《石油工程定向钻井专业讲座DirectionalDrillingLesson.ppt》由会员分享，可在线阅读，更多相关《石油工程定向钻井专业讲座DirectionalDrillingLesson.ppt（66页珍藏版）》请在三一文库上搜索。
1、Petroleum Engineering 406 Lesson 18 Directional Drilling,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,Homework:,READ. “Applied Drilling Enginee
2、ring” Ch. 8, pp. 351-363 REF. API Bulletin D20, “Directional Drilling Survey Calculation Methods and Terminology”,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-vertic
3、al direction. Even “vertical” hole sometimes require directional drilling techniques.,Examples: Slanted holes, high angle holes (far from vertical), and Horizontal holes.,North,Direction Angle,Direction Plane X,Inclination Angle,Z Axis (True Vertical Depth),Inclination Plane Y,q, a or I,f, e or A,No
4、n-Vertical Wellbore,Figure 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?,Lease Boundary,Surface Location for Well No. 1,Bottom Hole Location for Well 2,Surface Location for Well No. 2,Houses
5、,Oil-Water Contact,Figure 8.3 - Typical offshore development platform with directional wells.,NOTE: All the wells are directional,Top View,5 - 50 wells per platform,Figure 8.4 - Developing a field under a city using directionally drilled wells.,Drilling Rig Inside Building,Fig. 8.5 - Drilling of dir
6、ectional wells where the reservoir is beneath a major surface obstruction.,Why not drill from top of mountain?,Maximum lateral displ.?,Figure 8.6 - Sidetracking around a fish.,Sidetracked Hole Around Fish,Fish Lost in Hole and Unable to Recover,Cement Plug,Figure 8.7 - Using an old well to explore f
7、or new oil by sidetracking out of the casing and drilling directionally.,Possible New Oil,Sidetracked Out of Casing,Oil Producing Well Ready to Abandon,Old Oil Reservoir,Figure 8.8 - Major types of wellbore trajectories.,Build and Hold Type,Continuous Build,Build-hold Drop and/or Hold (Modified “S”
8、Type),Build-hold and Drop (“S Type”),Horizontal Departure to Target,Type I,Type III,Type II,Figure 8.10 - Geometry of the build section.,Build Section,Build Radius:,Build Section:,Build-hold-and drop for the case where:,Target,Drop Off,End of Build,Start of Buildup,Type II,Build-hold-and drop for th
9、e case where:,Kickoff,End of Build,Maximum Inclination Angle,Drop Off,Target,Type II,Fig. 8-14. Directional well used to intersect multiple targets,Target 1,Target 2,Target 3,Projected Trajectory,Projected Trajectory with Left Turn to Hit Targets,Fig. 8-15. Directional quadrants and compass measurem
10、ents,N18E,S23E A = 157o,N55W A = 305o,S20W,Figure 8-16: Plan View,Lead Angle,Lake,Surface Location for Well No. 2,Projected Well Path,Target at a TVD 9,659,Example 1: Design of Directional Well,Design a directional well with the following restrictions: Total horizontal departure = 4,500 ft True vert
11、ical depth (TVD) = 12,500 ft Depth to kickoff point (KOP) = 2,500 ft Rate of build of hole angle = 1.5 deg/100 ft,Example 1: Design of Directional Well,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 hol
12、e (MD)?,2500,10,000,Imax,Imax,TVD1,4,500,12,500,Type I: Build-and-Hold,HD1,Uniform 130” Increase in Drift per 100 ft of hole drilled,10,000 Vert. Depth,4,500 Horizontal Deviation,0,Try Imax = 27o ?,Solution,Type I Well 1.5 deg/100,2500,Available depth = 12,500-2,500 = 10,000,10,000,Imax,Imax,From Ch
13、art, Try = 27o,Imax,TVD1,HD1,Build Section,Imax,Imax,TVD1,HD1,MD1 = 1,800 (27/1.5) TVD1 = 1,734 HD1 = 416 Remaining vertical height = 10,000 - 1,734 = 8,266,From chart of 1.5 deg/100, with Imax = 27o In the BUILD Section:,8,266,Solution,Horizontally: 416 + 8,266 tan 27o = 4,628 We need 4,500 only: N
14、ext try Imax = 25 30 min,Imax,8,266,MD2 = 1,700 (25.5/1.5) TVD2 = 1,644 HD2 = 372,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 = 26 What is the size of target?,MD = MDvert + MDbuild + MDhold,Type II Patt
