Upload
anima1982
View
221
Download
0
Embed Size (px)
Citation preview
8/11/2019 4 Shear Waves
1/48
Seism ic Inv ers ion and A VO app l iedto L i tho log ic Pred ic t ion
Part 4 Shear Wave Analysis and
Inversion
8/11/2019 4 Shear Waves
2/48
4-2
Introduction
In our previous section on rock physics wediscussed fluid effects on P and S-wave velocity, anddensity.
We then looked at post-stack inversion applied to P-wave data.
In this section, we will look at various options foracquiring, analyzing, and inverting S-wave data.
We will start by analyzing the models that werecreated in the first section.
We will then look at the analysis of full S-wave data. Finally, we will discuss converted wave, or PS wave
data.
8/11/2019 4 Shear Waves
3/48
4-3
(a) Wet model (b) Gas model
Recall that, in the rock physics section, we analyzed the two modelsshown above. Model A consists of a wet sand, and Model B consists ofa gas-saturated sand. Specifically, we wanted to look at the effects ofthe gas on the density, P-wave velocity, and S-wave velocity of thesaturated sand.
Our Two Models
8/11/2019 4 Shear Waves
4/48
4-4
(a) P-wave motion (b) S-wave motion
Since the direction of particle motion for a P-wave is in the same direction asits wave movement, it will be more affected by a gas sand than the S-wave,since the direction of particle motion for the S-wave is at right angles to the
direction of its wave movement.
P- and S-waves
8/11/2019 4 Shear Waves
5/48
4-5
Model Values
This was indeed found to be the case when we computedthe wet and gas cases using the Biot-Gassmann equationsin Part 1 of the course. The values were as follows, wheretypical values for a shale have also been added.
Wet: V P = 2500 m/s, V S= 1250 m/s, = 2.11 g/cc, s = 0.33Vp/Vs = 2.0
Gas: V P = 2000 m/s, V S= 1310 m/s, = 1.95 g/cc, s = 0.13Vp/Vs = 1.53
Shale: V P = 2250 m/s, V S= 1125 m/s, = 2.0 g/cc, s = 0.33Vp/Vs = 2.0 Notice that the P-wave velocity drops dramatically in thegas sand, when compared to the wet sand, but the S-wavevelocity actually goes up.
8/11/2019 4 Shear Waves
6/48
4-6
VP VS
depth
Surface
Shale
Shale
Gas Sand
SeismicRaypath
As shown above, the seismic raypath is dependent on threephysical parameters: density ( ), P-wave velocity ( V P ), and S-wavevelocity ( V S), which were discussed in the rock physics section.
The Vertical Incidence Seismic Raypath
8/11/2019 4 Shear Waves
7/48
4-7
Exercise 4-1 Traveltimes
On the previous slide, the vertical units were in depth. If they had been in
time, the arrival times for P and S waves would have been different. Infact, as we will shortly see, there are three different traveltimes that wecan record: t PP , or P-wave down and P-wave up; t SS , or S-wave downand S-wave up; and t PS , or P-wave down and S-wave up (this is calledthe converted wave). Assuming that the gas sand in the previous slide isat a depth of 2000 m and has a thickness of 20 m , and using thevelocities on the slide before the previous one, work out the followingtraveltimes:
To base of shale: t PP 1 = To base of sand: t PP 2 =t
PS1 = t
PS 2 =
t SS1 = t SS 2 =
Isochron: t PP = t PP 2 - t PP 1 = t PS = t PS2 - t PS1 =t SS = t SS2 - t SS1 =
8/11/2019 4 Shear Waves
8/48
1-8
The reflection coefficient If the ray paths in the previous slide were at normal
incidence (i.e. vertical) the reflection coefficients for the Pand S-waves are as follows:
.
