles are depicted in Figure 10.16. CHAPTER 10 Drilling Bits

Page 10–13



Figure 10.14: Schematic and nomenclature of a PDC bit.

Figure 10.15: Nozzles in a PDC bit.

Figure 10.16: Back rake and side rake angles in PDC bits.


Page 10–14


10.2 10 .2


Bitt Cla Bi Class ssifi ifica catio tion n

In 1987, 1987, the IADC (International (International Association Association of Drilling Drilling Contractor Contractors) s) publishe published d a classification system for roller cone bits, revised in 1992 1 . The purpose of the system is provide a convenient method for categorizing rock bits according to their design features and intended application. The The 1992 1992 IADC IADC roll roller er bit bit clas classi sifica ficati tion on stand standar ard d defin defines es a 4–ch 4–char arac acte terr desi design gn– – relate related d code. code. The first 3 charac character ters s are numeric numeric and the charac character ter is alphaalphabetic. betic. The sequen sequence ce of numer numeric ic characte characters rs is define defined d as series , type , and bearing/gage . The alphabetic 4 th character describes features available . Roller Bits can be listed on a reference chart according to the digits in the IADC code. The chart form (see Figure 10.17) is explained as follows: 1. First Character – Cutting Structure Series (1 to 8): Eight categories or series numbers numbers describe general general formation formation characteristics characteristics.. Series 1 through through 3 refer refer to steel tooth (milled (milled tooth) bits. Series 4 through 8 refer refer to insert (tungsten carbide) bits. Within the steel tooth and insert groups, the formations become harder and more abrasive as the series numbers increase. 2. Second Character – Cutting Structure Types (1 to 4): Each Series is divided into 4 types or or degrees of hardness. Type 1 refers to bits designed for the softest formation in a particular series. Type 4 refers to the hardest formation within the series. 3. Third Character – Bearing/Gage: Seven categories of bearing design and gage protection are defined as bearing/gage . Categories 8 and 9 are reserved for future use. 4. Fourth Character – Features Available (Optional): Sixteen alphabetic characters are used to indicate features available as as shown in the chart. This includes includes special special cutting cutting structures, structures, bearings, bearings, hydraul hydraulic ic configuraconfigurations, and body gage protection. Series Series and and types types are arrang arranged ed in numer numerica ically lly–in –incre creasi asing ng rows rows from from top to bottom. Bearing/Ga Bearing/Gage ge categories categories are arranged arranged in numericall numerically–inc y–increasin reasing g columns from left to right. This creates 224 spaces with an additional 64 spaces reserved for future use. Thus, the code 111 indicates a steel tooth bit equipped with standard non– sealed roller bearings and a cutting structure designed to drill the very softest formations. At the opposite corner of the chart, an 847 code indicates an insert bit equipped with sealed friction bearings and gage protection, designed for the very hardest abrasive formations. Every roller cone bit can be assigned an exact position on the IADC classification fication chart. A comparison of the bits in each manufactur manufacturer’ er’s s product line is 1

SPE/IADC paper number 23937, February, 1992.


Page 10–15



Figure 10.17: IADC roller cone bit classification chart. thus obtained. It is the manufacturer’s responsibility to assign the most appropriate IADC code to each bit. The fact that each bit has a distinct IADC code does not mean that it is limited to drilling only the narrow range of formations defined by a single box on the chart. All bits will, within reason, drill effective effectively ly in both softer softer and harder formation formations s than specified specified by the IADC code. Also, Also, competitiv competitive e products with the same IADC code are built for similar applications but they may be quite different in design detail, quality, and performance. The optional 4 th character describe features of the roller cone that are not indicated by the first 3 characters in the IADC code. Such features are important since they can affect affect bit cost, application, application, and performanc performance. e. The current alphabet alphabetic ic characters characters are shown in Table able 10.1. 10.1. Some bit designs may have several several combination combination of features features availab available. le. In such cases, the most significant significant feature should be listed. Following are representative examples of four character IADC roller bit codes. (1) 124E – a soft formation, sealed roller bearing steel tooth bit with extended jets CHAPTER 10 Drilling Bits

Page 10–16



Table 10.1: IADC codes for roller cone bits. Code

Additional Feature

Description

A

Air Air Appl Applic ica ation tion

Ide Identifi ntifies es a bit spe specific cifical ally ly for appli pplica cati tion ons s with with air as the the dril drilli ling ng fluid fluid..

