Date post: | 19-Jul-2016 |

Category: |
## Documents |

Upload: | brahim-letaief |

View: | 108 times |

Download: | 6 times |

Share this document with a friend

Description:

Directional drilling

28

1 PETE 411 Well Drilling Lesson 33 Directional Drilling

Transcript

1

PETE 411Well Drilling

Lesson 33

Directional Drilling

2

Directional Drilling

◆ When is it used?◆ Type I Wells ◆ Type II Wells ◆ Type III Wells ◆ Directional Well Planning & Design◆ Survey Calculation Methods

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

I II III

3

Read ADE Ch.8 (Reference)

HW #17 due 11-27-02HW #18 due 12-06-02

4

Inclination Angleθ,θ,θ,θ,

αααα, I

Direction Angleφ,φ,φ,φ,

εεεε, A

5

6

Max.Horiz.

Depart.?

7

8

9

10

11

12

Type I Type II Type III

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

13

∆∆∆∆x

∆∆∆∆yI

I

r

r∆∆∆∆L

In the BUILDSection

∆∆∆∆x = r (1 - cos I)

∆∆∆∆y = r sin I

∆∆∆∆L = r ΙΙΙΙrad

degI r180

= L

π∆

BUR*000,18r

π=

14

15

Fig. 8.11

42131 xrr and xr <+<

1642131 xrr and xr >+<

17

N18E

N55WS20W

S23E

AzimuthAngle

18

19

Example 1: Design of Directional Well

Design a directional well with the following restrictions:• Total horizontal departure = 4,500 ft• True vertical depth (TVD) = 12,500 ft• Depth to kickoff point (KOP) = 2,500 ft• Rate of build of hole angle = 1.5 deg/100 ft• Type I well (build and hold)

20

Example 1: Design of Directional Well

(i) What is the maximum hole anglerequired.

(ii) What is the total measured depth (MD)?

(MD = well depth measured along the wellbore,not the vertical depth)

21

(i) Maximum Inclination

Angle

ft820,35.1000,18r1 ====

ππππ====

0r2 ====

(((( ))))

ft000,10500,2500,12

DD 14

====−−−−====

−−−−

ft500,4x4 ====

22

(i) Maximum Inclination Angle

−−−−−−−−++++−−−−

====500,4)820,3(2

500,4)820,3(2000,10500,4000,10tan 2 22

1-

�3.26max ====θθθθ

−−−−++++++++−−−−−−−−++++−−−−−−−−

====θθθθ −−−−

421

4212

1424141

max x)rr(2x)rr(2)DD(xDD

tan2

23

(ii) Measured Depth of Well

ft 395 )26.3 cos-3,820(1

)cos1(rx 1Build

========

θθθθ−−−−====�

ft 4,105 395500,4xHold

====−−−−====∴∴∴∴

ft 265,9L105,4sinL

Hold

Hold

====∴∴∴∴====θθθθ∴∴∴∴

24

(ii) Measured Depth of Well

Holdrad11 LrDMD ++++θθθθ++++====

ft 518,13MD ====

265,9180

26.33,8202,500 ++++

ππππ++++====

25

* The actual well path hardly ever coincides with the planned trajectory

* Important: Hit target within specified radius

26

What is known?I1 , I2 , A1 , A2 ,∆∆∆∆L = ∆∆∆∆MD1-2

Calculateββββ = dogleg angleDLS = β∗100/∆β∗100/∆β∗100/∆β∗100/∆L

DLS = dogleg severity

27

28(20)

Recommended