Petroleum Engineering 406 Lesson 18 Directional Drilling

Petroleum Engineering 406 Lesson 18 Directional Dril ing

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Lesson 10 – Directional Drilling  When is it used? Type I Wells (build and hold) Type II Wells (build, hold and drop) Type III Wells (build) Directional Well Planning & Design Survey Calculation Methods

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Homework: READ. “Applied Drilling Engineering” Ch. 8, pp. 351-363 REF. API Bulletin D20, “Directional Drilling Survey Calculation Methods and Terminology”

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What is Directional Drilling? Directional Drilling is the process of directing a wellbore along some trajectory to a predetermined target. Basically it refers to drilling in a non-vertical direction. Even “vertical” hole sometimes require directional drilling techniques. E xamples: Slanted holes, high angle holes (far from vertical), and Horizontal holes.

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Non-Vertical θ, α or I Wellbore Inclination Angle Z Axis (True Vertical Inclination Depth) Plane Y North Direction Angle φ, ε or A Direction Plane X

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Lease Boundary Surface Surface Location for Well No. 1 Location for Well No. 2 Bottom Hole Location for Well 2 Houses Oil-Water Contact Fi gure 8.2 – Plan view of a typical oil and gas structure under a lake showing how directional wells could be used to develop it. Best locations? Drill from lake?

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Top View NOTE: All the wel s are directional 5 – 50 wel s per platform Figure 8.3 – Typical offshore development platform with directional wel s.

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Drilling Rig Inside Building Figure 8.4 – Developing a field under a city using directionally drilled wells.

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Why not drill from Maximum top of lateral mountain? displ.? Fig. 8.5 – Drilling of directional wells where the reservoir is beneath a major surface obstruction.

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Cement Plug Fish Lost in Hole and Unable to Recover Sidetracked Hole Around Fish Figure 8.6 – Sidetracking around a fish.

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Figure 8.7 – Using an old Oil Producing Wel well to explore Ready to Abandon for new oil by Sidetracked sidetracking Out of Casing out of the casing and Possible drilling New Oil directionally. Old Oil Reservoir

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Horizontal Departure to Target Type II Build-hold and Drop (“S Type”) Build and Build-hold Drop and/or Hold Type I Hold Type (Modified “S” Type) Continuous Type III Build Figure 8.8 – Major types of wellbore trajectories.

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Figure 8.10 – Geometry of the build section. Build Section Build Radius:1 , r1 = 8 000 π * BUR

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Build Section: L engt o h f a rc , L = θ 1 r1 1 Vert icald epth = C D ‘ ‘= rs θ 1 in 1 Horiz. dev. = DD= ‘ r( θ 1 1 − cos 1) = L r 1 = 100 1 θ π 1 θ 1 *180 1 , 8 ↑ ↑ r1 = 000 r a d deg π * BUR

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Start of Buildup End of Build Type II Drop Off Target Build-hold-and drop for the case where: r < x a nd r +r < x 1 3 1 2 4

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Kickoff End of Build Type II Maximum Inclination Build-hold-and Angle drop for the case Drop Off where: Target r < x a n d r + r > x 1 3 1 2 4

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Projected Trajectory Projected Trajectory with Left Turn to Hit Targets Target 1 Target 2 Target 3 Fig. 8-14. Directional well used to intersect multiple targets

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N18E S23E A = 157o Fig. 8-15. Directional quadrants and compass measurements N55W S20W A = 305o

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Projected Wel Path Lead Angle Surface Location for Wel No. 2 Target at a Lake TVD 9,659 Figure 8-16: Plan View

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Example 1: Design of Directional Wel 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

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Example 1: Design of Directional Wel This is a Type I well (build and hold) (i) Determine the maximum hole angle (inclination) required. (ii) What is the total measured depth of the hole (MD)?

