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Petroleum Engineering 406 Lesson 20 Directional Drilling (continued) 1 Lesson 17 - Directional Drilling cont’d Tool-Face Angle Ouija Board Dogleg Severity Reverse Torque of Mud Motor Examples 2 Homework: READ: Applied Drilling Engineering”, Chapter 8 (to page 375) 3 Solution Tool Face (g) Fig. 8. 30: Graphical Ouija Analysis. 4 Over one drilled interval (bit run) GIVEN: a = 16 o Solution Tool Face (g) De = 12 o aN = 12 o De = 12o o Initial Inclination = 16 g=? o b=? o Fig. 8. 30: Graphical Ouija Analysis. 5 Fig. 8.33 b a aN De Basis of chart construction is a trigonometric relationship illustrated by two intersecting planes b = dogleg angle 6 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. 7 Problem 1 a=7 o e = 345 o (inclination) (azimuth) g = 45 o g = 45 (tool face angle) L = 10 m (course length) o d = 3 / 30 m (dogleg severity) o eN = ? o 8 Solution to Problem 1, part 1 I. Use Equation 8.43 to calculate b . The dogleg severity, b d (i) L d L 3 10 m o b 1 i 30 m 9 Solution to Problem 1, part 2 2. Use Equation 8.42 to calculate the direction change. De arc tan tan b sin g sin a tan b cos a cos g De tan tan 1 sin 45 1 sin 7 tan 1 cos 7 cos 45 De tan 1 0.092027 5.3 New direction =3450 +5.30 = 350.30 = N9.7W 10 Problem 2 Determine where to set the tool face angle, g 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. a=3 aN 5 e = 10 e Find g and d e N 30 L 60 ft 11 Solution to Problem 2, part 1 1. Find b using Equation 8.53 b cos 1cos De sin aN sin a cos aN cos a 1 cos cos 20 sin 5 sin 3 cos 5 cos 3 o o o o o cos 1 0.999116 2.4097o 12 Solution to Problem 2, part 2 2. Now calculate g from equation 8.48. g cos g cos 1 1 cos a cos b cos aN sin a sin b cos 3o cos 2.4097o cos 5o o o sin 3 sin 2.4097 g cos 1 0.7052 45.15 o 13 Solution to Problem 2, part 3 3. The dogleg severity, b 2.4097 d (i) 100 L 60 o d = 4.01o / 100 ft Alternate solution: Use Ouija Board 14 Fig. 8.31: Solution to Example 8.6. 15 Problem 3 Determine 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 a 4.3o e 89 o aN 7.1o L = 85 ft Da = 7.1 - 4.3 = 2.8. eN 100 De = 100 - 89 = 11 o 16 Solution to Problem 3, part 1 1. From Equation 8.55 2 Da 2 De 2 a aN b 2 sin sin sin sin 2 2 2 1/ 2 2 2.8 2 11 2 4.3 7.1 b 2 sin sin sin sin 2 2 2 1/ 2 1 1 b = 3.01 o 17 Solution to Problem 3, part 1 1. From Equation 8.43 the dogleg severity, b 3.01 d (i) 100 L 85 d 3.5 / 100 feet o 18 Solution to Problem 3, part 2 2. Construct line of length a (4.3o) Measure angle De (11o ) Construct line of length aN (7.1o) Measure length b 4.3 (Measure angle g) o b 11 Ragland Diagram 7.1 19 Some Equations to Calculate b Eq. 8.53 b arc cos(cos De sin aN sin a cos a cos aN ) Eq. 8.54 cosb cosDe sinaN sin a cos Da sin a sin aN Eq. 8.55 Da 2 De 2 a aN b 2 arc sin sin sin sin 2 2 2 2 20 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 b and the dogleg severity, d. 21 Fig. 8.32: Chart for determining dogleg severity 22 (aaN)/2 = 5.7 o aN a = 2.8 o b=3 o De = 11 o d = 3.5 /100 ft o 23 (aaN)/2 o = 5.7 De = 11 o 24 aN a = 2.8 o b=3 o 25 b=3 o d = 3.5 /100 ft o 26 (aaN)/2 = 5.7 o aN a = 2.8 o b=3 o De = 11 o d = 3.5 /100 ft o 27 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 bitgenerated 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? 28 Solution to Problem 4 From Equation 8.56, ML ML ML M GJ motor GJ BHA GJ drillstring ( ) ( ) 4 4 for Pipe, J OD ID 4.5 4 3.826 4 19.22 in4 32 32 ( ) 4 for Collars, J 7 2.813 4 229.6 in4 32 29 Solution to Problem 4, cont. M L L M G J collars J pipe in 1,000 ft.lbs 12 ft lbf 11.5 10 2 in radians 300 12in 3,650 12 in 4 4 19.22 in 229.6 in 0.001043 15.68 2,278.88 2.394 radians 180 deg 2.394 rad 137.2o rad 30 Solution to Problem 4, cont. If Length of drillpipe = 7,300 ft., M = 0.001043 15.68+2*2278.88] = 4.77 radians * M 273.3 o 180 deg rad ~ 3/4 revolution! 137.2 31 Example 8.10 Design a kickoff for the wellbore in Fig. 8.35. e = S48W = 228o eN = N53W = 307o a = 2o L = 150 ft aN = 6o De = 79o Find b, g and d o From Ouija Board, b 5.8 , g 97 o 32 New Direction Where to Set the Tool Face o b 5.8 o g 97 High Side Present Direction High Side Fig. 8.36: Solution for Example 8.10. 33 Dogleg Severity From Equation 8.43 the dogleg severity, b 5.8 d (i) 100 L 150 d 3.87 / 100 feet o 34 o o With jetting bit: 325 345 o M = 20 o 307 Fig. 8.36: Solution for Example 8.10. o 228 35 Tool Face Setting New Direction High Side Where to Set the Tool Face Compensating for Reverse Torque of the Motor Present Direction High Side Fig. 8.36: Solution for Example 8.10. 36