Transcript No Slide Title

Environmental and Exploration Geophysics II
Energy Loss and Partitioning
tom.h.wilson
[email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Tom Wilson, Department of Geology and Geography
Objectives for the day
•Review time-distance plots
•Discuss energy loss through divergence
and absorption mechanisms
•Present basic definition of reflection
coefficient and how it varies with offset or
incidence angle
•Basic definition of transmission coefficient
Tom Wilson, Department of Geology and Geography
Can you name the events?
Miller et al. 1995
Tom Wilson, Department of Geology and Geography
Ground roll = noise to the exploration geophysicist
Tom Wilson, Department of Geology and Geography
We start off with these noisy looking field records and with
some effort get a more geological look at the subsurface
Tom Wilson, Department of Geology and Geography
Diffraction (point source) Events
Tom Wilson, Department of Geology and Geography
Diffractions, like reflections are hyperbolic in a time-distance plot. They are
usually symmetrical about the apex. Diffractions usually arise from point-like
discontinuities and edges ( for example the truncated edges of stratigraphic
horizons across a fault).
Tom Wilson, Department of Geology and Geography
Note that the apex of the diffraction event is not located at the
source location (X=0). The apex is located over the discontinuity.
Tom Wilson, Department of Geology and Geography
A direct path
How do you figure out Xmin & tmin for the critical refraction?
What is the critical refraction time at longer offsets?
Tom Wilson, Department of Geology and Geography
1) label all plotted curves,
2) label all relevant points and
3) Describe the relationships that
you’ve graphed.
Tom Wilson, Department of Geology and Geography
What to you think the
influence of higher velocity
will be on the moveout?
•Do Exercise II and III using Excel.
•In exercise II comment on the origins of the
differences in the two reflection hyperbola? What is
their relationship to corresponding direct arrivals.
Tom Wilson, Department of Geology and Geography
t
x2
V12

4h12
V12
Tom Wilson, Department of Geology and Geography
Although the time intercept remains the
same - how do the shapes of the reflection
hyperbola differ in these two cases?
For Exercise III, explain the differences observed in the arrival
times of the reflection and diffraction observed in the shot record.
Why does the diffraction event drop below the reflection?
Tom Wilson, Department of Geology and Geography
Your plots should be
consistent with
expected relationships
discussed in class
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
First arrival picks and velocity
170 feet offset
•Can you pick out the
direct arrival?
•Can you pick out the
refraction arrivals?
•How many critical
refractions are there?
•How would you
determine the
refraction velocity?
•How would you
determine the air wave
or direct arrival
velocity?
Tom Wilson, Department of Geology and Geography
0.15 seconds
Velocities > VA, VB VC, Vairwave
VA=
VC=
VB=
Tom Wilson, Department of Geology and Geography
Section 2.3
What happens to the seismic energy generated by the
source as it propagates through the subsurface?
Basic ConceptsEnergy - The ability to do work. It comes in two forms - potential
and kinetic
Work expended (W) equals the applied force times the distance
over which an object is moved.
Power is the rate at which work is performed.
As a mechanical disturbance or wavefield propagates through the
subsurface it moves tiny particles back and forth along it path.
Particle displacements are continually changing. Hence, it is more
appropriate for us to consider the power, or the rate at which energy
is being consumed at any one point along the propagating
wavefront.
Tom Wilson, Department of Geology and Geography
W (the work)  Fx (force times displaceme nt)
dW
Power 
dt
The rate of change of work in unit time.
dW dFx 
dx

