Geology 228 Applied Geophysics Lecture 4dthomas/school_demo/G... · Seismic fundamentals ......
Transcript of Geology 228 Applied Geophysics Lecture 4dthomas/school_demo/G... · Seismic fundamentals ......
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Geology 228Applied Geophysics
Lecture 4
Seismic Refraction(Reynolds, Ch. 4-6)
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Seismic Methods• Some fundamentals of seismic waves
– One dimensional wave equation– the solution of a plane wave in unbounded uniform
medium– amplitude, phase, frequency, wave number, wave
length, ... • Huygens’ Principle and some simulations• Wavefront and ray: from physical wave to geometric
wave• Seismic refraction• Field examples
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Seismic Refraction• Snell’s law
– Incident angle, reflection angle, refraction angle– Reflection coefficient, transmission coefficient– Energy distribution
• Seismic refraction for a single horizontal layer• Seismic refraction for multiple horizontal layers• Seismic refraction for a single dipping layer• Seismic refraction tomography• Field examples
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Wave energy is dissipating in the media. There are three major ways in energy dissipation, or attenuation. They are
• Geometry spreading (total energy conservation)
• Intrinsic absorption caused by material imperfection (conversion to other types)
• Diffraction caused by material heterogeneity (reflection, refraction, reverberation, etc.)
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Seismic fundamentals• Some fundamentals of seismic waves
– One dimensional wave equation– Solution of a plane wave in unbounded
uniform medium– Amplitude, and phase – Frequency, and period– Wave number, and wave length
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A simplified case for the wave equation is the plane wave propagating in 1 direction, say the x-direction. In thiscase, the wave equation can be written as
2
2
22
2 1tu
vxu
∂∂
=∂∂ (3.1)
One solution for a plane wave propagating in an unbounded, uniform medium can be expressed as
)cos(0 kxtuu += ω (3.2)
This plane wave can be viewed as the wave generated by a plane source occupying the entire yz-plane togenerate wave propagation in the x-direction. In this equation, u0 is the amplitude, ω is the angular frequency; kis called the wave number. We will show the relationship of k with respect to angular frequency ω bydemonstrating Eqn (3.2) does satisfy the 1-D wave equation (3.1). Taking the secondary derivative of u withrespect to space, here the x-coordinate, is
)cos(02
2
2
kxtukxu
+−=∂∂ ω
and putting the second derivative of u with respect to time on the right hand side of Eqn (3.1) gives
)cos(102
2
2
2
2 kxtuvt
uv
+−=∂∂ ωω
comparing the last 2 equation leads to
vk ω=
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In time domain
Tf
fT
ff
112
2
==
==πωπω
In space domain
kk πλ
λπ 22
==
They are linked through the propagation velocity
vTv
k == λω
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A seismic wave field movie• Uniform medium
– 2D– point source– impulse time function – a rich frequency content
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Huygens’ Principle• In a wave field, all points with the same
phase construct a wave front;• Huygens’ Principle: Each point on a
particular wave front can be treated as a new source
• To illustrate this point let’s consider the following case.
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More complicated case to show the relationship between wavefront and ray
• 3 layer model• v3>v2>v1• v3:v2:v1 = 9:4:3
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Ray is the outward normal at each point of the wavefront
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Rays in a two-layer model: the velocity in the upper layer increases linearly from 4.0 – 5.5 km/sec, over a thickness of 10 km (gradient 0.15/km/sec/km). The velocity in the lower layer increases linearly from 8.0-8.5 km/sec, over a thickness of 4 km (gradient 0.125 km/sec/km).
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Snell’s law, Reflection coefficient, and
Transmission coefficient
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Refraction from multiple horizontal layers
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The refracted wave is the wave energy travels below the interface at the velocity on the second layer, but travel horizontally along the interface, then travel back to the geophone receiver planted on the surface. The refracted wave can only be received after the critical distance xcrit. To learn what is the critical distance, we have to know what is the critical angle icrit first. First we need introduce the Snell’s law.
2
2
1
1 sinsinv
iv
i=
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Snell’s law
2
2
1
1 sinsinv
iv
i=
At the critical incident angle, there is no transmitted energy radiated in the second layer, so the refraction angle is 90 degrees, so we have
2
1
221
1
sin
1)2/sin(sin
vvithen
vvvi
crit =
==π
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)arcsin(sin2
1
2
1
vvior
vvi critcrit ==
So that we can define the critical angle icrit as
and the critical distance xcrit is
2/12
1
2
1
2/12
2
1
2
1
1
11
]1)[(
2
])(1[(
)(2
cossin2tan2
−=
−=
==
vv
hvvvv
h
iihihx
crit
critcritcrit
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When a plane wave impinging at a flat interface, both reflection and transmission occur. The quantitative description of the reflection and transmission relies on the reflection coefficient and transmission coefficient. The Reflection Coefficient R and the Transmission Coefficient T are, respectively, defined as
incident
transm
incident
reflect
AAT
AA
R
=
=
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It can be demonstrated that the coefficients R and T are associated with the combination of the acoustic impedance. Acoustic impedance is defined as the product of the density and the velocity, I.e., Z = ρv. The reflection coefficient R in a general case is
i
i
nZZ
nZZR
α
α2
12
212
tan)1(1)/(
tan)1(1)/(
−−+
−−−=
where n = (v2 / v1)2 and αi is the angle of incidence of the wave ray.
