PETE 625 Well Control

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PETE 625Well Control

• Lesson 11
• Fracture Gradient Determination

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Fracture Gradient Determination

• Hubbert and Willis
• Matthews and Kelly
• Ben Eaton
• Christman
• Prentice
• Leak-Off Test (experimental)

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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

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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

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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.

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Cohesion, c 0 Angle of Internal Friction, w
30 deg.
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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

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The Hubbert Willis Equation

• pfp 1/3(sob - pp) pp
• pfp 1/3(sob 2pp) (minimum)
• pfp 1/2(sob pp) (maximum)

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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

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Matthews and Kelly

• seH KMKsev
• KMK matrix stress coefficient
• Including pore pressure,
• sH KMK(sob - pp) pp

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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

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Example 3.8

• Given Table 3.4 (Offshore LA)
• Estimate fracture initiation gradients at
  8,110 and 15,050 using Matthews and Kelly
  correlation

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TABLE 3.4
psi/ft
ft
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Example 3.8
Fig. 3.38
At 8,110 ft, KMK 0.69 gfi 0.69(1 - 0.465)
0.465 gfi 0.834 psi/ft
For the undercompacted interval at 15,050 ft, the
equivalent depth is determined by Eq.
3.68 15,050-(0.81515,050)/0.535 5,204 ft
KMK 0.69
KMK 0.61
Here KMK 0.61
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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

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Pennebakers Gulf Coast

• gfi Kp(gob - gp) gp
• where Kp is Pennebakers effective stress ratio

gfi Kp(gob - gp) gp
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Pennebakers overburden gradient from Gulf Coast
region
Depth where Dt 100 msec/ft
Well Depth, ft
Overburden Gradient, psi/ft
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Pennebakers Effective Stress Ratio
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Example 3.9

• Re-work Example 3.8 using Pennebakers
  correlations where the travel time of 100
  msec/ft is at 10,000 ft

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100 msec at 10,000 ft
8,110
15,050
0.77
0.94
0.945
0.984
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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)

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Eatons Gulf Coast Correlation

• Based on offshore LA in moderate water depths

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From Eaton
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Mitchells approximation for Eatons Overburden
Relationship for the Gulf Coast
Example 3.10 Estimate the fracture gradient at
8,110 ft using Eatons Method
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Mitchells approximation for Eatons Poissons
Ratio for the Gulf Coast
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Example 3.10 contd
At 15,050 ft, Michells approximation yields
gob 0.977 psi/ft, mE 0.468 and gfi 0.958
psi/ft.
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Summary

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

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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.

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Summary

• Ben Eaton considers variation in pore pressure
  gradient, overburden stress, and Poissons
  ratio.
• It is probably the most accurate of the three.

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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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Christmans approach

• Christman took into consideration the effect of
  water depth on overburden stress.

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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 Eatons overburden for the Santa Barbara
  Channel.

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Example 3.11 - Solution
From Fig. 3.42, next page
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Fig. 3.42- Christmans Correlation for Santa
Barbara Channel
1,490
0.451
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Example 3.12
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Fig. 3.45 - Procedure used to determine the
effective stress ratio in Example 3.12.
Effective stress ratio
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From Barker and Wood And Eaton and Eaton
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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.

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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

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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?

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Example

• Leak-off pressure PS DPHYD
• 1,000 0.052 9 4,000
• 2,872 psi
• Fracture gradient 0.718 psi/ft
• EMW ?

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PIT
??
How much surface pressure will be required to
test the casing seat to 14.0 ppg equivalent?
10.0 ppg
ps 0.052 (EMW - MW) TVDshoe ps 0.052
(14.0 - 10.0) 4,000 ps 832 psi
4,000
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LOT
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Rupture
Leak-off
Propagation
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Example 3.21
Interpret the leak-off test.
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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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What could cause this?
Poor Cement Job
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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.

9.5 ppg
1,500
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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

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Example

• The casing seat is tested to a leak off
  pressure of 367 psi. What EMW did the shoe
  actually hold?
• 367/(0.0521,500) 9.5
• EMW 14.2 ppg

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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?

10.2 ppg
1,500
4,500
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Example

• Max. CP (EMW - MW) 0.052 TVDshoe
• Max. CP (14.2 - 10.2) .052 1500
• Max. CP 312 psi

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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