15、ern,Given: KOP = 2,000 feet TVD = 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.,Procedure - Find:,a) The usable depth (8,000 feet) b) Maximum angle at completion of buildup (1
16、80) 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 /100 ft from chart.,Solve:,For the distances corresponding to the sides of the triangle in the middle. Add up the results. If not close enough, try a
17、different value for the maximum inclination angle, Imax,Example 1: Design of Directional Well,(i) Determine the maximum hole angle required. (ii) What is the total measured depth (MD)? (MD = well depth measured along the wellbore, not the vertical depth),(i) Maximum Inclination Angle,(i) Maximum Inc
18、lination Angle,(ii) Measured Depth of Well,(ii) Measured Depth of Well,We may plan a 2-D well, but we always get a 3D well (not all in one plane),Horizontal,Vertical,View,N,View,Fig. 8-22. A curve representing a wellbore between survey stations A1 and A2,MD, a1, e1,DMD,a2, e2,b = dogleg angle,Direct
19、ional 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,Directional Tools,(i) Whipstock (ii) Jet Bits (iii) Downhole motor and bent sub,Whipstocks,Standard retreivable Circulating Permanent Casi
20、ng,Setting a Whipstock,Small bit used to start Apply weight to: set chisel point & shear pin Drill 12-20 Remove whipstock Enlarge hole,Jetting Bit,Fast and economical For soft formation One large - two small nozzles Orient large nozzle Spud periodically No rotation at first,Small Jets,Jetting,Wash o
21、ut pocket Return to normal drilling Survey Repeat for more angle if needed,Mud Motors,Drillpipe,Non-magnetic Drill Collar,Bent Sub,Mud Motor,Rotating Sub,Increasing Inclination,Limber assembly Near bit stabilizer Weight on bit forces DC to bend to low side of hole. Bit face kicks up,Hold Inclination
22、,Packed hole assembly Stiff assembly Control bit weight and RPM,Decrease Inclination,Pendulum effect Gravity pulls bit downward No near bit stabilizer,Packed Hole Assemblies,Drillpipe,HW DP,String Stabilizer,Steel DC,String Stabilizer,String Stabilizer,MonelDC,Steel DC,NB Stab,Vertical Calculation,H
23、orizontal Calculation,3D View,Dog Leg Angle,Deflecting Wellbore Trajectory,0,90,180,270,Bottom Hole Location,Survey Calculation Methods,1. Tangential Method = Backward Station Method = Terminal Angle Method Assumption: Hole will maintain constant inclination and azimuth angles between survey points,
24、Poor accuracy!,A,B,IA,IB,IB,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) Simple enough for field use (ii) Much more accurate than “Tangential” Meth
25、od,A,B,IA,IB,IAVG,IAVG,Average Angle Method Vertical Plane:,A,B,IA,IB,IAVG,IAVG,Average Angle Method Horizontal Plane:,N,B,AA,AB,AAVG,E,DE,DN,A,Change in position towards the east: Change in position towards the north:,Change in depth:,Where L is the measured distance between the two stations A & B.
26、,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 wellbore survey shows that the inclination is 15 degrees from vertical, and the direction is 45 degrees east of north. The measured distance between this
27、station and the next is 300 ft.,Example,The coordinates of point 1 are: x1 = 1000 ft (easting) y1 = 2000 ft (northing) I1 = 15o z1 = 3000 ft (depth) A1 = 45o L12 = 300 ft At point 2, I2 = 25o and A2 = 65o Find x2 , y2 and z2,Solution,H12 = L12 sin Iavg = 300 sin 20 = 103 ft DE = H12 sin Aavg = 103 sin 55 = 84 ft DN = H12 cos Aavg = 103 cos 55 = 59 ft DZ = L12 cos Iavg = 300 cos 20 = 282 ft,Solution - contd,DE = 84 ft DN = 59 ft DZ = 282 ft x2 = x1 + DE = 1,000 + 84 ft = 1,084 ft y2 = y1 + DN = 2,000 + 59 ft = 2,059 ft z2 = z1 + DZ = 3,000 + 282 ft = 3,282 ft,