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For normal incidence, the reflection coefficient is
012
12 =+−
= iforZZZZR α
and the transmission coefficient is
1 2
2 1
20i
Z ZT for
Z Zα= =
+
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0)()(
212
212 =
+−
= iR forZZZZE α
and the transmission energy coefficient is
Recall that the energy of wave motion is proportional to the amplitude, so for the reflection energy coefficient we have
0)(
42
12
21 =+
= iT forZZZZE α
These relations are applicable for the case of the incidence angel less than 15 degrees. Apparently, we have
1=+ TR EE
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Seismic refraction• Seismic refraction only consider the first arrivals
- so simple and easy to use • The detection depth is about 1/4 to 1/10 of your
geophone spread
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Seismic Refraction
Vertical GeophonesSource(Plate)
Rock: Vp2
oscilloscope
ASTM D 5777
Soil: Vp1
x1x2x3x4
t1t2t3t4
Note: Vp1 < Vp2
zR
Determine depthto rock layer, zR
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Xcross
t
x
A piece of real data
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The travel time to each geophone for the direct wave in the first layer is simply
1vxtdirect =
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The travel time for the reflected wave for a 2-layer model can be derived as follows. Start with the ray path and the knowledge of Snell’s law we have
1
21
2
21
221
21
221
4
4)(
)2
()2
(
vhx
t
hxtv
hxtv
reflect
reflect
reflect
+=
+=
+=
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The second equation in the last slide can be re-formatted as
144
4)(
21
2
21
221
21
221
=−
+=
hx
htv
hxtv
reflect
reflect
The second equation is in a standard format for expressing a hyperbola curve.
t
x
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Another important concept is the cross-over distance Xcross. At Xcross, the refracted wave starts to take over to be the first arrive at a point.
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The ray path and the travel time for the refracted wave for a 2-layer model can be derived as
121 vCR
vBC
vSBtrefract ++=
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At the crossover distance xcross the travel times to the point are the same for the direct wave and the refracted wave, so we have
2/1
12
12
2/11212
21
2/121
22
2121
12
2
21
22
121
2
21
22
121
)(2
)])([(2)(2)(
2
2
vvvvxh
vvvvvvhvv
vvh
vvvvx
vvv
vh
vx
vx
vvv
vh
vx
vx
tt
cross
cross
crosscross
crosscross
refractdirect
+−
=
−+=−=−
−=−
−+=
=
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Seismic Refraction
0.000
0.005
0.010
0.015
0.020 Tr
avel
Tim
e (s
econ
ds)
0 10 20 30 40 50 Distance From Source (meters)
Horizontal Soil Layer over Rock
Vp1 = 1350 m/s
1
Vp2 = 4880 m/s
1z
x2
V VV Vc
c p2 p1
p2 p1=
−+
Depth to Rock:zc = 5.65 m
xc = 15.0 m
x values
t va
lues
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Travel times for the direct, reflected, and refracted waves for a 2-layer model
2
21
22
12
1
22
1
2
4
vvv
vh
vxt
vhxt
vxt
refract
reflect
direct
−+=
+=
=
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nn
n
nn
nnk
n
k k
knn
nk
n
k k
kn
Tvxt
vvhTh
vhT
+=
−=
=
++
−
=
+=
∑
∑
)1()1(
1
1
)1(1
cos]cos
2[
cos2
θθ
θ
For multiple layers, the thickness of each layer for n>1 can be calculated from
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For example, for the first 3 layers we have
34
324
2
214
1
13334
3
324
2
214
1
13
23
213
1
12223
2
213
1
12
2/121
22
211112
1
11
cos)]coscos(
2[]coscoscos[2
cos]cos
2[]coscos[2
)(cos2
θθθθθθ
θθθθ
θ
vvh
vhTh
vh
vh
vhT
vvhTh
vh
vhT
vvvvTh
vhT
+−=++=
−=+=
−==
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Procedure to get the stratigraphic structure from the refraction x-t plot :
1, from the slope get the velocity in each layer;2, from the velocity to get the critical angle of the k-th interface
1)1(sin
=+ =
n
knk v
vθ
3, get the interception time Tn;4, get the thickness hn.