B

Spec Specia iall Beari Bearing ng Seal Seal

Seal Seal confi configu gura ratio tion n that that prov provid ides es spec specia iall appl applic icat atio ion n adva advant ntag ages es such such as high RPM capability.

C

Center Jet

Larger diameter bits are sometimes equipped with center jets (Fig. 5) to provide a more uniform distribution of flow and hydraulic energy beneath beneath the bit. Almost all extended extended nozzle bits have center jets to provide a beneficial tooth–cleaning action that might otherwise be lost by concentrating all of the hydraulic energy on the bottomhole. Some manufacture manufacturers rs use diffuser–type diffuser–type center center jets while others others use standard standard rock bit jet nozzles. The pressure pressure drop through through these two types of jets is different and should be taken into consideration when doing hydraulic calculations for bits equipped with center jets.

D

Devi Deviat atio ion n Cont Contro roll

Cutt Cuttin ing g stru struct ctur ure e spec specifi ifica call lly y desi design gned ed to mini minimi mize ze devi deviat atio ion. n.

E

Extended Jets

Ext Extended jet jets (nozzles) are used mainly inly on soft forma rmation ion bits for improved improved bottomhole bottomhole cuttings removal. removal. Higher Higher jet impact energy is delivered delivered to the hole bottom by extended extended jets. Extended Extended jets (Fig. 6) are generally generally available available on bits larger larger than 9.5 inches. Miniature Miniature extended jets are not included in the “E” designation.

G

Gage Gage/B /Bod ody y Prot Protec ecti tion on

Welde elded d tung tungst sten en carb carbid ide e depo deposi sits ts (har (hardf dfac acin ing) g) or carb carbid ide e inse inserts rts added to the shirttail to protect the seal and/or body in special applications such as geothermal and directional drilling (Fig. 7).

H

Horizo Horizonta ntal/S l/Stee teering ring Applica Applicatio tion n

Design Designed ed specific specifically ally for for horizo horizonta ntall and steera steerable ble applica applicatio tions. ns.

J

Jet Deflection

These bits are used sed for making traject jecto ory cha changes where the forma rmations tions are soft soft enoughto enoughto be fluid-er fluid-erode oded. d. Such Such bits bits usuall usually y contai contain n two standard jet nozzles and one large jet nozzle and can be oriented to preferentially excavate the hole in a desired direction (Fig. 8).

L

Lug pads

Steel pads with tungsten carbide inser ts ts applied to the bit body. These pads generally are very close to gage diameter (Fig. 9).

M

Motor tor App Applica licati tio on

R

Reinf einfo orce rced welds lds

S

Stan Standa dard rd Stee Steell Tooth ooth

T

Two-Co o-Cone ne Bit Bits

Spec Specifi ifica call lly y desi desig gned ned for appl applic icat atio ion n on downho nhole moto otors. rs.

Two-co -cone bits bits are are rela relati tiv vely unco uncom mmon mon but som sometim etimes es util utiliz ize ed for obobtaining an acceptable combination of deviation control and penetration rate

W

Enha Enhanc nced ed Cutt Cuttin ing g Struc Structu ture re

X

Chis Chisel el Tooth ooth Inse Insert rt

Y

Coni Conica call Tooth ooth Inse Insert rt

Z

Othe Otherr Shap Shape e Inse Insert rts s


Page 10–17



(2) 437X – a soft formation, sealed friction bearing insert bit, with gage protection and chisel–shaped teeth.

