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TAMU - PemexWell Control
Lesson 9A
Fracture Gradients
- contd
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Fig
. 3.21 - Stress concentrations around a borehole
in a uniform stress field
Tension
Additional
compression
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5
Example 3.4
Given:
Formation depth = 10,000
Poissons ratio = 0.22 Pore pressure grad = 0.433 psi/ft
Hole diameter = 9.0 in
Mud density = 8.33 ppg
Overburden stress grad. = 1.00 psi/ft
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Example 3.4 - Part 1
1. Estimate the horizontal stress if therock behaves in an elastic manner
Solution:
H = ( /(1- ))*( ob - pp) + pp .Eq 3.21b
H = (0.22/(1 - 0.22)) * (1*10,000 -
0.433*10,000) + 4,330
H = 5,929 psi
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Example 3.4 Part 1 - contd
Effective horizontal stress
eH = H - *pp
If = 1.0 and pp = 4,330 psig
Then, He = 5,929 4,330
eH = 1,599 psi
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Example 3.4 Part 2
2. Estimate the tangential and radial stresses at theborehole wall if the horizontal stresses are equal.
r = pw Eq. 3.24
r = 4,330 psig
t= H1 + H2 - pw - 2( H1 - H2 ) cos2 ...Eq.
3.25
= 5,929 + 5,929 - 4,330 - 0 = 7,528 psig
rpw
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Example 3.4 Part 2 contd
The corresponding effective stresses
are:
re = 4,330 - 4,330 = 0 psi
te = 7,528 - 4,330 = 3,198 psi
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Example 3.4- Part 33. What are the radial and tangential stresses
2.0 ft from the wellbore centerline?
)psig929,5away,(far
psig985,5330,40.24
5.4
0.24
5.41929,5
pr
r
r
r1
H
2
2
2
2
ft2,t
w2
2
w
2
2
wHft2,t
=
=
+=
+=
psig873,5330,40.24
5.4
0.24
5.4
1929,5
pr
r
r
r1
2
2
2
2
ft2,r
w2
2w
2
2w
Hft2,r
=+
=
+
= Radial Stress: Eq. 3.31
Tangential Stress: Eq. 3.32
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Example 3.4 Part 4
4. Estimate the tangential and radialstresses at the borehole wall if 0.5 ppg
overbalance is provided by the mud
column.
pw = 4,330 + 0.052 * 0.5 * 10,000
pw = 4,590 psig = r
t = H1 + H2 - pw -2( H1 - H2 )cos2
= (2 * 5,929) - 4,590 = 7,268 psig
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Example 3.4 Part 4 contd
(0.5 lb/gal overbalance)
The corresponding effective stresses
are:
re = 4,590 - 4,330 = 260 psi
te = 7,268 - 4,330 = 2,938 psi
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Example 3.4 Part 5
( He1 = 3 * He2 )
5. What are the maximum and minimum
tangential stresses if
He2 =1,559 psi
He1 = 3 * 1,599 = 4,797 psi?
H1(max) = He1 + pp = 4,797 + 4,330
= 9,127 psig
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Example 3.4 Part 5 contt
t,max = 3 H1 - H2- pw
= 3 * 9,127 - (1,599 + 4,330) - 4,330
= 17,122 psig
t,min
= 3 H2
- H1-
pw
= 3 * 5,929 - 9,127 - 4,330
= 4,330 psig
Eq. 3.26
Eq. 3.27
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te,max = 17,122- 4,330= 12,792 psi
te,min = 4,330 4,330= 0 psi
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From example 3.4
The effective radial stress will be zero if
the wellbore and formation pressures are
balanced
The effective radial stress will increase in
compression as the wellbore pressureincreases
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From example 3.4
An underbalanced situation will lead to
inward tensile stress which may tend to
destabilize the walls of the hole.
A positive differential pressure reduces
the induced circumferential stress, whilea negative differential pressure
increases the compression.
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From example 3.4
In part (2) (isotropic), a wellbore
pressure in excess of 7,528 psi will
place the rock grains in tension and afracture will initiate if the tensile strength
is nil.
Being higher than the far field stress, this
fracture will propagate out into the rock
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From example 3.4
An effective horizontal stress ratio of
three exactly reduces the minimum
effective tangential stress to zero.
Any borehole pressure slightly over and
above the pore pressure in this case
would place the wellbore in tension and
likely create a fracture. No extension!
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Stress anisotropies
Borehole caliper logs
may help in determining
significant stress
anisotropies.
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Example 3.5 Part 1 - fractures
Eatons overburdent gradient at12,000 ft = 0.961 psi/ft. ts = 0
( )D
gggg tsppobfi+
=
1
2Fracture Initiation
Eq. 3.41
( ) ftpsigfi /698.00465.0465.0961.019.01
19.0*2 =
( ) ppobfp gggg + = 1( ) ftpsigfp /581.0465.0465.0961.0
19.01
19.0 =
Fracture Propagation
Eq. 3.42
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Example 3.5 Part 2: gp = 0.779 psi/ft
At 12,000 ft, gob = 0.961 psi/ft. ts = 0
( )D
gggg tsppobfi+
= 12 Fracture Initiation
Eq. 3.41
( ) ft/psi864.00779.0779.0961.019.01
19.0*2gfi =++
=
( ) ppobfp gggg + = 1( ) ftpsigfp /822.0779.0779.0961.0
19.01
19.0 =
Fracture Propagation
Eq. 3.42
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Example 3.5 Part 3:
gfi at 500 ft = ?