2
,
2
V V V ,
2
V V V
,,V V V ,V V V :where
,V
V
2
1
V V
V V R
,
V
V
2
1
V V
V V R
12 1S 2 S S
1P 2 P P
12 1S 2 S S 1P 2 P P
S
S
1S 12 S 2
1S 12 S 2 0 S
P
P
1P 12 P 2
1P 12 P 2 0 P
DDD
DD
DD
8/11/2019 4 Shear Waves
9/48
1-9
Exercise 4-2
Top Shale:V P1 = 2250 m/sV S1 = 1125 m/s 1 = 2.0 g/cc
Wet Sand:V P2 = 2500 m/sV S2 = 1250 m/s 2 = 2.11 g/cc
Base Shale:V P3 = 2250 m/sV S3 = 1125 m/s 3 = 2.0 g/cc
V P V S DV P DV S D R P0 R S0 D
P
P
V
V D
S
S
V
V D
Compute the parameters for the wet sand interfaces using the approximateformulae for the reflection coefficients:
8/11/2019 4 Shear Waves
10/48
1-10
Exercise 4-3
Top Shale:V P1 = 2250 m/sV S1 = 1125 m/s 1 = 2.0 g/cc
Gas sand:V P2 = 2000 m/sV S2 = 1300 m/s 2 = 1.94 g/cc
Base Shale:V P3 = 2250 m/sV S3 = 1125 m/s 3 = 2.0 g/cc
V P V S DV P DV S
D R P0 R S0 D
P
P
V
V D
S
S
V
V D
Compute the parameters for the gas sand interfaces using the approximateformulae for the reflection coefficients:
8/11/2019 4 Shear Waves
11/48
4-11
Model ValuesWe also found in an exercise that the P and S-impedances for thethree cases were:
Z Pgas = 3900 m /s*g/cc Z Sgas = 2555 m /s*g/cc
Z Pwet = 5275 m /s*g/cc Z Swet = 2638 m/s*g/cc
Z Pshale = 4500 m /s*g/cc Z Sshale = 2250 m /s*g /cc
Using the above values, the P and S reflection coefficients for the gasand wet cases, where the shale overlies the sand, are:
R Pgas = -0.071 R Sgas = 0.063
R Pwet = 0.079 R Swet = 0.079
An interesting thing to note about the reflection coefficients is that thegas and wet cases for the P-waves show opposite polarity, whereas
the gas and wet cases for the S-waves show the same polarity.
8/11/2019 4 Shear Waves
12/48
4-12
The four figures on the next two slides show syntheticzero-offset models of the four cases we haveconsidered: the P and S-wave responses of both the wetcase (Model A) and the gas case (Model B). (Note thatthe parameter values have been changed slightly)
We have used a 25 Hz Ricker wavelet as the seismicwavelet, and that this wavelet has a wavelength that isless than the time thickness of the sand. Thus, we areseeing tuning of the top and base responses.
The key thing to note is that the P-wave responsechanges polarity in going from a wet to a gas sand, butthe S-wave response remains the same polarity.
Synthetic Models
8/11/2019 4 Shear Waves
13/48
4-13
(a) P-wave log, density and synthetic from model A
(b) S-wave log, density and synthetic from model A (note the different traveltimes).
8/11/2019 4 Shear Waves
14/48
4-14
(a) P-wave log, density and synthetic from model B
(b) S-wave log, density and synthetic from model B
8/11/2019 4 Shear Waves
15/48
4-15
The above diagram shows a schematic diagram of (a) P, or compressional,waves, (b) SH, or horizontal shear-waves, and (c) SV, or vertical shear-waves,where the S-waves have been generated using a shear wave source. Thisrecording approach, using multi-component geophones, was used over a gassand in Alberta to look for the presence of a gas sand. (Ensley, 1984)
P- and S-wave recording
(a) (b) (c)
8/11/2019 4 Shear Waves
16/48
4-16
The above diagram shows recorded and processed seismic sections of (a) P, orcompressional, waves, and (b) SH, or horizontal shear-waves, over the Myrnhamgas field in Alberta. As predicted by the theory, the P-waves respond to the gassand whereas the S-waves do not, allowing us to predict the presence of thegas. Note the different time arrivals in the two sections. The arrows indicate thesame events and the ellipses outline the anomaly. (Ensley, 1984)
P and SH-waves Gas Sand Example
(a) (b)
8/11/2019 4 Shear Waves
17/48
4-17
P and SH-waves Coal Example
(a) (b)
The above diagram shows recorded and processed seismic sections of (a) P, orcompressional, waves, and (b) SH, or horizontal shear-waves, over a falsebright spot due to a coal near the gas field in the previous slide. Note that theP-waves and the S-waves both respond to the coal, allowing us to predict thatthe bright -spot is not due to the presence of gas. Again, the arrows showequivalent events, and the ellipses show the zone of interest. (Ensley, 1984)
8/11/2019 4 Shear Waves
18/48
4-18
Converted S-waves
The previous example used full S-wave recording, in
which S-waves were generated at the surface of theearth using an S-wave vibrator, and the reflectionswere recorded using multi-component geophones.
