PETE 661 Drilling Engineering

1. PETE 661Drilling Engineering Lesson 19 Kick Detection and Control

2. 2 Kick Detection and Control Primary and Secondary Well Control What Constitutes a Kick Why Kicks Occur Kick Detection Methods Kicks while Tripping

3. 3 Kick Detection and Control Shut-in Procedures Soft Shut-in Hard Shut-in Water Hammer

4. 4 Kick Detection and Control Read: ADE Ch. 6 Reference: Advanced Well Control Manual, SPE Textbook, ~2003... Homework # 11 - due November 25

5. 5 Kick Detection and Control The focus of well control theory is to contain and manage formation pressure. Primary well control involves efforts at preventing formation fluid influx into the wellbore. Secondary well control involves detecting an influx and bringing it to the surface safely.

6. 6 Kicks A kick may be defined as an unscheduled influx of formation fluids. Fluids produced during underbalanced drilling are not considered kicks Fluids produced during a DST are not considered kicks

7. 7 Kicks For a kick to occur, we need: Wellbore pressure < pore pressure A reasonable level of permeability A fluid that can flow

8. 8 Kicks Kicks may occur while: Drilling Tripping Making a connection Logging Running Casing Cementing N/U or N/D BOP, etc.

9. 9 Causes of Kicks Insufficient wellbore fluid density Low drilling or completion fluid density Reducing MW too much Drilling into abnormally pressured formations Temperature expansion of fluid Excessive gas cutting

10. 10 Causes of Kicks - cont�d Reduction of height of mud column Lost circulation because of excess static or dynamic wellbore pressure Fluid removal because of swabbing Tripping pipe without filling the hole

11. 11 Causes of Kicks - cont�d Excessive swab friction pressure while moving pipe Wellbore collision between a drilling and producing well Cement hydration

12. 12 Kick indicators Indicator Drilling break Increase in mud return rate Pit gain Flow w/ pumps off Significance Medium High High Definitive

13. 13 Kick indicators Indicator Pump pressure decrease / rate increase Increase in drillstring weight Gas cutting or salinity change Significance Low Low Low

14. 14 Kick Influx Rate This equation would rarely be strictly applicable in the event of a kick since fluid compressibility is not considered and transient relationships better describe influx flow behavior.

15. 15 Kick Influx Rate Extremely important to detect a kick early, to minimize its size. If a kick is suspected,

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19. 19 Kick size Under most conditions a 10 bbl kick can be handled safely. An exception is slimhole drilling, where even a small kick occupies a large height in the annulus. In floating drilling, where the vessel moves, small kicks are more difficult to detect

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21. 21 Acoustic kick detection

22. 22 Minimum kick size that can be detected by an acoustic system

23. 23 Delta flow indicator

24. 24 Delta flow indicator

25. 25 Delta flow indicator Field Examples of Kick Detection and Final Containment Volumes using the Delta Flow Method Hole Depth Influx Volume Volume Size ft. Rate Detected Contained in. gal/min bbl bbl

26. 26 BOP stack

27. 27 BOP Control Panel

28. 28 Choke Manifold

29. 29 Choke panel

30. 30 If a kick is suspected Lift the drillstring until a tool joint is just above the rotary table Shut down the mud pumps Check for flow

31. 31 If a kick is suspected If flowing - shut the annular, open the HCR valve, and close the choke Record SIDPP and SICP Record pit gain and depth (MD and TVD) Note the time

32. 32 Hard Shut-In Assure beforehand the choke manifold line is open to preferred choke and choke is in closed position. After a kick is indicated, hoist the string and position tool joint above rotary table. Shut off pump Observe flowline for flow.

33. 33 Hard Shut-In 5. If flow is verified, shut the well in by using annular preventer and open the remote-actuated valve to the choke manifold. 6. Notify supervisor (company drilling supervisor, toolpusher or rig manager). 7. Read and record shut-in drillpipe pressure (SIDPP).

