Transcript Flow equations in various cases

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Stress, Strain, Pressure, Deformation,
Strength, etc.
Maurice Dusseault
Intro to Petroleum Geomechanics
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Stresses (I)
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Stresses in a solid sediment arise because of
gravity and geological history
Stresses are different in diffferent directions
Three principal stresses are orthogonal, and
the vertical direction is usually one of them
Overburden weight is = v (+/- 5%)
The lateral stresses, hmin and HMAX (or h
and H) are at 90 degrees to one another
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Stresses (II)
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Normal stresses () act orthogonal to a plane
and cause the material to compress
Shear stresses () act parallel to a plane and
cause the- material to distort
y
yx
xy
x+
xy
yx

+
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y
x -
Static equilibrium:
xx
yy
xy = - yx
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Pressures
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Pressures refer to the fluid potential (p)
Pressures can be hydrostatic, less than
hydrostatic (rare) or greater (common).
Called underpressured or overpresssured
Pressures at a point are the same in all
directions because they are within the fluid
We assume that capillary effects are not
important for large stresses and pressures
Differences in pressures lead to flow
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Pressures at Depth
~10 MPa
pressure (MPa)
Fresh water: ~10 MPa/km
Sat. NaCl brine: ~12 MPa/km
Hydrostatic pressure distribution: p(z) = rwgz
1 km
Underpressured case:
underpressure ratio = p/(rwgz),
a value less than 1
underpressure
Overpressured case:
overpressure ratio = p/(rwgz),
a value greater than 1.2
overpressure
depth
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Normally pressured range:
0.95 < p(norm) < 1.2
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Effective Stress () Principle
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The famous Terzaghi concept (1921)
Only “effective” stresses [’ ] affect strength
and deformation behaviour
Effective stress is the stress component
transmitted through the solid rock matrix
Total stresses are the sum of the effective
stresses and the pressures:  =  + p, or:
ij = []ij + [p]
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Effective Stresses
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Pressure is the same in
all directions (a fluid)
Effective stress is the
sum of the grain-tograin (matrix) forces
The sum of p and 
gives total stresses, 
Usually, v = r(z)dz
hmin , HMAX must be
measured or estimated
f1
f2
f4
po
f3
v + po = v (or Sv)
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h + po = h (or Sh)

h + po = h (or Sh)
v + po = v (or Sv)
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Rock Strength (I)
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Strength is the resistance to shear stress (shear
strength), compressive normal stress (crushing
strength), tensile stress (tensile strength), or
bending stress (beam strength).
All of these depend on effective stresses (),
therefore we must know the pore pressure (p)
Rock specimen strength is usually very
different than rock mass strength because of
joints, bedding planes, fissures, etc.
Which to use? Depends on the problem scale.
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Rock Strength (II)
Tensile strength (To) is
extremely difficult to measure:
it is direction-dependent, flawdependent, sample sizedependent, ...
 To is used in fracture models
(HF, thermal fracture, tripping
or surge fractures)
 For a large reservoir, To may
be assumed to be zero because
of joints, bedding planes, etc.
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
F
Prepared
rock
specimen
A
To = F/A F
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Rock Strength (III)
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Shear strength is a vital geomechanics
strength aspect, often critical for design
Shearing is associated with:
•Borehole instabilities, including breakouts, failure
•Reservoir shear and induced seismicity
•Casing shear and well collapse
•Reactiviation of old faults, creation of new ones
•Hydraulic fracture in soft, weak reservoirs
•Loss of cohesion and sand production
•Bit penetration, particularly PCD bits
n
n is normal effective stress
is the shear stress  plane
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slip plane

rock
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Rock Strength (IV)
a
slip
planes

r
r
stress difference
a
1 3

peak
strength
a
axial strain
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Shear strength depends
on the frictional
behaviour and the
cohesion of the rock
Carry out a series of
triaxial shearing tests at
different 3, plot each as
a stress-strain curve,
determine peak strengths
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1 - 3
Curves
stress difference
peak
strength
damage
starts
massive damage,
shear plane develops
cohesion
breaking
sudden stress drop (brittle)
continued damage
“elastic” part of  curve
ultimate or
residual
strength
seating, microcrack closure
axial strain
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a
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Rock Strength (V)

