9A. Fracture Gradients - Cont'd

8/9/2019 9A. Fracture Gradients - Cont'd

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TAMU - PemexWell Control

Lesson 9A

Fracture Gradients

- contd


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3

Fig

. 3.21 - Stress concentrations around a borehole

in a uniform stress field

Tension

Additional

compression


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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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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

Documents

9A. Fracture Gradients - Cont'd