Tech Drilling Directional Drilling

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Ouija Board Calculations for directional Drilling.

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<p> 1Petroleum Engineering 406Lesson 20Directional Drilling(continued) 2Lesson 17- Directional Drilling contdTool-Face AngleOuija BoardDogleg SeverityReverse Torque of Mud MotorExamples 3Homework:READ:Applied Drilling Engineering, Chapter 8(to page 375) 4Fig. 8. 30: Graphical Ouija Analysis.SolutionTool Face () 5Fig. 8. 30: Graphical Ouija Analysis.SolutionTool Face ()Initial Inclination = 16oGIVEN: = 16o = 12o = 12o = ? o = ? oNew Inclination - 12o = 12oOver one drilled interval (bit run) 6Fig. 8.33Basis of chart construction is a trigonometric relationship illustrated by two intersecting planes = dogleg angle 7Problem 1Determine the new direction () for a whipstock set at 705 m with a tool-face setting of 450 degrees right of high side for a course length of 10 m. The inclination is 70 and the direction is N15W. The curve of the whipstock will cause a total angle change of 30/30 m. 8Problem 1 = 7o (inclination) = 345o (azimuth) = 45o (tool face angle)L = 10 m (course length) = 3o/ 30 m (dogleg severity) = ? o = 45o 9Solution to Problem 1, part 1I. Use Equation 8.43 to calculate . The dogleg severity, o1m 30m 10 3iL) i (L 10Solution to Problem 1, part 22. Use Equation 8.42 to calculate the direction change.New direction =3450 +5.30 = 350.30 = N9.7W + cos cos tan sinsin tantan arc 45 cos 7 cos 1 tan 7 sin45 sin 1 tantan1+ 3 . 5 092027 . 0 tan1 11Problem 2Determine where to set the tool face angle, for a jetting bit to go from a direction of 100 to 300 and from an inclination of 30 to 50. Also calculate the dogleg severity, assuming that the trajectory change takes 60 ft. = 3 = 10Find ft 60 L305NN and 12Solution to Problem 2, part 11. Find using Equation 8.53[ ][ ]o 1o o o o o 1N N14097 . 2 999116 . 0 cos3 cos 5 cos 3 sin 5 sin 20 cos coscos cos sin sin cos cos + + 13Solution to Problem 2, part 22. Now calculate from equation 8.48.o 115 . 45 7052 . 0 cos 1]1</p> <p> o oo o o14097 . 2 sin 3 sin5 cos 4097 . 2 cos 3 coscos1]1</p> <p> sin sincos cos coscosN1 14Solution to Problem 2, part 33. The dogleg severity, = 4.01o / 100 ftAlternate solution: Use Ouija Board 100604097 . 2(i)Lo 15Fig. 8.31: Solution to Example 8.6. 16Problem 3Determine the dogleg severity following a jetting run where the inclination was changed from 4.3o to 7.1o and the direction from N89E to S80E over a drilled interval of 85 feet.1. Solve by calculation.2. Solve using Ragland diagram L = 85 ft oNooNo100 89 7.1 3 . 4 = 7.1 - 4.3 = 2.8. = 100 - 89 = 11 17Solution to Problem 3, part 11. From Equation 8.552 / 12 2 2 121 . 7 3 . 4sin211sin28 . 2sin sin 21]1</p> <p>,_</p> <p>++ 2 / 1N2 2 2 12sin2sin2sin sin 21]1</p> <p>,_</p> <p> + + = 3.01o 18Solution to Problem 3, part 1 1. From Equation 8.43 the dogleg severity,1008501 . 3(i)L feet 100 / 5 . 3o 19Solution to Problem 3, part 22. Construct line of length (4.3o) Measure angle (11o ) Construct line of length N (7.1o) Measure length (Measure angle )4.37.111oRagland Diagram 20Some Equations to Calculate ) cos cos sin sin cos(cos arcN N + N Nsin sin cos sin sin cos cos + Eq. 8.53Eq. 8.54</p> <p>,_</p> <p> + </p> <p>,_</p> <p> +</p> <p>,_</p> <p> 2sin2sin2sin sin arc 2N2 2 2Eq. 8.55 21Overall Angle Change and Dogleg SeverityEquation 8.51 derived by Lubinski is used to construct Figure 8.32, a nomograph for determining the total angle change and the dogleg severity, . 22Fig. 8.32: Chart for determining dogleg severity 23 = 11o(+)/2 = 5.7o = 2.8o = 3o = 3.5o/100 ft 24 = 11o(+)/2 = 5.7o 25 = 2.8o = 3o 26 = 3o = 3.5o/100 ft 27 = 11o(+)/2 = 5.7o = 2.8o = 3o = 3.5o/100 ft 28Problem 4 - Torque and Twist1.Calculate the total angle change of 3,650 ft. of 4 1/2 inches (3.826 ID) Grade E 16.60 #/ft drill pipe and 300 ft. of 7 drill collars (2 13/16 ID) for a bit-generated torque of 1,000 ft-lbf. Assume that the motor has the same properties as the 7 drill collars. Shear modules of steel, G = 11.5*106 psi.2. What would be the total angle change if 7,300 ft. of drill pipe were used? 29Solution to Problem 4From Equation 8.56,g drillstrin BHA motorMGJMLGJMLGJML</p> <p>,_</p> <p>+</p> <p>,_</p> <p>+</p> <p>,_</p> <p> ( ) ( )4 4 4 4 4in 22 . 19 826 . 3 5 . 432ID OD32J Pipe, for ( )4 4 4in 6 . 229 813 . 2 732J Collars, for 30Solution to Problem 4, cont.[ ]o4 42pipe collarsM2 . 137raddeg 180rad 394 . 2radians 394 . 2 88 . 278 , 2 68 . 15 001043 . 0in 22 . 19in 12 650 , 3in 6 . 229in 12 300inlbf10 5 . 11ftin12 lbs . ft 000 , 1JLJLGM + 11]1</p> <p>+11]1</p> <p>,_</p> <p>+</p> <p>,_</p> <p> radians 31Solution to Problem 4, cont.If Length of drillpipe = 7,300 ft., M = 0.001043 [15.68+2*2278.88] = 4.77 radians * raddeg 180oM3 . 273 ~ 3/4 revolution!137.2 32Example 8.10Design a kickoff for the wellbore in Fig. 8.35. From Ouija Board, 5.8, 97</p> <p> = S48W = 228o N = N53W = 307o = 2o L = 150 ft N = 6o = 79o Find , and 33Fig. 8.36: Solution for Example 8.10.Where to Set the Tool FaceHigh SidePresent DirectionHigh SideNew Direction 5.8 97</p> <p> 34Dogleg SeverityFrom Equation 8.43 the dogleg severity,1001508 . 5(i)L feet 100 / 87 . 3o 35Fig. 8.36: Solution for Example 8.10.307oWith jetting bit: 325o345o228oM = 20o 36Fig. 8.36: Solution for Example 8.10.Where to Set the Tool FaceHigh SidePresent DirectionNew DirectionTool Face SettingCompensating for Reverse Torque of the MotorHigh Side</p>