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Petroleum Engineering 406

Petroleum Engineering 406. Lesson 20 Directional Drilling (continued). Lesson 17 - Directional Drilling cont’d. Tool-Face Angle Ouija Board Dogleg Severity Reverse Torque of Mud Motor Examples. Homework:. READ: Applied Drilling Engineering”, Chapter 8 (to page 375). Tool Face ( g ).

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Petroleum Engineering 406

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  1. Petroleum Engineering 406 Lesson 20 Directional Drilling (continued)

  2. Lesson 17- Directional Drilling cont’d • Tool-Face Angle • Ouija Board • Dogleg Severity • Reverse Torque of Mud Motor • Examples

  3. Homework: READ: • Applied Drilling Engineering”, Chapter 8 (to page 375)

  4. Tool Face (g) Solution Fig. 8. 30: Graphical Ouija Analysis.

  5. Over one drilled interval (bit run) GIVEN: a = 16o De = 12o aN = 12o Tool Face (g) Solution New Inclination - 12o De = 12o g = ? ob = ? o Initial Inclination = 16o Fig. 8. 30: Graphical Ouija Analysis.

  6. Fig. 8.33 Basis of chart construction is a trigonometric relationship illustrated by two intersecting planes b a aN b = dogleg angle De

  7. Problem 1 Determine the new direction (eN) 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.

  8. Problem 1 a = 7o (inclination) e = 345o (azimuth) g = 45o (tool face angle) L = 10 m (course length) d = 3o/ 30 m (dogleg severity) eN = ? o g = 45o

  9. Solution to Problem 1, part 1 I. Use Equation 8.43 to calculate . The dogleg severity,

  10. Solution to Problem 1, part 2 2. Use Equation 8.42 to calculate the direction change. New direction =3450 +5.30= 350.30 = N9.7W

  11. Problem 2 Determine where to set the tool face angle, for a jetting bit to go from a direction of 100to300 and from an inclination of 30 to 50. Also calculate the dogleg severity, assuming that the trajectory change takes 60 ft. a = 3 e = 10 Find

  12. Solution to Problem 2, part 1 1. Find b using Equation 8.53

  13. Solution to Problem 2, part 2 2. Now calculate from equation 8.48.

  14. Solution to Problem 2, part 3 3. The dogleg severity,  = 4.01o / 100 ft Alternate solution: Use Ouija Board

  15. Fig. 8.31: Solution to Example 8.6.

  16. Problem 3 Determine the dogleg severity following a jetting run where the inclination was changed from 4.3oto 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 Da = 7.1 - 4.3 = 2.8. De = 100 - 89 = 11

  17. Solution to Problem 3, part 1 1. From Equation 8.55 b = 3.01o

  18. Solution to Problem 3, part 1 1. From Equation 8.43 the dogleg severity,

  19. Solution to Problem 3, part 2 2. Construct line of length  (4.3o) Measure angle  (11o ) Construct line of length N (7.1o) Measure length  (Measure angle ) 4.3 11o b Ragland Diagram 7.1

  20. Some Equations to Calculate b Eq. 8.53 Eq. 8.54 Eq. 8.55

  21. Overall Angle Change and Dogleg Severity Equation 8.51 derived by Lubinski is used to construct Figure 8.32, a nomograph for determining the total angle change and the dogleg severity, .

  22. Fig. 8.32: Chart for determining dogleg severity

  23. (a+aN)/2 = 5.7o aN - a = 2.8o b = 3o De = 11o d = 3.5o/100 ft

  24. (a+aN)/2 = 5.7o De = 11o

  25. aN - a = 2.8o b = 3o

  26. b = 3o d = 3.5o/100 ft

  27. (a+aN)/2 = 5.7o aN - a = 2.8o b = 3o De = 11o d = 3.5o/100 ft

  28. Problem 4 - Torque and Twist 1. 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?

  29. Solution to Problem 4 From Equation 8.56,

  30. Solution to Problem 4, cont. radians

  31. Solution to Problem 4, cont. If Length ofdrillpipe = 7,300 ft., M =0.001043 15.68+2*2278.88] = 4.77 radians * ~ 3/4 revolution! 137.2

  32. Example 8.10 Design a kickoff for the wellbore in Fig. 8.35. e = S48W = 228oeN = N53W = 307o a = 2oL = 150 ft aN = 6o De = 79oFind b, g and d From Ouija Board, b = 5.8o, g = 97o

  33. New Direction Where to Set the Tool Face b = 5.8o g = 97o High Side High Side Present Direction Fig. 8.36: Solution for Example 8.10.

  34. Dogleg Severity From Equation 8.43 the dogleg severity,

  35. With jetting bit: 325o 345o qM = 20o 307o Fig. 8.36: Solution for Example 8.10. 228o

  36. Tool Face Setting Where to Set the Tool Face Compensating for Reverse Torque of the Motor New Direction Present Direction High Side High Side Fig. 8.36: Solution for Example 8.10.

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