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) Solution Fig. 8. 30: Graphical Ouija Analysis.

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

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

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.

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

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

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

Problem 2 Determine 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. a = 3 e = 10 Find

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

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

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

Fig. 8.31: Solution to Example 8.6.

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 L = 85 ft Da = 7.1 - 4.3 = 2.8. De = 100 - 89 = 11

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

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

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

Some Equations to Calculate b

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, .

Fig. 8.32: Chart for determining dogleg severity

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

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

aN - a = 2.8o b = 3o

b = 3o d = 3.5o/100 ft

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

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?

Solution to Problem 4 From Equation 8.56,

Solution to Problem 4, cont. radians

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

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 From Ouija Board, b = 5.8o, g = 97o

Fig. 8.36: Solution for Example 8.10. 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.

Dogleg Severity From Equation 8.43 the dogleg severity,

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

Fig. 8.36: Solution for Example 8.10. 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.