Fracture Gradient Determination
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Transcript of Fracture Gradient Determination
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TAMU - PemexWell Control
Lesson 9B
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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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.
min_emax_e *3
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Cohesion, c = 0Angle of Internal Friction, = 30 deg.
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The Hubbert & Willis Equation
If the overburden is the maximum stress, the assumed horizontal stress is:
H = 1/3(ob - pp) + pp
Equating fracture propagation pressure to minimum stress gives
pfp = 1/3(ob - pp) + pp
min_emax_e *3
obemin_e *3
1
pobpmin p*3
1p
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The Hubbert & Willis Equation
pfp = 1/3(ob - pp) + pp
pfp = 1/3(ob + 2pp) (minimum)
pfp = 1/2(ob + 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
He = KMKve
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Matthews and Kelly
eH = KMKev
KMK = matrix stress coefficient
Including pore pressure,
H = KMK(ob - pp) + pp
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Matthews and Kelly
Equating fracture initiation pressure to the minimum in situ horizontal
stress gives
pfi = KMK(ob - 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/ftft
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Example 3.8
KMK = 0.69
At 8,110 ft, KMK = 0.69:
gfi = 0.69(1 - 0.465) + 0.465
gfi = 0.834 psi/ft
KMK = 0.61
For the undercompacted interval at 15,050 ft, the equivalent depth is determined by Eq. 3.68:
=[15,050-(0.815*15,050)]/0.535
= 5,204 ftHere KMK = 0.61
Fig. 3.38
nob
pob
nob
Veeq gg
p
ggD
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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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Pennebaker’s Gulf Coast
gfi = Kp(gob - gp) + gp
where Kp is Pennebaker’s effective stress ratio
gfi = Kp(gob - gp) + gp
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Pennebaker’s overburden
gradient from Gulf Coast
regionDepth where
t = 100 sec/ftWel
l D
epth
, f
t
Overburden Gradient, psi/ft
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Pennebaker’s Effective Stress Ratio
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Example 3.9
Re-work Example 3.8
using Pennebaker’s correlations
where the travel time of 100 sec/ft
is at 10,000 ft
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8,110
15,050
100 sec at 10,000 ft
0.77 0.940.945 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)
gfi = Kp(gob - gp) + gp
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Eaton’s Gulf Coast Correlation
Based on offshore LA in moderate water depths
ppobE
Efi ggg
1g
ratio stress effective an is
term ratio sPoisson' bracketed the Note
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From Eaton
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Mitchell’s approximation for Eaton’s Overburden Relationship for the Gulf Coast:
2
ob 000,1
D0006.0
000,1
D01494.084753.0g
a9.2...............000,1
D10*199.1
35
2ob 11.80006.011.801494.084753.0g
35 11.810*199.1
ft/psi936.0gob
Example 3.10 Estimate the fracture gradient at 8,110 ft using Eaton’s Method
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Mitchell’s approximation for Eaton’s Poisson’s Ratio for the Gulf Coast
2
E 000,1
D00668.0
000,1
D05945.023743.0
70.3.....000,1
D10*71.6
000,1
D00035.0
46
3
438.0E
2E 11.800668.011.805945.023743.0
463 11.810*71.611.800035.0
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Example 3.10 – cont’d
69.3............ggg1
g ppobE
Efi
465.0465.0936.0438.01
438.0gfi
ft/psi832.0gfi
At 15,050 ft, Michell’s approximation yields:
gob = 0.977 psi/ft, E = 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 Poisson’s 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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Christman’s 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 ftAir gap = 75 ftSea Water Gradient = 0.44 psi/ft
Assume Eaton’s overburden for the Santa Barbara Channel.
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Example 3.11 - Solution
From Fig. 3.42, next page
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1,490
0.451
Fig. 3.42- Christman’s Correlation for Santa Barbara Channel
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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
AndEaton 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 + PHYD
= 1,000 + 0.052 * 9 * 4,000
= 2,872 psi
Fracture gradient = 0.718 psi/ft
EMW = ?
ft
psi
000,4
872,2
D
P OFFLEAK
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PIT
4,000’
10.0 ppg
??
How much surface pressure will be required to test the casing seat to 14.0 ppg equivalent?
ps = 0.052 * (EMW - MW) * TVDshoe
ps = 0.052 * (14.0 - 10.0) * 4,000
ps = 832 psi
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LOT
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Leak-off
Rupture
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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Poor Cement Job
What could cause this?
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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/(0.052*1,500) + 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