 F  Fv
dt
dt
dt
This force is the force exerted by the seismic wave at
specific points along the propagating wavefront.
Tom Wilson, Department of Geology and Geography
and since force is pressure (p) x area (A), we have
dW
Power 
 Fv  pvA
dt
The power generated by the source is PS. This
power is distributed over the total area of the
wavefront A so that
PS  pvA
We are more interested to find out what is going
on in a local or small part of the wavefield.
* Note we are trying to quantify the effect of wavefront
spreading at this point and are ignoring heat losses.
Tom Wilson, Department of Geology and Geography
PS
pv 
A
Over what surface area is the energy
generated by the source distributed?
Tom Wilson, Department of Geology and Geography
Area of a hemisphere is 2R2 hence -
pv 
PS
2R 2
This is the power per unit area being dissipated along
the wavefront at a distance R from the source. (Recall
p is pressure, v is the particle velocity, PS magnitude of
the pressure disturbance generated at the source and R
is the radius of the wavefront at any given time.
Tom Wilson, Department of Geology and Geography
In the acoustic wave equation a quantity Z which is called
the acoustic impedance. Z = V, where  is the density of
the medium and V is the interval velocity or velocity of
the seismic wave in that medium.
Z is a fundamental quantity that describes reflective
properties of the medium.
We also find that the pressure exerted at a point along the
wavefront equals Zv or Vv, where v is the particle
velocity - the velocity that individual particles in the
disturbed medium move back and forth about their
equilibrium position.
Tom Wilson, Department of Geology and Geography
Particle velocity versus interval velocity
Tom Wilson, Department of Geology and Geography
pv   Vv  v
Combining some of these
ideas, we find that v, the
particle displacements vary
inversely with the distance
traveled by the wavefield R.
pv  Zv 2
2
Zv 
v
Ps
2R 2
Ps
Z 2R 2
1
Ps
v
R Z 2
Tom Wilson, Department of Geology and Geography
•We are interested in the particle velocity variation with
distance since the response of the geophone is proportional to
particle (or in this case) ground displacement. So we have
basically characterized how the geophone response will vary
as a function of distance from the source.
•The energy created by the source is distributed over an ever
expanding wavefront, so that the amount of energy available
at any one point continually decreases with distance traveled.
•This effect is referred to as spherical divergence. But in fact,
the divergence is geometrical rather than spherical since the
wavefront will be refracted along its path and its overall
geometry at great distances will not be spherical in shape.
Tom Wilson, Department of Geology and Geography
The effect can differ with wave type. For example, a
refracted wave will be confined largely to a cylindrical
volume as the energy spreads out in all directions
along the interface between two intervals.
z
R
The surface area along the leading edge of the
wavefront is just 2Rz.
Tom Wilson, Department of Geology and Geography
Hence (see earlier discussion for the hemisphere)
the rate at which source energy is being expended
(power) per unit area on the wavefront is
Ps
Zv 
A
2
Remember Zv2 is just
Ps
Zv 
2Rz
2
Following similar lines of reasoning as before,
we see that particle velocity
v
Tom Wilson, Department of Geology and Geography
Ps
R
v
Ps
R
The dissipation of energy in the wavefront
decreases much less rapidly with distance traveled
than does the hemispherical wavefront.
This effect is relevant to the propagation of waves in
coal seams and other relatively low-velocity intervals
where the waves are trapped or confined. This effect
also helps answer the question of why whales are
able to communicate over such large distances
using trapped waves in the ocean SOFAR channel.
Visit
http://www.beyonddiscovery.org/content/view.page.asp?I=224
for some info on the SOFAR channel.
Tom Wilson, Department of Geology and Geography
Sound Surveillance System
See also -
http://www.beyonddiscovery.org/Includes/Dialogs/ExternalLink.asp?ID=3101&URL=
Tom Wilson, Department of Geology and Geography
Divergence losses decrease particle velocity or the
amplitude of a seismic wave recorded by the
geophone - i.e. the amplitude of one of the wiggles
observed in our seismic records.
Unless we correct for this effect – turn up the volume – the
signal will become too weak for us to hear.
Tom Wilson, Department of Geology and Geography
Since amplitude (geophone response) is proportional to
the square root of the pressure, we can rewrite our
divergence expression as
Ps
Ar 
r
As
Ar 
r
Thus the amplitude at distance r from the source (Ar)
equals the amplitude at the source (AS) divided by the
distance traveled (r).
This is not the only process that acts to decrease the
amplitude of the seismic wave.
Tom Wilson, Department of Geology and Geography
Absorption
When we set a spring in motion, the spring oscillations
gradually diminish over time and the weight will cease to move.
In the same manner, we expect that as s seismic wave
propagates through the subsurface, energy will be consumed
through the process of friction and there will be conversion of
mechanical energy to heat energy.
We guess the following - there will be a certain loss of amplitude
dA as the wave travels a distance dr and that loss will be
proportional to the initial amplitude A.
i.e.
dA  AS dr
How many of you remember how to solve such
an equation?
Tom Wilson, Department of Geology and Geography
dA(r )  AS dr
dA(r )
 dr
AS
 is a constant referred to
as the attenuation factor
In order to solve for A as a function of distance traveled (r)
we will have to integrate this expression In the following
discussion,let
A0  AS
A dA
r
A0 A  0dr
ln A  ln A0  r
Tom Wilson, Department of Geology and Geography
ln A  ln A0  r
A
ln
 r
A0
ln
e
A
A0
 e r
A
 e r
A0
A  A0er
Tom Wilson, Department of Geology and Geography
Mathematical Relationship
A(r )  A0er
Graphical Representation
Tom Wilson, Department of Geology and Geography
A(r )  A0er
 - the attenuation factor is also a function of additional terms  is wavelength, and Q is the absorption constant


