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PETE 625 Well Control

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PETE 625 Well Control

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    1. PETE 625 Well Control Lesson 11 Fracture Gradient Determination

    2. 2

    3. Assignments HW # 6: Ch 3, Problems 1- 10 due Wednesday, June 23 HW # 7: Ch 3, Problems 11- 20 due Monday, June 28 Read: Chapter 3

    4. 4 Well Planning Safe drilling practices require that the following be considered when planning a well: Pore pressure determination Fracture gradient determination Casing setting depth Casing design H2S considerations Contingency planning

    5. 5

    6. 6

    7. 7 The Hubbert & Willis Equation Provides the basis of fracture theory and prediction used today. Assumes elastic behavior. Assumes the maximum effective stress exceeds the minimum by a factor of 3.

    8. 8

    9. 9 The Hubbert & Willis Equation If the overburden is the maximum stress, the assumed horizontal stress is: sH = 1/3(sob - pp) + pp Equating fracture propagation pressure to minimum stress gives pfp = 1/3(sob - pp) + pp

    10. 10 The Hubbert & Willis Equation pfp = 1/3(sob - pp) + pp pfp = 1/3(sob + 2pp) (minimum) pfp = 1/2(sob + pp) (maximum)

    11. 11 Matthews and Kelly Developed the concept of variable ratio between the effective horizontal and vertical stresses, not a constant 1/3 as in H & W. Stress ratios increase according to the degree of compaction sHe = KMKsve

    12. 12 Matthews and Kelly seH = KMKsev KMK = matrix stress coefficient Including pore pressure, sH = KMK(sob - pp) + pp

    13. 13 Matthews and Kelly Equating fracture initiation pressure to the minimum in situ horizontal stress gives pfi = KMK(sob - pp) + pp and gfi = KMK(gob - gp) + gp

    14. 14 Example 3.8 Given: Table 3.4 (Offshore LA) Estimate fracture initiation gradients at 8,110’ and 15,050’ using Matthews and Kelly correlation

    15. 15

    16. 16 Example 3.8

    17. 17 Example 3.8 At 15,050 ft, KMK = 0.61: gfi = 0.61*(1-0.815)+0.815 = 0.928 psi/ft 0.928 / 0.052 = 17.8 lb/gal! Note: Overburden gradient was assumed to be 1.0 psi/ft

    18. 18 Pennebaker’s Gulf Coast gfi = Kp(gob - gp) + gp where Kp is Pennebaker’s effective stress ratio

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    20. 20

    21. 21 Example 3.9 Re-work Example 3.8 using Pennebaker’s correlations where the travel time of 100 msec/ft is at 10,000 ft

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    23. 23 Example 3.9 At 8,110’ gfi = 0.77(0.945 - 0.465) + 0.465 gfi = 0.835 psi/ft At 15,050’ gfi = 0.94(0.984 - 0.815) + 0.815 gfi = 0.974 psi/ft (18.7 ppg)

    24. 24 Eaton’s Gulf Coast Correlation Based on offshore LA in moderate water depths

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    26. 26

    27. 27 Mitchell’s approximation for Eaton’s Overburden Relationship for the Gulf Coast:

    28. 28 Mitchell’s approximation for Eaton’s Poisson’s Ratio for the Gulf Coast

    29. 29 Example 3.10 – cont’d

    30. 30 Summary Note that all the methods take into consideration the pore pressure gradient. As the pore pressure increases, so does the fracture gradient.

    31. 31 Summary Hubbert and Willis apparently consider only the variation in pore pressure gradient… Matthews and Kelly also consider the changes in rock matrix stress coefficient and the matrix stress.

    32. 32 Summary Ben Eaton considers variation in pore pressure gradient, overburden stress, and Poisson’s ratio. It is probably the most accurate of the three.

    33. 33 Summary The last two are quite similar and yield similar results. None of the above methods considers the effect of water depth.

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    36. 36

    37. 37 Christman’s approach Christman took into consideration the effect of water depth on overburden stress.

    38. 38

    39. 39 Example 3.11 Estimate the fracture gradient for a normally pressured formation located 1,490’ BML. Water depth = 768 ft Air gap = 75 ft Sea Water Gradient = 0.44 psi/ft Assume Eaton’s overburden for the Santa Barbara Channel.

    40. 40 Example 3.11 - Solution

    41. 41

    42. 42 Example 3.12

    43. 43 Fig. 3.45 - Procedure used to determine the effective stress ratio in Example 3.12.

    44. 44

    45. 45 Experimental Determination Leak-off test, LOT, - pressure test in which we determine the amount of pressure required to initiate a fracture Pressure Integrity Test, PIT, pressure test in which we only want to determine if a formation can withstand a certain amount of pressure without fracturing.

    46. Experimental Determination of Fracture Gradient The leak-off test Run and cement casing Drill out ~ 10 ft below the casing seat Close the BOPs Pump slowly and monitor the pressure

    47. Experimental Determination of Fracture Gradient Example: In a leak-off test below the casing seat at 4,000 ft, leak-off was found to occur when the standpipe pressure was 1,000 psi. MW = 9 lb/gal. What is the fracture gradient?

    48. Example Leak-off pressure = PS + DPHYD = 1,000 + 0.052 * 9 * 4,000 = 2,872 psi Fracture gradient = 0.718 psi/ft EMW = ?

    49. 49 PIT

    50. 50 LOT

    51. 51

    52. 52

    53. 53 Example 3.21

    54. 54 Solution pfi = 1,730 + 0.483 * 5,500 - 50 1,730 psi = leak off pressure 0.483 psi/ft = mud gradient in well 5,500’ depth of casing seat 50 psi = pump pressure to break circulation pfi = 4,337 psi = 0.789 psi/ft = 15.17 ppg

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    56. 56

    57. 57 Example Surface hole is drilled to 1,500’ and pipe is set. About 20’ of new hole is drilled after cementing. The shoe needs to hold 14.0 ppg equivalent on a leak off test. Mud in the hole has a density of 9.5 ppg.

    58. 58 Example What surface pressure do we need to test to a 14.0 ppg equivalent? (14.0 - 9.5) * 0.052 * 1,500 = 351 psi

    59. 59 Example The casing seat is tested to a leak off pressure of 367 psi. What EMW did the shoe actually hold? 367/(0.052*1,500) + 9.5 EMW = 14.2 ppg

    60. 60 Example After drilling for some time, TD is now 4,500’ and the mud weight is 10.2 ppg. What is the maximum casing pressure that the casing seat can withstand without fracturing?

    61. 61 Example Max. CP = (EMW - MW) * 0.052 * TVDshoe Max. CP = (14.2 - 10.2) * .052 * 1500 Max. CP = 312 psi

    62. 62 Example Now we are at a TD of 7,500 with a mud weight of 13.7 ppg. What is the maximum CP that the shoe can withstand? Max. CP = (14.2 - 13.7) * 0.052 * 1,500 Max. CP = 39 psi

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