At 500 ft, gob = 0.855 psi/ft. (Eaton)
Shallow, so the fracture gradient is that
required to lift the overburden
ftpsiggg obfpfi /855.0
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Example 3.5 Part 4: ts = 300 psig
Impact of tensile strength on gfi is what?
At 500 ft,
Predicted breakdown pressure at 500 ft is ~ 1.5psi/ft! Lifting the overburden is probably easier
At 12,000 ft:
( )D
gggg tsppobfi+
= 12
ftpsiD
ts /025.0000,12
300 =
ftpsiDts /60.0
500
300 = ft/psi46.1gfi =
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Offshore/On Land
Higher overburden gradient
Higher fracture gradientLower overburden gradient
Lower fracture gradient
See Eatons paper
gob = (psw+ s)/D= (psw+ s)/(Da+Dsw+Ds)
gob = s/D= s/(Da+Ds)
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From Eatons paper
Try to duplicate this
0.90.6
7,0
00
1,0
00
Water Depth, ftFig. 3
Overburden Gradient, psi/ft
TVD
bel o
w
RKB,
1,000
ft
TVD Water ob
Depth
11,000 1,000 0.9
11,000 7,000 0.6
0.9 psi/ft = 17.3 lb/gal
0.6 psi/ft = 11.5 lb/gal
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Poissonsratio
fromEaton
Deep Water
Original
Poissons Ratio
TVD
bel o
w
RKB ,
1,000ft
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Inclined wellbores
In general, such wellbores tend to have
lower fracture gradients.
They often have more compressive
stability problems than comparable,
near-vertical wells in the same area.
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Max. prin. Stress dir.
Wellbore direction
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E
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Example 3.6 Inclination = 70 deg
Azimuth = N88E TVD = 14,000
Maximum in situ stress
is vertical
Minimum horiz. Stress
grad. = 0.739 psi/ft
Salt dome intrusion has
created an incrementalhorizontal stress of 0.061
psi/ft in the S42E direction
Pore Pressure
= 12.0 ppg
Poissons ratio = 0.25
Assume elastic
behavior
Estimate the
fracture pressure
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Example
3.6
AWELL = N88E
S42E
IWELL = 70 deg
= 50 deg
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Example 3.6
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Example 3.6 contd
Eq. 3.47
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Example 3.6 contd
Eq. 3.48
Eq. 3.49
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Example 3.6 contd
Eq. 3.50
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Example 3.6 contd
Eq. 3.51
Eq. 3.52
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Example 3.6 contd
is the hole position (Fig. 3.28)
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*
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Continued iterations show that gfi = 0.695 psi/ft
In a vertical well gfi = 0.794 psi/ft
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43Vary to find the minimum value of 3
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Next, gradually increase the wellbore pressure
until the effective minimum principal stress at
the wellbore vanishes.
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Borehole
instability
Lost
circulation
Maximum (theoretical) safe inclination
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Effect of mud filtrate loss
Pore Pressure near wellbore is increased
Effective tangential stress is reduced
Therefore fracture initiation
pressure is reduced
Eff f d fil
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Effect of mud filtrate 1. Most of the poccurs across the filter
cake
2. Lower pore
pressure than in high
flow case
3. Effective hoop
stress is higher
4. Fracture
resistance is higher
5. High quality mud isimportant
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Oil Base Muds
Are more prone to lost circulation than
water based muds!
Why?
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Oil Base Muds
Rheology - Oil base muds tend to be
more viscous at low shear rates than
comparable water base muds, thus
yielding higher ECD
Oil muds tend to be more compressible,
thereby yielding higher densities andhigher pressures
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Oil Base Muds
But
oil muds also have higher relativethermal expansions
temperature changes when circulation
stops
so, wellbore pressures change
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Mud Temperature < Rock Temperature
in lower part of hole
Mud Temperature > Rock Temperature
in upper part of hole
Mud
Temperature in
Annulus
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As the oil mud heats up, the density drops
This results in a drop in BHP
Better filter cake increases the
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Better filter cake increases the
fracture resistance of the rock.
E l 3 7
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Example 3.7
At 80
o
F formation temperature thefracture gradient is 0.864 psi/ft
= 0.19 D = 500
While drilling deeper, the temperature in
the vicinity of the wellbore increases to 90
deg F.
What is the new fracture gradient?
Assume E = 2.5* 106 = 8.0*10-6/oF
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Example 4.7
( et )T = E T/(1- )
( et )T =(2.5*106)(8.0*10-6)(90-80)/(1-0.19)
( et )T = 247 psi
New gfi = 0.864+(247/500) = 1.358 psi/ft
This helps explain why gfi may increase with
time.
Eq. 3.62