However, there is a simpler, and cheaper, way to
record S-wave information, as shown in the next slide. If we use a P-wave source, and record the data at
different offsets using multi-component geophones,we can record converted S-waves, and reflected P-
waves which contain some influence from the S-waves.
8/11/2019 4 Shear Waves
19/48
4-19
ReflectedP-wave = R P
ReflectedS-wave
TransmittedP-wave
IncidentP-wave
TransmittedS-wave
Mode Conversion of an Incident P-wave
V P1
, V S1
, 1
V P2 , V S2 , 2
Consider the interface between two geologic horizons of differing Pand S-wave velocity and density and an incident P-wave at angle i .This will produce both P and S reflected and transmitted waves, as
shown above. These are SV waves in the in-line direction.
q i
r
q r
q t
t
8/11/2019 4 Shear Waves
20/48
4-20
Utilizing mode conversion
But how do we utilize mode conversion?
There are actually two ways: Record the converted S-waves using multi-component
receivers (in the X and Z direction). Interpret the amplitudes of the P-waves as a function of offset,
or angle, which contain implied information about the S-waves.This is called the AVO (Amplitude versus Offset) method, andwill be discussed in subsequent parts of the course.
When we record the converted waves, we need to bevery careful in their processing and interpretation, as
will be shown next. In the AVO method, we can make use of the Zoeppritz
equations, to extract pseudo S-wave information fromP-wave reflections at different offsets.
8/11/2019 4 Shear Waves
21/48
4-21
Converted wave analysis
Before looking at a converted wave interpretation, we
will discuss the steps involved in converted waveanalysis, using a dataset from Alberta. The most difficult part of converted wave interpretation
is in interpreting events on the PP and PS sections that
come from the same geological horizon but havedifferent arrival times and amplitudes. As we will see, there are two ways to correct for these
problems:
(1) Use the well log velocities and perform modeling at thewells. (2) Use seismic pick analysis.
8/11/2019 4 Shear Waves
22/48
4-22
Initial multi-component display
(a) (b)Let us consider the data shown above, where (a) shows PP data and(b) shows PS data. Although this data is over the same part of thesubsurface, it is hard to correlate between the two sections due to timeand amplitude differences.
8/11/2019 4 Shear Waves
23/48
4-23
Converted display assuming Vp/Vs =2
(a) (b)
This slide again shows (a) PP data and (b) PS data. However, nowthe PS data has been converted to PP time assuming that the Vp/Vs
ratio is equal to two. The fit is better, but still not very good.
8/11/2019 4 Shear Waves
24/48
4-24
P wave log correlation
We have now correlated the P -wave log at the log intersection on thePP data. Notice the good tie on the right, where the blue trace is thesynthetic, and the red trace is the seismic trace.
8/11/2019 4 Shear Waves
25/48
4-25
PS log correlation
We have now correlated the P and S -wave logs at the log intersectionon the PS data. Again, notice the good tie on the right, where the bluetrace is the synthetic, and the red trace is the seismic trace.
8/11/2019 4 Shear Waves
26/48
4-26
PP and PS extracted wavelets
(a) (b)
(c) (d)
The wavelets on theprevious synthetics wereextracted from theseismic data and areshown on the left, where(a) shows the waveletextracted from the PPsection, (b) shows theamplitude spectrum ofthe PP wavelet, (c)shows the waveletextracted from the PS
section, and (d) showsthe amplitude spectrumof the PS wavelet.Notice the difference infrequency content.
