美国石油学校定向井课件(英文)

PetroleumEngineering406Lesson18DirectionalDrillingLesson10-DirectionalDrillingWhenisitused?TypeIWells(buildandhold)TypeIIWells(build,holdanddrop)TypeIIIWells(build)DirectionalWellPlanning&DesignSurveyCalculationMethodsHomework:READ.“AppliedDrillingEngineering”Ch.8,pp.351-363REF.APIBulletinD20,“DirectionalDrillingSurveyCalculationMethodsandTerminology”WhatisDirectionalDrilling?DirectionalDrillingistheprocessofdirectingawellborealongsometrajectorytoapredeterminedtarget.Basicallyitreferstodrillinginanon-verticaldirection.Even“vertical”holesometimesrequiredirectionaldrillingtechniques.Examples:Slantedholes,highangleholes(farfromvertical),andHorizontalholes.NorthDirectionAngleDirectionPlaneXInclinationAngleZAxis(TrueVerticalDepth)q,aorIf,eorANon-VerticalWellboreFigure8.2-Planviewofatypicaloilandgasstructureunderalakeshowinghowdirectionalwellscouldbeusedtodevelopit.Bestlocations?Drillfromlake?LeaseBoundarySurfaceLocationforWellNo.1BottomHoleLocationforWell2SurfaceLocationforWellNo.2HousesFigure8.3-Typicaloffshoredevelopmentplatformwithdirectionalwells.NOTE:AllthewellsaredirectionalTopView5-50wellsperplatformFigure8.4-Developingafieldunderacityusingdirectionallydrilledwells.DrillingRigInsideBuildingFig.8.5-Drillingofdirectionalwellswherethereservoirisbeneathamajorsurfaceobstruction.Whynotdrillfromtopofmountain?Maximumlateraldispl.?Figure8.6-Sidetrackingaroundafish.SidetrackedHoleAroundFishFishLostinHoleandUnabletoRecoverCementPlugFigure8.7-Usinganoldwelltoexplorefornewoilbysidetrackingoutofthecasinganddrillingdirectionally.PossibleNewOilSidetrackedOutofCasingOilProducingWellReadytoAbandonOldOilReservoirFigure8.8-Majortypesofwellboretrajectories.BuildandHoldTypeContinuousBuildBuild-holdDropand/orHold(Modified“S”Type)Build-holdandDrop(“SType”)TypeITypeIIITypeIIFigure8.10-Geometryofthebuildsection.BuildSectionBuildRadius:BUR*,00018r1BuildSection:degrad180*100Lr)cos(1rDD'dev.Horiz.sinrD'C'depthVerticalrLarc,ofLength11111111111πθθθθθBUR*,00018r1Build-hold-anddropforthecasewhere:42131xrrandxrTargetDropOffEndofBuildStartofBuildupTypeIIBuild-hold-anddropforthecasewhere:KickoffEndofBuildMaximumInclinationAngleDropOffTarget42131xrrandxrTypeIIFig.8-14.DirectionalwellusedtointersectmultipletargetsTarget1Target2Target3ProjectedTrajectoryProjectedTrajectorywithLeftTurntoHitTargetsFig.8-15.DirectionalquadrantsandcompassmeasurementsN18ES23EA=157oN55WA=305oS20WFigure8-16:PlanViewLeadAngleLakeSurfaceLocationforWellNo.2ProjectedWellPathTargetataTVD9,659Example1:DesignofDirectionalWellDesignadirectionalwellwiththefollowingrestrictions:Totalhorizontaldeparture=4,500ftTrueverticaldepth(TVD)=12,500ftDepthtokickoffpoint(KOP)=2,500ftRateofbuildofholeangle=1.5deg/100ftExample1:DesignofDirectionalWellThisisaTypeIwell(buildandhold)(i)Determinethemaximumholeangle(inclination)required.(ii)Whatisthetotalmeasureddepthofthehole(MD)?2500’10,000’ImaxImaxTVD14,500’12,500’TypeI:Build-and-HoldHD1Uniform1’30”IncreaseinDriftper100ftofholedrilled10,000’Vert.Depth4,500’HorizontalDeviation0’TryImax=27o??SolutionTypeIWell1.5deg/100’2500’Availabledepth=12,500-2,500=10,000’10,000’ImaxImaxFromChart,Try=27oImaxTVD1HD1BuildSectionImaxImaxTVD1HD1MD1=1,800’(27/1.5)TVD1=1,734’HD1=416’Remainingverticalheight=10,000-1,734=8,266’Fromchartof1.5deg/100’,withImax=27oIntheBUILDSection:8,266’SolutionHorizontally:416+8,266tan27o=4,628Weneed4,500’only:NexttryImax=25’30minImax8,266’MD2=1,700’(25.5/1.5)TVD2=1,644’HD2=372’Solution:Remainingverticaldepth=10,000-1644=8,356ft.Horizontaldeviation=372+8,356tan25.5=4,358ft.{4500}Approx.maximumangle=26Whatisthesizeoftarget?410MD=MDvert+MDbuild+MDhold13,500'MD13,458'25.5cos8,3561,7002,50025.5atMD13,577'27cos266,8'800,1'500,227atMDTypeIIPatternGiven:KOP=2,000feetTVD=10,000feetHoriz.Depart.=2,258feetBuildRate=20per100feetDropRate=1030’per100feetThefirstpartofthecalculationisthesameaspreviouslydescribed.Procedure-Find:a)Theusabledepth(8,000feet)b)Maximumangleatcompletionofbuildup(180)c)Measureddepthandverticaldepthatcompletionofbuildup(M.D.=900ft.andTVD=886)d)Measureddepth,horizontaldepartureandTVDfor1/100ftfromchart.021Solve:Forthedistancescorrespondingtothesidesofthetriangleinthemiddle.Adduptheresults.Ifnotcloseenough,tryadifferentvalueforthemaximuminclinationangle,ImaxExample1:DesignofDirectionalWell(i)Determinethemaximumholeanglerequired.(ii)Whatisthetotalmeasureddepth(MD)?(MD=welldepthmeasuredalongthewellbore,nottheverticaldepth)(i)MaximumInclinationAngler11800015,.0r2D4112500250010000Dft,,,(i)MaximumInclinationAngle500,4)820,3(2500,4)820,3(2000,10500,4000,10tan2x)rr(2x)rr(2)DD(xDDtan2221-42142121424141maxq3.26maxq(ii)MeasuredDepthofWellft265,9L105,4sinLft4,105395500,4xft395)26.3cos-3,820(1)cos1(rxHoldHoldHold1Buildqq(ii)MeasuredDepthofWell265,918026.33,8202,500LrDMDHoldrad11qft518,13MDWemayplana2-Dwell,butwealwaysgeta3Dwell(notallinoneplane)HorizontalVerticalViewNViewFig.8-22.Acurverepresentingawellborebetweensur