34. 34 Hard Shut-In 8. Read and record shut-in casing pressure (SICP). 9. Rotate the drillstring though the closed annular preventer if feasible. 10. Measure and record pit gain.

35. 35 Hard Shut-In

36. 36 Soft Shut-In Assure beforehand choke manifold line is open to preferred choke and choke in in open position. After kick is indicated, hoist string & position tool joint above rotary table. Shut off pump.

37. 37 Soft Shut-In Observe flowline for flow. If flow is verified, open remote-actuated valve to choke manifold and close annular preventer. Shut well in by closing choke. Notify supervisor (company drilling supervisor, toolpusher, rig manager).

38. 38 Soft Shut-In Read and record SIDPP. Read and record SICP. Rotate drillstring through closed annular preventer if feasible. Measure and record pit gain.

39. 39 Soft Shut-In

40. 40 Example 5.1 A kick is detected while drilling at 13,000 ft. The well is shut-in by the ram preventer in 5 seconds. Determine water hammer load at surface if influx flow rate is 3.0 bbl/min, the mud�s acoustic velocity is 4,800 ft/s and mud density is 10.5 lbm/gal

41. 41 Example 5.1, continued For the same conditions: Compute velocity assuming the annulus flow area corresponds to 5.0 in. drillpipe inside 8.921 in. inner diameter casing. Ignore effect of influx properties on wave travel time and amplitude.

42. 42 Example 5.1, continued

43. 43 Example 5.1, continued The relationship is only valid if valve is fully closed before the shock wave has time to make the round trip from surface to total depth. If this condition is not met, closure is defined as �slow� as opposed to �rapid� and resultant pressure surge will be lower. Regardless of method, some pressure increase, however minor, cannot be avoided and the soft shut-in procedure may in fact be considered rapid in some cases.

44. 44 Example 5.1, cont�d Solution: The time for the pressure wave to traverse the system is ?t = dist/vel = (2)(13,000)/4,800 = 5.4 sec Hence this would be characterized as a rapid shut-in and Equation 5.2 is appropriate.

45. 45 Example 5.1 cont�d 2. The velocity change in the annulus is computed as:

46. 46 Example 5.1 cont�d The surface pressure increase is given by equation 5.2

47. 47 Off Bottom Kicks Slugging of drillpipe Hole fill during trips Surge and Swab pressures Kick detection during trips Shut-In Procedures Blowout Case History

48. 48 Off Bottom Kicks

49. 49 Failure to keep the hole full

50. 50 Nominal Dimensions-Displacement Factors for API Drillpipe Outside Nominal Nominal Average Displacement Diameter Inside Weight Approximate Factor in. Diameter, in. lbm/ft Weight bbl/ft 2-3/8 1.995 4.85 5.02 0.00182 1.815 6.65 6.80 0.00247 2-7/8 2.441 6.85 7.09 0.00258 2.151 10.40 10.53 0.00383 3-1/2 2.992 9.50 10.15 0.00369 2.764 13.30 13.86 0.00504 2.602 15.50 16.39 0.00596

51. 51 Nominal Dimensions-Displacement factors for API Drillpipe Outside Nominal Nominal Average Displacement Diameter Inside Weight Approximate Factor in. Diameter, in. lbm/ft Weight bbl/ft 4 3.476 11.85 12.90 0.00469 3.340 14.00 15.14 0.00551 3.240 15.70 17.13 0.00623 4-1/2 3.958 13.75 14.75 0.00537 3.826 16.60 17.70 0.00644 3.640 20.00 21.74 0.00791 3.500 22.82 24.33 0.00885

52. 52 Nominal Dimensions-Displacement factors for API Drillpipe Outside Nominal Nominal Average Displacement Diameter Inside Weight Approximate Factor in. Diameter, in. lbm/ft Weight bbl/ft 5 4.276 19.50 21.58 0.00785 4.000 25.60 27.58 0.01003 5-1/2 4.778 21.90 23.77 0.00865 4.670 24.70 26.33 0.00958 6-6/8 5.965 25.20 27.15 0.00988 5.901 27.70 29.06 0.01057