10.2.1 10.2.1

PDC Bit Cla Classifi ssificati cation on System System

Four characters are used in a prescribed order to indicate seven fixed cutter bit design features: cutter type, body material, bit profile, fluid discharge, flow distribution distribution,, cutter size, and cutter density density.. These These design design traits were selected selected as being most descriptive of fixed cutter bit appearance. The four–character bit code is entered on an IADC–API Daily Drilling Report Form as shown in Fig. 2. The space requirements are consistent with the four–cha four–characte racterr IADC roller bit classification classification code. The two codes are readily readily distinguished from one another by the convention that diamond bit code begin with a letter, while roller bit codes begin with a number. Each of the four characters in the IADC fixed cutter bit classification code are further described as follows: First Character – Cutter Type and Body Material: The The first charact character er of the fixed cutter classification code describes the primary cutter type and body material (Fig. 1). Five letters are presently defined:

• D – natural diamond/matrix diamond/matrix body, body, • M – PDC/ matrix matrix body, body, • S – PDC/steel PDC/steel body, body, • T – TSP/matrix TSP/matrix body, body, • 0 – other other.. The term PDC is defined defined as “polyc “polycrysta rystallin lline e diamon diamond d compac compactt .” The term TSP is defined defined as “thermally “thermally stable polycrystalline polycrystalline”” diamond . TSP material material are composed of man–made polycrystalline diamond which has the thermal stability of natural diamond. This is accomplished through the removal of trace impurities and in some cases the filling of lattice structure pore spaces with a material of compatible thermal expansion coefficient. The distinction of primary cutter types is made because fixed cutter bits often contain a variety of diamond materials. Typically one type of diamond is used as the primary cutting element while another type is used as backup material. Second Character – Profile: The numbers numbers 1 through 9 in the second character of the fixed cutter classification code refer to the bit’s longitudinal–sectional profile profile (Fig. (Fig. 3). The term profile profile is used here here to describ describe e the longitud longitudina inal– l– section section of the cutter/bott cutter/bottomho omhole le pattern. This distinction distinction is made because because the CHAPTER 10 Drilling Bits

Page 10–18



Table 10.2: Range for IADC bit profile. G – Gage height

High: G > 3D/8 Med: 3D/8 G D/8 D /8 > G Low: D/

≥ ≥

High C > D /4 1 4 7

C – Cone height Medium D/4 C D/8 2 5 8

≥ ≥ ≥

Low D/8 > C 3 6 9

cutter/bottomhole profile is not necessarily identical to the bit body profile. Nine basic profiles are defined by arranging two profile parameters – outer taper (gage height) and inner concavity (cone height) – in a 3 x 3 matrix (see Table 10.2). The rows and columns of the matrix are assigned high, medium, and low values for each parameter. Gage height systematically decreases from top to bottom. Cone height systematically decreases from left to right. Each profile is assigned a number. The ranges are based on the bit diameter D . Third Character – Hydraulic Design: The number numbers s 1 through 9 in the the third character of the fixed cutter classification code refer to the hydraulic design of the bit. The hydrauli hydraulic c design is described by two componen components: ts: the type of fluid fluid outle outlett and the flow distrib distributi ution. on. A 3 x 3 matrix of orifice types types and and flow flow distribution distributions s defines defines 9 numeric numeric hydrau hydraulic lic design codes (see Tabl Table e 10.3. 10.3. The orifice type varies from changeable jets to fixed ports to open throat from left to right in the matrix. The flow distribution distribution varies from bladed bladed to ribbed ribbed to open face from top to bottom. There is usually a close correlation between the flow distribution and the cutter arrangement.

