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# 33 Directional Drilling

Date post: 19-Jul-2016
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Directional drilling
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1 PETE 411 Well Drilling Lesson 33 Directional Drilling
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PETE 411Well Drilling

Lesson 33

Directional Drilling

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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

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

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Inclination Angleθ,θ,θ,θ,

αααα, I

Direction Angleφ,φ,φ,φ,

εεεε, A

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6

Max.Horiz.

Depart.?

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8

9

10

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Type I Type II Type III

Build and Hold

Build-Hold and Drop

ContinuousBuild

KOP

EOC

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∆∆∆∆x

∆∆∆∆yI

I

r

r∆∆∆∆L

In the BUILDSection

∆∆∆∆x = r (1 - cos I)

∆∆∆∆y = r sin I

degI r180

= L

π∆

BUR*000,18r

π=

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Fig. 8.11

42131 xrr and xr <+<

1642131 xrr and xr >+<

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N18E

N55WS20W

S23E

AzimuthAngle

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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)

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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)

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(i) Maximum Inclination

Angle

ft820,35.1000,18r1 ====

ππππ====

0r2 ====

(((( ))))

ft000,10500,2500,12

DD 14

====−−−−====

−−−−

ft500,4x4 ====

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(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

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(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∴∴∴∴

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(ii) Measured Depth of Well

ft 518,13MD ====

265,9180

26.33,8202,500 ++++

ππππ++++====

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

* Important: Hit target within specified radius

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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

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