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)11(211
2 du vvv+=
True velocity in the second layer:
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Homework• Exercise: refraction over a 2 layer model
Xs (m) Ys (m) Xr (m) Yr (m) T (sec)36 8 36 8 0.00000036 8 38 8 0.00658736 8 41 8 0.01349436 8 44 8 0.01429836 8 47 8 0.01510136 8 50 8 0.01590436 8 53 8 0.01686836 8 56 8 0.01767136 8 59 8 0.01831436 8 62 8 0.01992036 8 65 8 0.02056336 8 68 8 0.02104536 8 71 8 0.022009
Seismic Travel Time (sec)
0.000000
0.005000
0.010000
0.015000
0.020000
0.025000
Tim
e (s
ec)
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Seismic Refraction Surveys
• Measure: propagation of elastic wave through layers of the Earth
• Results: depth structure and velocities of elastic waves • Equipment: 48-channel portable StrataView
Geophones
Shots
Distance (m)
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Intercept – Time MethodProcedure:
- First arrivals pick up.- Plot travel time versus distance - Calculate depths to layer interfaces:
21
12
1211 2 ⎟⎟
⎠
⎞⎜⎜⎝
⎛+−
=VVVVTZ
22
23
23
13
21
231
222
2
VV
VVVV
VVZTZ
−⎥⎥⎦
⎤
⎢⎢⎣
⎡ −−=
DISTANCE (m)
TRA
VEL
TIM
E (m
s)
Xc1 Xc2
1/V1
1/V21/V3
T1
T2
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Delay Time Method – SIPwin Program
First arrivals pick up
Layer assignment Create depth-velocity model
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Intercept Time Method
X-T plot of EE'
0102030405060
0 100 200 300 400 500
Offset (m)
Trav
el T
ime
(ms)
Travel time- Distance plot of line EE’ (top) and DD’ (bottom)
X-T plot of DD'
0
10
20
30
40
50
60
70
80
0 100 200 300 400 500 600 700 800 900 1000
Offset (m)
Trav
el time (m
s)
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05
1015202530
0 100 200 300 400 500
Distance (m)
Depth (m
) 1963 m/s
4951 m/s
433 m/s
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0
10
20
3040
50
60
70
0 100 200 300 400 500 600 700 800 900 1000
DISTANCE (m)
5800 m/s
1600 m/s
400 m/s
DE
PTH
(m)
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Seismic Refraction ResultsProfile Parallel to the Tennis Courts
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The Project
In central Japan, a 2300m long mountain tunnel was planned to be built in a Tertiary mudstone area. The rock condition was found to be quite different from the result of prior investigation, and the cutting face collapsed after construction was started. A High Resolution Seismic Refraction analysis is applied to evaluate the rock condition of a non-excavated section in detail. The principal objective of the HRSRanalysis was to detect the distribution and extent of weathered rock at the non-excavated section in order to modify the tunnel design and ensure construction safety.
Result and Interpretation
The existing data used in the HRSR analysis were collected at receiver intervals of 10m and maximum exploration depth of 150m. The figure shows the final velocity model and construction record. It can be seen that the collapsed zones correspond to the lower velocity areas (the green zones in the figure) relative to the surrounding area. This result of the HRSR analysis allowed precise prediction of the weathered rock zone and was very useful for modifying the tunnel design and ensuring construction safety at the non-excavated section.
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Example of Geotechnical Applications
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Seismic refraction tomography example: Locating DNAPL Traps in a Complex Shallow
Aquifer, Hill Air force Base, Utah
• Project of Dept of Geophysics, Rice University
• Reference: Zelt, C. A., "Lateral velocity resolution from 3-D seismic refraction data," Geophys. J. Int., 135 (1998): 1101-1112.
• http://terra.rice.edu/department/staff/morozov/emsp2000/60115.html
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Seismic refraction tomography example: Locating DNAPL Traps in a Complex Shallow
Aquifer, Hill Air force Base, Utah
• Project of Dept of Geophysics, Rice University
• Reference: Zelt, C. A., "Lateral velocity resolution from 3-D seismic refraction data," Geophys. J. Int., 135 (1998): 1101-1112.
• http://terra.rice.edu/department/staff/morozov/emsp2000/60115.html
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Map of the depth to the confining clay layer that underlies the area is based on geologic information collected at the site from monitoring wells and soil tests. Each data point is indicated with a blue dot. The map was constructed by interpolation between the data points. DNAPLscollect in the deep depression running through the center of the map. This area is the focus of the seismic imaging effort. The location of the three profiles from the 2-D survey are shown. The pipe is part of the remediation facilities.
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First-arrival picks from the Line 2 slide hammer data plotted as a function of receiver position. There are 4173 picks from 62 shots. The picks from 3 shots have been colored to highlight the time-offset characteristics of the data; the position of the corresponding shots is indicated by the large colored circles on the distance axis.
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Top: Final 2-D velocity model for the Line 2 sledge hammer data.
Bottom: Raypathsthrough the final model for the Line 2 sledge hammer data.
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Final 2-D velocity model for the Line 2 combined rifle and shotgun data. Contour interval is 100 m/s; the 500, 1000 and 1500 m/s black contours are labeled. The known water table depth in the channel is indicated by the brown arrows. The pink dot indicates the depth (13.2 m) to the clay aquiclude from a well at 25 m distance.
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Relative difference between the final model for the slide hammer data and the 1-D starting model. The approximate position of the channel inferred from the reflection images is indicated above, and the depth to the water table is also indicated. There is a good correlation between the low-velocity region in the tomographic model and the extent of the channel and its depth. Note the large velocity perturbations from the starting model, up to 65% in magnitude.