53. 53 Displacement Factors for High Strength Drillpipe Outside Nominal Average Displacement Diameter Weight Approximate Factor in. lbm/ft Weight, lbm/ft. bbl/ft 2-3/8 6.65 6.95 0.00253 2-7/8 10.40 11.01 0.00400 3-1/2 13.30 14.51 0.00528 15.50 17.02 0.00619 4 14.00 15.85 0.00577 15.70 17.50 0.00637 4-1/2 16.60 18.65 0.00678 20.00 22.40 0.00815 22.82 25.21 0.00917

54. 54 Displacement Factors for High Strength Drillpipe Outside Nominal Average Displacement Diameter Weight Approximate Factor in. lbm/ft Weight, lbm/ft. bbl/ft 5 19.50 22.34 0.00813 25.60 28.60 0.01040 5-1/2 21.90 25.14 0.00914 24.70 28.13 0.01023 6-5/8 25.20 28.33 0.01031 27.70 30.58 0.01112

55. 55 Displacement Factors for Heavy-Wall Drillpipe Outside Nominal Connection Approx. Displacement Diameter Inside Weight Factor in. Diameter, in. lbm/ft bbl/ft 3-1/2 2.063 NC38 23.20 0.00844 2.250 NC38 25.30 0.00920 4 2.563 NC40 29.70 0.01080 4-1/2 2.750 NC46 41.00 0.01491 5 3.00 NC50 49.30 0.01793

56. 56 Example 5.2 Drill a well to 9,500 total depth with a 10.0 lbm/gal mud. 8.097 in. ID casing has been set at 1,500 ft. Determine the hydrostatic pressure loss if ten 90 ft stands of 4 1/2 in., 16.60 lbm/ft Grade E drillpipe are pulled without filling the hole. Also determine the losses after pulling ten stands of drillpipe if the bit is plugged and after pulling one stand of 6 1/4 x 2 1/2 in drill collars.

57. 57 Example 5.2 Solution The displacement factor for open drillpipe is obtained from Table 5.5 and the displacement volume is computed as: Vd = (0.00644) (10) (90) = 5.80 bbl

58. 58 Example 5.2 To determine the drop in fluid level, we must have capacity factors for the drillpipe and annulus. These can be obtained directly from a published table or by calculation. Inside Drillpipe: Ci = 3.8262/1,029.4 = 0.1422 bbl/ft. and Inside Annulus: Cc = (8.0972 - 4.52)/1,029.4 = 0.04402 bbl/ft.

59. 59 Example 5.2 These values are only approximate since the effect of the pipe upsets and tool joints are not considered. The mud level will fall by ?h = 5.80/(0.01422 + 0.04402) = 99.6 ft. and the corresponding hydrostatic pressure loss is ?p = 99.6(10.0/19.25) = 52 psi.

60. 60 Example 5.2 Tripping out with a plugged bit implies the string is pulled wet and, if no mud falls back in the hole, the drillstring inner capacity is being evacuated along with the steel. The volume removed after pulling ten stands wet is V = Vi + Vd = (0.00644 + 0.01422)(10)(90) = 18.59 bbl (inside drillpipe + steel in drillpipe)

61. 61 Example 5.2 The mud level drop in the annulus and pressure loss are thus ?h = 18.59/0.04402 = 422.3 ft. and ?p = (422.3)(0.519) = 219 psi.

62. 62 Example 5.2 For drill collars, we compute the displacement factor and displacement volume as Cd = (6.252 - 2.52)/1,029.4 = 0.03188 bbl/ft. and Vd = (0.0318) (1)(90) = 2.87 bbl.

63. 63 Example 5.2 The pressure loss is determined in the same manner as the open drillpipe case. Ci = 2.52/1,029.4 = 0.00607 bbl/ft Ca = (8.0972- 6.252)/1,029.4 = 0.02574 bbl/ft ?h = 2.87/(0.00607 + 0.02574) = 90.2 ft and ?p = (0.519) (90.2) = 47 psi