8/11/2019 4 Shear Waves
27/48
4-27
Synthetic to seismic correlation
The display above shows the offset synthetics computed from the welllogs and using the wavelets shown in the previous slide. We will bediscussing offset synthetics in the next section, but for now simplynotice that the PS -wave synthetic has zero amplitude at zero offset.
PS-w ave offse t sy nthet ic PP-w ave offse t sy nthet ic
8/11/2019 4 Shear Waves
28/48
4-28
Seismic tie assuming that Vp/Vs = 2.0
(a) (b)
This slide again shows (a) PP data and (b) PS data, converted to PP time assuming that the Vp/Vs ratio is equal to 2. We have spliced in
the synthetics using the correct velocities. Notice the misfit.
P PS i i d h i i i h ll
8/11/2019 4 Shear Waves
29/48
4-29
P - PS seismic and synthetic ties with welllog derived velocities
(a) (b)This slide again shows (a) PP data and (b) PS data. However, now the PSdata has been converted to PP time using the Vp/Vs ratio from the logs.
The fit is very good at the wells but the sections dont match laterally.
8/11/2019 4 Shear Waves
30/48
H i hi
8/11/2019 4 Shear Waves
31/48
4-31
Horizon matching
This slide again shows the (a) PP data and (b) PS data. Now, the horizonshave been matched by computing a laterally varying Vp/Vs ratio.
(a)
(b)
8/11/2019 4 Shear Waves
32/48
4-32
Vp/Vs ratio from horizon match
This slide shows the laterally varying Vp/Vs ratio that was computed usingthe horizon picks in the previous slide.
8/11/2019 4 Shear Waves
33/48
4-33
Vp/Vs Ratio maps
By applying this technique to all of the lines in the 3D volume, a map ofVp/Vs ratios can be computed. The maps above show the change inVp/Vs ratio between different pairs of events shown in the previous slides.
8/11/2019 4 Shear Waves
34/48
4-34
Converted-wave case study
Let us now see how the previous analysis can beapplied in a field example.For our case study, we will go back to theBlackfoot example considered in the last part ofthe course.
Recall that this case study involved thedelineation of a Lower Cretaceous channel sandsystem.We will start by re-displaying several of the slidesfrom the previous section, including the PP
section.We will then look at the PS converted wave datato see what can be added to the interpretation.
8/11/2019 4 Shear Waves
35/48
8/11/2019 4 Shear Waves
36/48
4-36
Blackfoot case study
Another look at the index map from the previous section showingseismic cross-line 95, and two east-west cross-sections. The
wells are also indicated.
8/11/2019 4 Shear Waves
37/48
4-37
Blackfoot case study
A repeat from the last section of seismic cross-line 95 from the PP data, showing a clear indication of the three valleys. (Dufour et al.)
8/11/2019 4 Shear Waves
38/48
4-38
Blackfoot case study
Seismic cross-line 95 from the PS data. Note that resolution is not asgood as the PP data and shows only a single valley. (Dufour et al.)
8/11/2019 4 Shear Waves
39/48
4-39
Blackfoot case study
A comparison of the (a) PP data, and (b) PS data from line 95. The lackof resolution in the PS data is now clear. (Dufour et al.)
(a) (b)
8/11/2019 4 Shear Waves
40/48
4-40
Blackfoot case study
In this case study, seismic amplitude inversion wasnot performed on the PS data.
Instead, the authors extracted information about theV P /V S ratio using the seismic time picks, which can bethought of as a type of inversion. The formula used
was:V P /V S = 2( t PS / t PP ) 1 , where t PS is thePS isochron and t PP is the PP isochron.
From our earlier discussion of P and S-waves, weknow to expect that the V P /V S ratio should go downwhen we encounter a gas sand, since V P goes downbut V S goes up slightly.
8/11/2019 4 Shear Waves
41/48
4-41
Exercise 4-4 Vp/Vs ratio
Using the isochrons computed in exercise 4-1, and the
formula on the previous slide, compute the Vp/Vs ratiofor the gas sand example of slide 4-5, and show that thismethod gives an accurate answer.