Q
1/Q is the amount of energy dissipated in one
wavelength () - that is the amount of mechanical
energy lost to friction or heat.
Tom Wilson, Department of Geology and Geography


Q
 is also a function of interval velocity, period and frequency
Tom Wilson, Department of Geology and Geography

r
A(r )  A0e Q



A(r )  A0e QV
r
 is just the reciprocal of the frequency so we
can also write this relationship as

f
A(r )  A0e QV
Tom Wilson, Department of Geology and Geography
r
Smaller Q translates into higher energy loss or amplitude decay.
Tom Wilson, Department of Geology and Geography
A(r )  A0e

f
QV
r
increase f and decrease A
Higher frequencies are attenuated to a much
greater degree than are lower frequencies.
Tom Wilson, Department of Geology and Geography
When we combine divergence and absorption we get the
following amplitude decay relationship
A0 r
A(r ) 
e
r
The combined effect is rapid amplitude decay as the
seismic wavefront propagates into the surrounding
medium.
We begin to appreciate the requirement for high
source amplitude and good source-ground coupling to
successfully image distant reflective intervals.
Tom Wilson, Department of Geology and Geography
But we are not through - energy continues to be
dissipated through partitioning - i.e. only some of the
energy (or amplitude) incident on a reflecting surface
will be reflected back to the surface, the rest of it
continues downward is search of other reflectors.
The fraction of the incident amplitude of the
seismic waves that is reflected back to the surface
from any given interface is defined by the reflection
coefficient (R) across the boundary between layers
of differing velocity and density.
Z 2  Z1
R

Ainc Z1  Z 2
Arefl
Tom Wilson, Department of Geology and Geography
Z 2  Z1
R

Ainc Z1  Z 2
Arefl
Z1 and Z2 are the impedances of the bounding layers.
 2V2  1V1

1V1   2V2
Tom Wilson, Department of Geology and Geography
http://www.crewes.org/ResearchLinks/ExplorerPrograms/ReflEx/REcrewes.htm
Tom Wilson, Department of Geology and Geography
You can also have a look at AVO.xls linked on the class site
Tom Wilson, Department of Geology and Geography
AVA response predicted for shallow strata in the Central
Appalachians, Marshall Co., WV
Tom Wilson, Department of Geology and Geography
Z 2  Z1
R
Z1  Z 2
The transmitted wave amplitude T is
T 1 R
Z1  Z 2  Z 2  Z1 

T
 
Z1  Z 2  Z1  Z 2 
2 Z1
T
Z1  Z 2
Tom Wilson, Department of Geology and Geography
At a distance of 100 m from a source, the
amplitude of a P-wave is 0.1000 mm, and at a
distance of 150 m the amplitude diminishes to
0.0665 mm. What is the absorption coefficient
of the rock through which the wave is
traveling?
(From Robinson and Coruh, 1988)
Any Questions?
Tom Wilson, Department of Geology and Geography
 Please read through Chapter 3, pages 65
through 95; Chapter 4, pages 149 to 164.
 Hand in problems 2.3 and 2.6 (Today)
 Questions about problems 2.7 and 2.12?
Hand those in on Wednesday.
 Start working Exercises I-III and bring
questions to class on Wednesday.
 Any questions about the attenuation
problem? We’ll probably end up discussing
this some more in class on Wednesday. This
will be due next Monday.
Tom Wilson, Department of Geology and Geography