Table 10.3: Range for IADC bit hydraulic design. change changeab able le jets jets fixed fixed ports open open throat throat bladed 1 2 3 ribbed 4 5 6 open face 5 8 9 The IADC has not explicitly defined the difference between the three flow distribution distribution categories categories but but working working definitions definitions are offered offered as follows follows.. The term bladed refers refers to raised, continuous flow restrictors with a standoff distance from the bit body body of more than 1.0 inch. inch. In most cases cases cutter cutters s are affixed affixed to the blades blades so that the cutter arrangement arrangement may also be described as bladed. bladed. The term ribbed refers refers to raised continuous flow restrictors with a standoff distance from the bit body of 1.0 inch or less. Cutters are usually affixed to most of the ribs so that the cutter arrangemen arrangementt may also be described described as ribbed. ribbed. The term open face refers refers to non-restricted flow arrangements. Open face flow designs generally have a more even distribution of cutters over the bit face than with bladed or ribbed designs. A special case is defined: the numbers 6 and 9 describe the crowfoot/water CHAPTER 10 Drilling Bits

Page 10–19



course design of most natural diamond and many TSP bits. Such designs are further described as having either radial flow, crossflow (feeder/collector), or other hydraulics. Thus, the letters R (radial flow), X (crossflow), or O (other) are used as the hydraulic design code for such bits.

Cutter Size and Placement Density The number numbers s 1 through through 9 and 0 in the 4th character of the fixed cutter classification code refer to the cutter size and placem placemen entt density density on the bit. A 3 x 3 matrix of cutter cutter sizes sizes and placem placement ent densities defines 9 numeric codes (see Table 10.4.

Table 10.4: Range for IADC cutter size and density. Density ligh lightt medi medium um hea heavy large 1 2 3 medium 4 5 6 small 5 8 9 The placement density varies from light to medium to heavy from left to right in the matrix. The cutter size varies from large to medium to small from top to bottom. The ultimate combination of small cutters set in a high density pattern is the impregnated bit , designated by the number 0.

10.3 10. 3

Drill Dri ll Bit Select Selection ion and Evalu Evaluati ation on

Since a well is drilled only once and each well penetrated the formations at different locations with different drilling parameters, a selection of a “best bit” can not be performed. The next best way to find an “optimum bit” is to compare bit performances of drilling bits when they were run under similar conditions. Then Then a cost– cost–pe per– r–ffoot oot valu value e of each each bit bit appl applica icati tion on can can be calcu calcula late ted. d. Along Along with with this criteria, the individual bit wear are evaluated. This knowledge is applied to the well to be drilled (length, inclination inclination,, drillability drillability,, abrasive abrasiveness ness,, etc of the different sections). In practice, when the well is planned, bits that have been used previously in this area (by this drilling team) are evaluated according to their applicability. Sometimes when a bit manufacturer has developed a new bit, it is introduced to the industry with an expected minimum performance. Thus, when such a new bit is applied and the proposed performance is met (usually better than ones of already applied bits), the operator has increased the pool of possible bits to use for future wells. In case the performance performance proposed proposed by the manufactur manufacturer er is not met, agreements that the bit is given at reduced cost to the operator are common. Another way of bit evaluation is the determination of the specific energy CHAPTER 10 Drilling Bits

Page 10–20



Figure 10.18: Tooth wear diagram for milled tooth bits. using Equation (10.2).

e =

W OB 2πN T + , Ah Ah ROP

(10.2)

where W OB is the weight applied to the bit, Ah is the cross–sectional area of ROP P is the rate the hole, N is the rotary speed, T is the torque at the bit, and RO of penetration. Here the cutting–performance of various bits are compared to each other. For this, the mechanical energy of the bit is related to the drilled rock volume. It should be noted that a bit selection considering the specific energy may not lead to the finding of the most economic bit. In all practical cases, to evaluate previously applied bits, the so called bit record records s are studied. studied. These These bit record records s includ include e all avail availab able le informa informatio tion n (bit (bit size, type, manufacturer, nozzles used, rotation time, applied WOB, applied RPM, etc) about the bits applied within drilled wells.