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Type I: Build-and-Hold 2500’ Imax TVD1 12,500’ 10,000’ HD1 Imax 4,500’

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0’ Uniform 1’30” Increase in Drift per 100 ft of hole drilled 10,000’ Try I = 27o ?? max Vert. Depth 4,500’ Horizontal Deviation

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Solution Type I Well 1.5 deg/100’ 2500’ Available depth I = 12,500-2,500 max TVD1 = 10,000’ 10,000’ From Chart, Imax Try = 27o Imax HD1

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Build Section From chart of 1.5 deg/100’, with I = 27o max In the BUILD Section: MD = 1,800’ (27/1.5) I 1 max TVD TVD = 1,734’ 1 1 HD = 416’ 1 8,266’ Imax Remaining vertical height HD = 10,000 – 1,734 = 8,266’ 1

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Solution Horizontally: 416 + 8,266 tan 27o = 4,628 We need 4,500’ only: Next try I = 25’ 30 min max 8,266’ Imax MD = 1,700’ (25.5/1.5) 2 TVD = 1,644’ 2 HD = 372’ 2

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Solution: Remaining vertical depth = 10,000-1644 = 8,356 ft. ∴ Horizontal deviation = 372+8,356 tan 25.5 = 4,358 ft. { <4500 } Approx. maximum angle = 26104 What is the size of target?

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MD = MD + MD + MD vert build hold  2 , 8 66 MD a t 2 7 = 5 , 2 00 ‘+ 8 , 1 00 ‘+cos 2 7  1 = 3,577′  8,356 MD a t 2 5.5 =2,500 1 + ,700 +cos 2 5.5 1 = 3,458′ M ∴ D 1 ≅ 3,500′

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Type II Pattern Given: KOP = 2,000 feetTVD = 10,000 feet Horiz. Depart. = 2,258 feet Build Rate = 20 per 100 feet Drop Rate = 10 30’ per 100 feet The first part of the calculation is the same as previously described.

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Procedure – Find:  a) The usable depth (8,000 feet) b) Maximum angle at completion of buildup (180)  c) Measured depth and vertical depth at completion of build up (M.D.=900 ft. and TVD = 886)  d) Measured depth, horizontal departure and TVD for 1 1 / 0 100 ft from chart. 2

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Solve:  For the distances corresponding to the sides of the triangle in the middle.  Add up the results.  If not close enough, try a different value for the maximum inclination angle, Imax

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Example 1: Design of Directional Wel (i) Determine the maximum hole angle required. (ii) What is the total measured depth (MD)? (MD = well depth measured along the wellbore, not the vert ical depth)

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(i) Maximum Inclination Angle , r = 18 000 1 15 . π r = 0 2 ( D − D 4 1 ) = 12 5 , 00 − 2 5 , 00 = 10 0 , 00 ft

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(i) Maximum Inclination Angle D D x2 (D D )2 2(r r )x  1 4 − 1 − 4 + 4 − 1 − 1 + θmax = − 2 tan  2 4   2(r1 + r ) 2 − x 4    1 , 0 000 5 , 4 002 1 , 0 0002 2 8 , 3 ( 20) 5 , 4 00  1 – − + − = 2 tan    2 8 , 3 ( 20) − 5 , 4 00     θ = 2 . 6 3 max

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(i ) Measured Depth of Wel x = r 1 ( − cosθ ) Build 1 = 3,820(1- cos 26 .3 ) = 395 f t ∴x = 5 , 4 00 − 395 Hold = 4,105 f t ∴L sinθ = 10 , 4 5 Hold ∴L = , 9 265 ft Hold

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(i ) Measured Depth of Wel MD = D + rθ + L 1 1 rad Hold  26.3π  = 2,500 + 3,820  + , 9 265  180  MD =13 5 , 18 f t

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Horizontal N View Vertical View We may plan a 2-D well, but we always get a 3D well (not all in one plane)

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MD, α , ε 1 1 ∆MD β = dogleg angle α , ε 2 2 Fig. 8-22. A curve representing a wellbore between survey stations A and A 1 2

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Directional Drilling  1. Drill the vertical (upper) section of the hole.  2. Select the proper tools for kicking off to a non-vertical direction  3. Build angle gradually

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Directional Tools  (i) Whipstock (ii) Jet Bits (iii) Downhole motor and bent sub

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Whipstocks Standard retreivable Circulating Permanent Casing

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Setting a Whipstock  Small bit used to start Apply weight to: – set chisel point &– shear pin  Drill 12’-20’ Remove whipstock Enlarge hole

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Jetting Bit  Fast and economical Small Jets  For soft formation One large – two smal nozzles  Orient large nozzle Spud periodically No rotation at first

