石油工程定向钻井专业讲座DirectionalDrillingLesson.ppt

Petroleum Engineering 406Lesson 18 Directional DrillingLesson 10 - Directional Drilling4 When is it used?4 Type I Wells (build and hold)4 Type II Wells (build, hold and drop)4 Type III Wells (build)4 Directional Well Planning & Design4 Survey Calculation Methods Homework:READ. “Applied Drilling Engineering”Ch. 8, pp. 351-363REF. 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-vertical direction. Even “vertical” hole sometimes require directional drilling techniques.Examples: Slanted holes, high angle holes (far from vertical), and Horizontal holes.NorthDirection AngleDirection Plane XInclination AngleZ Axis (True Vertical Depth)Inclination Plane Yq, aq, a or If, ef, e or ANon-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 BoundarySurface Location for Well No. 1Bottom Hole Location for Well 2Surface Location for Well No. 2HousesOil-Water ContactFigure 8.3 - Typical offshore development platform with directional wells.NOTE: All the wells are directionalTop View5 - 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 directional 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 FishFish Lost in Hole and Unable to RecoverCement PlugFigure 8.7 - Using an old well to explore for new oil by sidetracking out of the casing and drilling directionally.Possible New OilSidetracked Out of CasingOil Producing Well Ready to AbandonOld Oil ReservoirFigure 8.8 - Major types of wellbore trajectories. Build and Hold TypeContinuous BuildBuild-hold Drop and/or Hold (Modified “S” Type)Build-hold and Drop (“S Type”)Horizontal Departure to TargetType I Type III Type II Figure 8.10 - Geometry of the build section.Build SectionBuild Radius:Build Section:Build-hold-and drop for the case where:TargetDrop OffEnd of BuildStart of BuildupType II Build-hold-and drop for the case where:KickoffEnd of BuildMaximum Inclination AngleDrop OffTargetType II Fig. 8-14. Directional well used to intersect multiple targetsTarget 1Target 2Target 3Projected TrajectoryProjected Trajectory with Left Turn to Hit TargetsFig. 8-15. Directional quadrants and compass measurementsN18ES23EA = 157oN55WA = 305oS20WFigure 8-16: Plan ViewLead AngleLakeSurface Location for Well No. 2Projected Well PathTarget at a TVD 9,659Example 1: Design of Directional WellDesign a directional well with the following restrictions:4Total horizontal departure = 4,500 ft4True vertical depth (TVD) = 12,500 ft4Depth to kickoff point (KOP) = 2,500 ft4Rate of build of hole angle = 1.5 deg/100 ftExample 1: Design of Directional Well4This 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)?2500’10,000’ImaxImaxTVD14,500’12,500’Type I: Build-and-HoldHD1Uniform 1’30” Increase in Drift per 100 ft of hole drilled10,000’Vert.Depth4,500’ Horizontal Deviation0’Try Imax = 27o ??SolutionType I Well 1.5 deg/100’2500’Available depth = 12,500-2,500 = 10,000’10,000’ImaxImaxFrom Chart,Try = 27oImaxTVD1HD1Build SectionImaxImaxTVD1HD1MD1 = 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’SolutionHorizontally:416 + 8,266 tan 27o = 4,628 We need 4,500’ only: Next try Imax = 25’ 30 minImax8,266’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 PatternGiven: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.Procedure - Find:4a) The usable depth (8,000 feet)4b) Maximum angle at completion of buildup (180)4c) Measured depth and vertical depth at completion of build up (M.D.=900 ft. and TVD = 886)4d) Measured depth, horizontal departure and TVD for 1 /100 ft from chart.Solve:4For the distances corresponding to the sides of the triangle in the middle.4Add up the results.4If not close enough, try a 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 Inclination 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)HorizontalVerticalViewNViewFig. 8-22. A curve representing a wellbore between survey stations A1 and A2MD, a a1 1, e, e1 1D DMDa a2 2, e, e2 2b b = dogleg angleDirectional Drilling41. Drill the vertical (upper) section of the hole.42. Select the proper tools for kicking off to a non-vertical direction43. Build angle graduallyDirectional Tools4(i) Whipstock4(ii) Jet Bits4(iii) Downhole motor and bent subWhipstocksStandard retreivable CirculatingPermanent CasingSetting a Whipstock4Small bit used to start4Apply weight to:–set chisel point &–shear pin4Drill 12’-20’4Remove whipstock4Enlarge holeJetting Bit4Fast and economical4For soft formation4One large - two small nozzles4Orient large nozzle4Spud periodically4No rotation at firstSmall JetsJetting4Wash out pocket4Return to normal drilling4Survey4Repeat for more angle if neededMud MotorsDrillpipeNon-magnetic Drill CollarBent SubMud MotorRotating SubIncreasing Inclination4Limber assembly4Near bit stabilizer4Weight on bit forces DC to bend to low side of hole.4Bit face kicks upHold Inclination4Packed hole assembly4Stiff assembly4Control bit weight and RPMDecrease Inclination4Pendulum effect4Gravity pulls bit downward4No near bit stabilizerPacked Hole AssembliesDrillpipeHW DPString StabilizerSteel DCString StabilizerString StabilizerMonelDCSteel DCNB StabVertical CalculationHorizontal Calculation3D ViewDog Leg AngleDeflecting Wellbore Trajectory090180270Bottom Hole LocationSurvey Calculation Methods1. Tangential Method= Backward Station Method= Terminal Angle Method4Assumption: Hole will maintain constant inclination and azimuth angles between survey pointsPoor accuracy!!ABIAIBIBAverage Angle Method = Angle Averaging MethodAssumption: 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” MethodABIAIBIAVGIAVGAverage Angle MethodVertical Plane:ABIAIBIAVGIAVGAverage Angle MethodHorizontal Plane:NBAAABAAVGED DED DNA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.ExampleThe 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 station and the next is 300 ft….ExampleThe coordinates of point 1 are:x1 = 1000 ft (easting)y1 = 2000 ft (northing) I1 = 15oz1 = 3000 ft (depth) A1 = 45o L12 = 300 ft At point 2, I2 = 25o and A2 = 65oFind x2 , y2 and z2SolutionH12 = L12 sin Iavg = 300 sin 20 = 103 ftDE = H12 sin Aavg = 103 sin 55 = 84 ftDN = H12 cos Aavg = 103 cos 55 = 59 ftDZ = L12 cos Iavg = 300 cos 20 = 282 ftSolution - cont’dDE = 84 ftDN = 59 ftDZ = 282 ftx2 = x1 + DE = 1,000 + 84 ft = 1,084 fty2 = y1 + DN = 2,000 + 59 ft = 2,059 ftz2 = z1 + DZ = 3,000 + 282 ft = 3,282 ft。