10.3. 10. 3.1 1

Tooth ooth Wear

With tooth wear, the reduction of tooth height is graded after a bit was run. The grading is reported in nearest eighth, thus a bit whose teeth are worn out to half of its original height, is reported as T–4. Normally the tooth wear of a bit is not even distributed over the bit, some teeth are worn more than others, some are broken out. Broken teeth are generally remarked as “BT”. The reported wear is an average one based on the most severely worn teeth. Reporting of the tooth wear is possible when the teeth are measured before and after the bit was run. In general, tooth wear has no direct relationship with the drilling rate realizable. For insert bits, tooth wear occurs, due to the hardness of the teeth, as breaking or losing of them. Thus a T-4 graded insert bit may have half of its teeth broken or lost. A diagram of tooth wear for milled bits is shown in Figure 10.18.

10.3. 10. 3.2 2

Beari Be aring ng Wear

Evaluation of bearing wear in the field is difficult since the bit would need to be disassemble disassembled d for inspection. inspection. Thus it is mainly mainly determined if the bearings bearings are intact intact or faile failed. d. Failed ailed bearin bearings gs is the situatio situation n that that the cones cones are stuck stuck (no rotation possible), or that they are worn out and the bearings are exposed. The classification is similar to the tooth wear, using a B instead of T. Thus a bit which bearings are worn to 7 is marked as B–7. CHAPTER 10 Drilling Bits

Page 10–21



Often the bearing wear is reported based on the total bit running hours. Thus Thus,, when when a bit bit is expec xpecte ted d to hav have a rota rotati tion on time time of 40 hour hours s and and was rota rotati ting ng on bottom for 10 hours, the bearing wear is reported as B-2.

10.3 10 .3.3 .3

Gage Gage Wea earr

When the gauge diameter of a bit is worn, the drilled hole will be undergage (and tapered) with may lead to damage of the next bit and stuck pipe. Measurement of the gauge wear is performed with the help of a calliper and a ruler. The loss of diameter in eighth of inches is reported, denoting with the letter “O” for “out of gage”. In this way, a bit which diameter is reduced by 0.5 in is repo reporte rted d as G–O– G–O–4 4 . Whe When n the the bit bit is in gaug gauge e, it is repo reporte rted d usin using g the the lett letter er “I”. “I”. In addition to the wear gradings listed above, the bit record commonly includes a column of comments. Here the bit conditions are commonly remarked.

10.4 10. 4

Facto Factors rs that that Affect Affect the Rate Of Pene Penetra tratio tion n

Although throughout the text various aspects that influence the ROP are mentioned whenever appropriate, the following considerations are often applied to determine the recommended drilling parameters.

10.4 10 .4.1 .1

Bitt Type Bi ype

The type of bit used to drill a certain formation has a large impact on the achieved achieved penetration rate. Roller cutting bits with long teeth exhibit commonly the highest penetration rates but they are only applicable at soft formations. At hard formations where PDC bits dominate, the realized ROP is mainly a function of size and amount of cutters, along with an optimum combination of drilling parameters.

10.4.2 10.4.2

Format Formation ion Charact Characteris eristics tics

The most important formation properties that determine the penetration rate are the elastic limit and the ultimate rock strength. The strength of a formation is usually estimated using the Mohr failure criteria. When drilling is initiated, a threshold force or bit weight

 W db

t

has to be

overcom overcome. e. This threshold threshold force can be found found when plotting plotting drilling rates as a function of bit weight per diameter and then extrapolated to zero drilling rate. A correlation between threshold and shear strength is shown in Figure 10.19. CHAPTER 10 Drilling Bits

Page 10–22



Figure 10.19: Correlation between rock strength and threshold WOB. Another formation property that has a large influence to the realized ROP is the permeability permeability.. In rocks with high permeability permeability,, the drilling mud filtrates filtrates into the rock ahead of the bottom of the hole and thus reduces the differential pressure. Other rock properties like its abrasiveness and gummy clay minerals content contribute indirectly to the ROP by influencing the drilling bit (wear, dulling, etc).