To plot a yield criterion from triaxial tests, plot
1, 3 at failure on equally scaled n axes,
join with a semicircle, then sketch tangent (= Y)
Y

cohesion
c
To
3
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1
n
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1  3 Plotting Method
Plot 1, 3 values at
peak strength on axes
 Fit a curve or a straight
line to the data points
 The y-intercept is the
unconfined strength
Y (1 - 3tan2 - C0 = 0)
(straight line approximation)

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1
Curved or
linear fit
tan2
Uniaxial
compressive
strength, C0
3
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Rock Strength (VI)
Strength of joints
or faults require
shear box tests
 Specimen must
be available and
aligned properly
in a shear box
 Different stress
values (N) are
used
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
Area - A
N - normal force
S
Shear box
S - shear
force
N
S
A
Linear “fit”
Curvilinear “fit”
data point
c
cohesion
N
A
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Yield Criterion

This type of plot is called a Mohr-Coulomb
plot. Y is usually called a Mohr-Coulomb
“yield” or “failure” criterion
•It represents the shear
strength of the rock (S/A) at
various normal stresses
(N/A), A is area of plane
•For simplicity, a straight
line fit is often used
S
=
A
Y
Linear “fit”
Curvilinear “fit”
data point
c
cohesion
N
= n
A
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Cohesion
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Bonded grains
Crystal strength
Interlocking grains
Cohesive strength
builds up rapidly
with strain
But! Permanently
lost with fabric
damage and
debonding of grains
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1 - 3
complete - curve
stress difference

cohesion
mobilization
friction
mobilization
a
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Friction
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Frictional resistance to slip between surfaces
Must have movement () to mobilize it
Slip of microfissures can contribute
Slip of grains at their contacts develops
Friction is not destroyed by strain and damage
Friction is affected by normal effective stress
Friction builds up more slowly with strain
mob = cohesion + friction
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f = c + n
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Estimating Rock Strength
Laboratory tests OK in some cases (salt, clay),
and are useful as indicators in all cases
 Problems of fissures and discontinuities
 Problems of anisotropy (eg: fissility planes)
 Often, a reasonable guess, tempered with data,
is adequate, but not always
 Size of the structure (eg: well or reservoir) is a
factor, particularly in jointed strata
 Strength is a vital factor, but often it is
difficult to choose the “right” strength value
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Strength Anisotropy
UCS
UCS
Vertical
core
0°
30°
60°

90°


Bedding
inclination
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Crushing Strength (I)
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Some materials (North Sea Chalk, coal,
diatomite, high porosity UCSS) can crush
Crushing is collapse of pores, crushing of
grains, under isotropic stress (minimal )
Tests involve increasing all-around effective
stress (’) equally, measuring V/’
Tests can involve reducing p in a highly
stresses specimen (ie: ’ increases as p drops)
UCSS = unconsolidated sandstone
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Crushing Strength (II)
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Apply p,  ( > p), allow
to equilibrate ( =  - p)
Increase  by increasing 
or dropping p
 =  - p

Record volumetric strain,
plot versus effective stress
The curve is the crushing
behavior with +
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
V


p

LE
crushing
material
V
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Rock Stiffness
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To solve any ’-problem, we have to know
how the rock deforms in response to a stress
change
This is often referred to as the “stiffness”
For linear elastic rock, only two parameters
are needed: Young’s modulus, E, and
Poisson’s ratio, (see example)
For more complicated cases, more parameters
are required
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Rock Stiffness Determination
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Stiffness controls stress changes
Estimate stiffness using correlations based on
geology, density, porosity, lithology, ....
Use seismic velocities (vP, vS) for an upperbound limit (invariably an overestimate)
Use measurements on laboratory specimens
(But, there are problems of scale and joints)
In situ measurements (THE tool, others ...)
Back-analysis using monitoring data
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What are E and ?

deformation
L
Young’s modulus (E):
E is how much the material
compresses under a change
in effective stress
Poisson’s ratio (
 is how much rock expands
laterally when compressed.
If = 0, no expansion (eg: sponge)
In  = 0.5, complete expansion,
therefore volume change is zero
Intro to Petroleum Geomechanics
radial
dilation
r
L
strain () =
L
E = 