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Jetting  Wash out pocket Return to normal dril ing  Survey Repeat for more angle if needed

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Mud Motors Drillpipe Non-magnetic Drill Collar Bent Sub Mud Motor Rotating Sub

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Increasing Inclination  Limber assembly Near bit stabilizer Weight on bit forces DC to bend to low side of hole.  Bit face kicks up

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Hold Inclination  Packed hole assembly  Stiff assembly Control bit weight and RPM

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Decrease Inclination  Pendulum effect Gravity pul s bit downward  No near bit stabilizer

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Packed Hole Assemblies Drillpipe String String String NB Stabilizer Stabilizer Stabilizer Stab Monel HW DP Steel DC Steel DC DC

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Vertical Calculation Horizontal Calculation

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3D View Dog Leg Angle

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Deflecting Wellbore Trajectory 0 270 90 180

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Bottom Hole Location Direction : 5 N 3  E Distance : 2,550 f t TVD : 10,000 E = 5 , 2 50 s in 5 3  = 2,037 f t N = 2,550 c os 5 3  =1,535 f t 2 2 Closure = 2,550 = E +N E 1 –  o Closure D irection = tan   = 53 N 

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Survey Calculation Methods 1. Tangential Method = Backward Station Method= Terminal Angle Method Assumption: Hole will maintain constant inclination and azimuth angles between survey points

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A Known : L ocationo f A D istanc e AB Ang e l s I I , I A B A Angle s A ,A A B I Calculati n o : V = ABcosI B AB B H = AB sinI AB B B Poor accuracy!! I B

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Average Angle Method = Angle Averaging Method Assumption: Borehole is parallel to the simple average drift and bearing angles between any two stations. Known: Location of A, Distance AB, Angles I I , ,A ,A A B A B

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A (i) Simple enough for field use(ii) Much more accurate than “Tangential” Method IA IB IA + Iavg = IB   I  2  AVG B  AA + Aavg = AB   2 I   AVG

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A Average Angle MethodVertical Plane: IA IA +IB  I Iavg =   B  2  IAVG V = AB cos I AB avg B H = AB sinI AB avg I AVG

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N Average Angle MethodHorizontal Plane: H = AB sin I A AB avg B B E ∆ = AB sinI sin A avg avg N A ∆ = AB sinI cos A avg avg AVG ∆N AA Z ∆ = AB cos Iavg ∆E E A

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Change in position towards the east: I +I   A + A  x ∆ = E ∆ = L sin A B  sin A B  .  .( ) 1  2   2  Change in position towards the north: I +I   A + A  y ∆ = N ∆ = L sin A B  cos A B  .  .(2)  2   2  Change in depth: I +I  Z ∆ = L cos A B   ..(3)  2  Where L is the measured distance between the two stations A & B.

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Example The coordinates of a point in a wellbore are:x = 1000 ft (easting)y = 2000 ft (northing)z = 3000 ft (depth) At this point (station) a wel bore survey shows that the inclination is 15 degrees from vertical, and the direction is 45 degrees east of north. The measured distance between this station and the next is 300 ft….

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Example The coordinates of point 1 are:x = 1000 ft (easting) 1 y = 2000 ft (northing) I = 15o 1 1 z = 3000 ft (depth) A = 45o 1 1 L = 300 ft 12 At point 2, I = 25o and A = 65o 2 2 Find x , y and z 2 2 2

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Solution I +I  15 + 25  I 1 2 =   =   = 20 avg  2   2   A + A   45 + 65  A 1 2 =   =   = 55 avg  2   2  H = L sin I = 300 sin 20 = 103 ft 12 12 avg ∆E = H sin A = 103 sin 55 = 84 ft 12 avg ∆N = H cos A = 103 cos 55 = 59 ft 12 avg ∆Z = L cos I = 300 cos 20 = 282 ft 12 avg

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Solution – cont’d ∆E = 84 ft∆N = 59 ft∆Z = 282 ft x = x + ∆E = 1,000 + 84 ft = 1,084 ft 2 1 y = y + ∆N = 2,000 + 59 ft = 2,059 ft 2 1 z = z + ∆Z = 3,000 + 282 ft = 3,282 ft 2 1

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