10.4.3 10.4.3

Drilli Dri lling ng Fluid Fluid Prope Properties rties

Among the various drilling fluid properties, the following are identified as influencing the penetration rate: [(a)] 1. drilling fluid density density,, 2. rheologica rheologicall flow properties, properties, 3. filtration filtration characteristi characteristics, cs, 4. solids content content and distribution distribution,, CHAPTER 10 Drilling Bits

Page 10–23



Figure 10.20: Variation of ROP with different fluid properties. 5. chemical chemical composition composition.. Penetration rate in general decreases with increasing fluid density, viscosity, and solids content, content, and increases with increasing increasing filtration filtration rate. This latter is mainly caused by the reduction of the differential pressure in the formation right below below the bottom bottom of the hole. The drilling fluid viscosity viscosity controls the frictional pressure losses along the drillstring and thus reducing the available hydraulic energy energy at the bit. Solids Solids particles with size less than 1 µm (colloid size) influence the ROP dramatically since they tend to plug off the porous of the rock, reducing reducing the filtration filtration below the bit. The effects effects of these various various parameters parameters are loosely shown in Figure 10.20. The penetration rate is largely dependent on the differential pressure as seen in Figure 10.21. The effective differential pressure at the bottom has several implications in reducing reducing the ROP. The first is the chip hold–down hold–down effect. effect. The second is the increase of the confining pressure, which increases the strength of the rock. If the rate of penetration versus pressure differential is plotted in a semi– log paper, a reasonable linear relationship can be obtained, as seen in Figure 10.22. An expression for the relationship can then be written as:

log

R = R0

−m ( pbh − pf ) ,

where R is the rate of penetration at a particular overbalance, R 0 is the rate of penetration for zero overbalance, overbalance, m is the slope or the regressed line, p bh is the CHAPTER 10 Drilling Bits

Page 10–24



Figure 10.21: Effect of differential pressure in the ROP. bottom hole pressure, and p f is the formation pore pressure at the bottom hole depth.

10.4.4 10.4.4

Operati Operating ng Condit Condition ions s

The effects of changes in the operating conditions, namely WOB and rotary speed, are shown in Figure 10.23. Ideally, the ROP should increase linearly with the WOB (for a fixed rotary speed), as shown in the segment s egment a–b–c in the graph . However However,, field tests t ests show that above a given value the response departs from the linear behavior, and an increase in WOB does not correspond to the expected increase in ROP, as in the segment segment c–d. In situations situations as in the segment d–e, d–e, the rate of penetration penetration may even reduce. This behavior is called “floundering”. Two factors contribute to the floundering behavior. One is the reduction of hole cleaning capacity due to the increase of ROP (assumed the hydraulics is kept constant). The second is the complete complete imbedding imbedding of the cutters (teeth or inserts) into the formation. formation. It is important, therefore to find the onset of the floundering region. Drillers conduct conduct a variety variety of tests to optimize performan performance. ce. The most common is the drill rate test, test, which consists of simply experimenting with various WOB and CHAPTER 10 Drilling Bits

Page 10–25



Figure Figure 10.22: 10.22: Exponent Exponential ial relationsh relationship ip between between of differen differential tial pressure and ROP.

(a)

(b)

Figure 10.23: Effect of WOB (a) and rotary speed (b) in the ROP. RPM settings settings and observing the results. results. The paramete parameters rs are then used that resulted in the highest ROP. In some sense, all optimization schemes use a similar similar comparati comparative ve process. process. That is, they they seek to identify identify the paramete parameters rs that yield the best results relative to other settings. Another scheme is the drilloff . In the drilloff test, the driller applies a high WOB and locks the brake to prevent the top of the string from advancing while continuing to circulate and rotate the string. As the bit drilled ahead, the string elongated and the WOB declined. ROP was calculated from the change in the rate of drill string elongation as the weight weight declined. declined. The point at which the ROP stops responding responding linearly linearly with increasing increasing WOB WOB is defined defined as the flounder flounder point. This is taken to be the optimum WOB. This process has enhanced performance, but does not provide an objective assessment of the true potential drill rate, only the flounder point of the current system. CHAPTER 10 Drilling Bits