 = r
L
L
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Rock Properties from Correlations
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A sufficient data base must exist
The GMU must be properly matched to the
data base; for example, using these criteria:
•Similar lithology
•Similar depth of burial and geological age
•Similar granulometry and porosity
•Estimate of anisotropy (eg: shales and laminates)
•Correlation based on geophysical properties (KBES)
Use of a matched analogue advised in cases
where core cannot be obtained economically
GMU = geomechanical unit
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What is a Matched Analogue?
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Rocks too difficult to sample without damage,
or too expensive to obtain
Study logs, mineralogy, even estimate the
basic properties (f, E, …)
Find an analogue that is closely matched, but
easy to sample for laboratory specimens
Use the analogue material as the basis of the
test program
Don’t push the analogue too far!!
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Seismic Wave Stiffness
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vP, vS are dynamic responses affected by rock density
and elastic properties
Because seismic strains are tiny, they do not compress
microcracks, pores, or contacts
Thus, ED and D are always higher than the static test
moduli, ES and S
The more microfissures, pores, point contacts, the
more ED > ES, x 1.3, even to x 10
If porosity ~ 0,  very high, ES approaches ED
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Laboratory Stiffness
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Cores and samples are microfissured; these open
when stress is relieved, E may be underestimated
In microfissured or porous rock, crack closure,
slip, and contact deformation dominate stiffness
ES and S under confining stress are best values
Joints are a problem: if joints are important in
situ, their stiffness may dominate rock response,
but it is difficult to test in the laboratory
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Cracks and Grain Contacts (I)
E1
E
2
E3
Microflaws can
close, open, or
slip as  changes
Flaws govern
rock stiffness
The nature of the grain-to-grain
contacts and the overall porosity
govern the stiffness of porous SS
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E1
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Cracks and Grain Contacts (II)
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Point-to-point contacts are much more
compliant than long (diagenetic) contacts
Large open microfissures are compliant
Oriented contacts or microfissures give rise
to anisotropy of mechanical properties
Rocks with depositional structure or exposed
to differential stress fields over geological
time develop anisotropy through diagenesis
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How Do We “Test” This Rock Mass?
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Joints and fractures can
be at scales of mm to
several meters
Large f core: 115 mm
Core plugs: 20-35 mm
If joints dominate,
small-scale core tests
are “indicators” only
This issue of “scale”
enters into all Petroleum
Geomechanics analyses
A large core specimen
A core “plug”
1m
Machu Picchu, Peru, Inca Stonecraft
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Induced Anisotropy
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UCSS# subjected to large stress differences
develops anisotropy (contacts form in 1
direction and break in 3 direction)
A brittle isotropic rock develops microcracks
mainly parallel to 1 direction
Now, these rocks have developed anisotropy
because of their -history (i.e.: damage)
This is a challenging area of analysis
# UCSS = unconsolidated sandstone
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In Situ Stiffness Measurements
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Pressurization of a packer-isolated zone,
with measurement of radial deformation
Direct borehole jack methods (mining)
Geotechnical pressuremeter modified for
high pressures (membrane inflated at high
pressure, radial deformation measured)
Correlation methods (penetration,
indentation, others?)
These are not widely used in Petroleum Eng.
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Back-Analysis for Stiffness
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Apply a known effective stress change,
measure deformations (eg: uplift, compaction)
Use an analysis model to back-calculate the
rock properties (best-fit approach)
Includes all large-scale effects
Can be confounded by heterogeneity,
anisotropy, poor choice of GMU, ...
Often used as a check of assumptions
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Direct Borehole Stability Problems

Stuck drill pipe
 differential
pressure sticking
 wedging in the borehole
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Stuck casing during installation
Lost circulation (in many cases)
Mudrings, cuttings build-up in washouts
Borehole squeeze
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Indirect Stability Problems

Slow advance rates in drilling
 Longer
hole exposure = greater costs
 Longer exposure = greater chance of instability
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Solids build-up and loss of mud control
Blowouts
 Washouts,
sloughing cause tripping and drilling
difficulties, swabbing
 A blowout eventually develops as control is lost
Intro to Petroleum Geomechanics