fracture gradient - part ii
TRANSCRIPT
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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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Fracture Gradient Determination
Read AWC Chapter 4 all
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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 andprediction used today.
Assumed elastic behavior.
Assumed effective stress exceeds the
minimum by a factor of 3.
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The Hubbert & Willis Equation
If the overburden is maximum, the assumedhorizontal stress is:
H = 1/3(ob - pp) + pp Equating fracture propagation pressure to
minimum stress gives
pfp = 1/3(ob - pp) + pp
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The Hubbert & Willis Equation
pfp = 1/3(ob - 2pp) (minimum)
pfp = 1/2(ob - pp) (maximum)
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Matthews and Kelly
Developed the concept of variable ratiobetween the effective horizontal and vertical
stresses, not a constant 1/3 as in H & W.
Stress ratios increase according to the
degree of compaction
eH = KMKev
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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 theminimum in situ horizontal stress gives
pfi = KMK(ob - pp) + pp and
gfi = KMK(gob - gp) + gp
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Example 4.8
Given: Table 4.4 (Offshore LA) Estimate fracture initiation gradient at 8110
and 15,050 using Matthews and Kelly
correlation
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Example 4.8
KMK = 0.69
For 8110
gfi = 0.69(1 - .465) + .465
gfi = 0.834 psi/ft
KMK = 0.61
For the undercompacted
interval at 15,050, the
equivalent depth is
determined by:
De=
[15050-(.815*15050)]/.535
= 5204
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Example 4.8
gfi = 0.61*(1-.815)+.815 = .928 psi/ft
Note: Overburden gradient was assumed to
be 1.0 psi/ft
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Penebakers Gulf Coast
gfi = Kp(gob - gp) + gp
where Kp
is Penebakers effective stress
ratio
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Penebakers overburden
gradient from Gulf
Coast region
Depth where
t = 100 sec/ft
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Penebakers Effective Stress Ratio
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Example 4.9
Re-work Example 4.8 using Penebakerscorrelations where the travel time of 100
sec/ft is at 10,000
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Example 4.9
At 8110 gfi = 0.77(0.945 - 0.465) + 0.465
gfi
= 0.835 psi/ft
At 15050
gfi = 0.94(0.984 - 0.815) + 0.815
gfi = 0.974 psi/ft
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Eatons Gulf Coast Correlation
Based on offshore LA in moderate waterdepths
( )
ratiostresseffectiveanis
termratiosPoisson'bracketedtheNote
1ppob
E
E
fi gggg +
=
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Mitchells approximation
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Mitchells approximation
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Mitchells approximation
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Example 4.10
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Example 4.10
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Summary
Note that all the methods take intoconsideration the pore pressure gradient.
As the pore pressure increases, so does the
fracture gradient
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Summary
Hubbert and Willis apparently consider onlythe 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 porepressure 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 yieldsimilar results.
None consider the effect of water depth.
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Christmans approach
Christman took into consideration the effectof water depth on overburden stress.
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Example 4.11
Estimate the fracture gradient for aformation located 1490 BML. Water depth
is 768, air gap is 75.
Repeat for water depth of 1500
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Example 4.11
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Example 4.11
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Christman
Christman also noted that anomalously lowfracture gradients seemed to be associated
with formation having low bulk densities
for the burial depth. He then developed thecorrelation in Fig 4.45
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Example 4.12
Re-work the first part of Example 4.11 ifthe logged bulk density at 1490 BML is
2.08 g/cc
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gfi = 0.6 * (0.73-0.452) + 0.452
gfi = 0.619 psi/ft
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Prentice method
Water depth of 1000
Total depth = 4000
Water gap = 200
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Prentice method
Convert the water depth to an equivalentsection of formation.
E.g. 1000 * 0.465 psi/ft = 465 psi
From Eatons overburden stress chart the
stress gradient at 4000 equals 0.89 psi/ft
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Prentice method
465 psi/0.89 psi/ft = 522 equivalent depth
Calculate and convert apparent fracture
gradient to actual fracture gradient
522 + 3000 = 3522 equivalent
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Prentice method
From Eatons fracture gradient chart, thegradient at 3522 = 13.92 ppg
or
Fracture pressure = 0.052 * 13.92 * 3522
= 2549 psi
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Prentice method
The effective fracture gradient from themud flow line at the drill ship deck to the
casing seat is:
2549 * 19.23/(200 + 1000 + 3000)
= 11.67 ppg
F = 2549/4200 = 0.607 psi/ft 0.607/.052 = 11.67 ppg
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Experimental Determination
Leak-off test, LOT, - pressure test in whichwe determine the amount of pressurerequired to initiate a fracture
Pressure Integrity Test, PIT, pressure test inwhich we only want to determine if a
formation can withstand a certain amount ofpressure without fracturing.
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PIT
4000
10.0 ppg
??
How much surface pressure will be
required to test the casing seat to a
14.0 ppg equivalent?
ps = (EMW - MW) * 0.052 * TVDshoe
ps = (14.0 - 10.0) * 0.052 * 4000
ps = 832 psi
LOT
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LOT
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Leak-off
Rupture
Propagation
Example 4 22 - 2
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Example 4.22 - 2Interpret the leak-off test.
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Solution
Pfi = 1730 + .483*5500 - 50 1730 psi = leak off pressure
0.483 psi/ft = mud gradient in well
5500 depth of casing seat
50 psi = pump pressure to break circulation
Pfi = 4337 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 drilledto 1500 and pipe isset. About 20 of newhole is drilled after
cementing. The shoeneeds to hole 14.0 ppgequivalent on a leakoff test. Mud in the
hole has a density of9.5 ppg.
9.5 ppg
1500
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Example
What surface pressure do we need to test toa 14.0 ppg equivalent?
(14.0 - 9.5) * .052 * 1500 = 351 psi
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Example
The casing seat is tested to a leak offpressure of 367 psi. What EMW did the
shoe actually hole?
367/.052/1500 + 9.5 = 14.2 ppg EMW
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Example
After drilling for some
time, TD is now 4500
and the mud weight is
10.2 ppg. What is the
maximum casing
pressure that the
casing seat can
withstand withoutfracturing?
10.2 ppg
1500
4500
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Example
Max. CP = (EMW - MW) * .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 7500 with a mudweight of 13.7 ppg. What is the maximum
CP that the shoe can withstand?
Max. CP = (14.2 - 13.7) * .052 * 1500
Max. CP = 39 psi