Page 10–26



10.4. 10 .4.4.1 4.1 Bit Wear

As the bit is worn during drilling, the penetration rate decreases. This reduction of ROP is generally less severe for insert bits as for milled tooth bits. 10.4. 10 .4.4.2 4.2 Bit Hy Hydra drauli ulics cs

Practice has shown that effective bit hydraulics can improve the penetration rate dramatically dramatically.. The enhanced enhanced jetting action promotes promotes a better better cleaning cleaning of the teeth as well as the bottom of the hole. hole. To improve improve the cleaning cleaning capacity capacity of the bit extended nozzles are often used where the discharging nozzle ends are closer to the hole bottom. bottom. Extended Extended nozzles nozzles usually requires requires the use of a 4th central nozzle, to guarantee a suitable cleaning of the cones, particularly in “gummy” formations As discussed in well hydraulics, maximum hydraulic horsepower and maximum jet impact force are the most used criteria to optimize hydraulics. When a low WOB is applied and drilling rates are low, the required hydraulics for efficient hole cleaning cleaning is small. small. When the WOB is increased increased and the well is drilled faster, efficient hydraulic programs have to be followed to realize the higher penetration rates.


Page 10–27


CHAPTER Drilling Bits


Page –28



Appendix A Drill Pipe Dimensions (as in API RP7C)

CHAPTER A Drill Pipe Dimensions

Page A–1



Table A.1: New Drill Pipe Dimensional Data (1) Size OD in. D

(2) Nominal Weight Threads and Couplings, lb/ft 4.85 6.65

(3)

(4)

(5) ID in. d 1. 1 .995 1.815

(6) Section Area Body of Pipe sq. in. A 1.3042 1.8429

(7) Polar Sectional Modulus cu. in. Z 1.321 1.733

Plain end Weight lb/ft 4.43 6.26

Wall Thickness i n. 0.190 0.280

2 78

6.85 10.40

6.16 9.72

0.217 0.362

2. 2 .441 2.151

1.8120 2.8579

2.241 3.204

3 12

9.50 13.30 15.50

8.81 12.31 14.63

0.254 0.368 0.449

2. 2 .992 2.764 2.602

2.5902 3.6209 4.3037

3.923 5.144 5.847

4

11.85 14.00 15.70

10.46 12.93 14.69

0.262 0.330 0.380

3.476 3.340 3.240

3.0767 3.8048 4.3216

5.400 6.458 7.157

4 12

13.75 16.60 20.00 22.82

12.24 14.98 18.69 21.36

0.271 0.337 0.430 0.500

3.958 3.826 3.640 3.500

3.6004 4.4074 5.4981 6.2832

7.184 8.543 10.232 11.345

5

16.25 19.50 25.60

14.87 17.93 24.03

0.296 0.362 0.500

4.408 4.276 4.000

4.3743 5.2746 7.0686

9.718 11.415 14.491

5 12

19.20 21.90 24.70

16.87 19.81 22.54

0.304 0.361 0.415

4.892 4.778 4.670

4.9624 5.8282 6.6296

12.221 14.062 15.688

6 58

25.20 27.70

22.19 24.22

0.330 0.362

5.965 5.901

6.5262 7.1227

19.572 21.156

2 38


Page A–2



Table A.2: New Drill Pipe Torsional and Tensile Data. Courtesy API


Page A–3



Table A.3: New Drill Pipe Collapse and Internal Pressure Data. Courtesy API


Page A–4



Table A.4: Premium Drill Pipe Torsional and Tensile Data. Courtesy API


Page A–5



Table A.5: Premium Drill Pipe Collapse and Internal Pressure Data. Courtesy API


Page A–6

Drilling Engineering Fundamentals

Recommend Documents