8/11/2019 4 Shear Waves
42/48
4-42
Blackfoot case study
Extracted amplitude slices from the (a) PP data, extracted from theupper valley (40), and (b) PS data, extracted from the Glauconiticchannel. The white outlines shown the outline of the valley and the
anomalous amplitudes are defined by the red outlines. (Dufour et al.)
(a) (b)
f
8/11/2019 4 Shear Waves
43/48
4-43
Blackfoot case study
Computed V P /V S ratio slices the (a) Mannville-Wabamun interval, and (b)top of Glauconitic-incised valley-Wabamun interval. The white outlinesshown the outline of the valley. Notice the good match of theanomalously low V P /V S ratios to the productive wells. (Dufour et al.)
(a) (b)
l
8/11/2019 4 Shear Waves
44/48
4-44
Conclusions
In this section, we have discussed the use ofrecorded shear wave sections for the computation ofreservoir parameter change.Our first example showed how we could differentiatea gas sand bright -spot from a coal bright -spotusing SH wave generation and multi-componentrecording.We then discussed the use of converted wave data,where the PS conversion (which is an SV wave) isrecorded using multi-component geophones.
We showed how to integrate the PP and PS recorded section to produce a Vp/Vs estimate andthen showed a case study in which this techniquewas used to explore for channel sands.
8/11/2019 4 Shear Waves
45/48
4-45
Exercise 4-1 Answers
To base of shale: t PP1 = 1778 ms To base of sand: t PP2 =1798 m st PS1 = 2667 m s t PS 2 = 2692 m s
t SS1 = 3556 m s t PS2 = 3586 ms
Isochron: t PP = t PP 2 - t PP 1 = 20 m s t PS = t PS2 - t PS1 = 25 mst SS = t SS2 - t SS1 = 31 ms
E ercise 4 2 Ans ers
8/11/2019 4 Shear Waves
46/48
4-46
Exercise 4-2 Answers
Top Shale:V P1 = 2250 m/sV S1 = 1125 m/s 1 = 2.0 g/cc
Wet Sand:V P2 = 2500 m/sV S2 = 1250 m/s 2 = 2.11 g/cc
Base Shale:V P3 = 2250 m/sV S3 = 1125 m/s 3 = 2.0 g/cc
V P V S DV P DV S D R P0 R S0 D
P
P
V
V D
S
S
V
V D
2375 1187.5 2.06 250 125 0.11 0.105 0.105 0.05 0.079 0.079
2375 1187.5 2.06 - 250 - 125 - 0.11 -.105 -.105 -.05 -.079 -.079
In the following table, we have computed the parameters for the wet sandinterfaces using the approximate formulae for the reflection coefficients:
Question:Why do you think R P0 and R S0 are identical?
Exercise 4 3 Answers
8/11/2019 4 Shear Waves
47/48
4-47
Exercise 4-3 Answers
Top Shale:V P1 = 2250 m/sV S1 = 1125 m/s 1 = 2.0 g/cc
Gas sand:V P2 = 2000 m/sV S2 = 1300 m/s 2 = 1.94 g/cc
Base Shale:V P3 = 2250 m/sV S3 = 1125 m/s 3 = 2.0 g/cc
V P V S DV P DV S
D R P0 R S0 D
P
P
V
V D
S
S
V
V D
2125 1212.5 1.97 - 250 175 -0.06 -.118 0.144 -.03 -.074 0.057
2125 1212.5 1.97 250 - 175 0.06 0.118 -.144 0.03 0.074 -.057
In the following table, we have computed the parameters for the gas sandinterfaces using the approximate formulae for the reflection coefficients:
Questions:(1) Why are R P0 and R S0 different now?(2) How can the polarity of the two reflection
coefficients help us identify the gas sand?
E i 4 4 A
8/11/2019 4 Shear Waves
48/48
4-48
Exercise 4-4 Answer
Recall that V P /V S = 2000/1310 =1.53. Also, recall that t PS= 25 m s and t PP = 20 ms .
V P /V S = 2( t PS / t PP ) 1 = 2(25/20) -1 = 1.5
Note that the fact that we computed a value of 1.5 ratherthan 1.53 is due to the fact that we rounded -off thetraveltimes to the closest millisecond. If we had usedmore accuracy, the velocity ratio would have beencomputed as 1.53