Transcript
S e i s m i c IIn n v e r s i o n a n d A V O ap a p p l ie ied t o L i tth h o l o g i c P r ed ed i c t i o n Part 4 – Shear Wave Analysis and Inversion
Introduction • In our previous section on rock physics we discussed fluid effects on P and S-wave velocity, and density. • We then looked at post-stack inversion applied to Pwave data. • In this section, we will look at various options for acquiring, analyzing, and inverting S-wave data. • We will start by analyzing the models that were created 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. 4-2
Introduction • In our previous section on rock physics we discussed fluid effects on P and S-wave velocity, and density. • We then looked at post-stack inversion applied to Pwave data. • In this section, we will look at various options for acquiring, analyzing, and inverting S-wave data. • We will start by analyzing the models that were created 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. 4-2
Our Two Models
(a) Wet model
(b) Gas model
Recall that, in the rock physics section, we analyzed the two models shown above. Model A consists of a wet sand, and Model Model B consists of a gas-saturated sand. Specifically, Specifically, we wanted to look at the effects of the gas on the density, P-wave velocity, and S-wave velocity of the saturated sand.
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P- and S-waves
(a) P-wave motion
(b) S-wave motion
Since the direction of particle motion for a P-wave is in the same direction as its wave movement, it will be more affected by a gas sand than the S-wave, S -wave, since the direction of particle motion for the t he S-wave is at right angles to the direction of its wave movement.
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Model Values This was indeed found to be the case when we computed the wet and gas cases using the Biot-Gassmann equations in Part 1 of the course. The values were as follows, where typical values for a shale have also been added. Wet: V P = 2500 m/s, V S= 1250 m/s, = 2.11 g/cc, s = 0.33 Vp/Vs = 2.0 Gas: V P = 2000 m/s, V S= 1310 m/s, = 1.95 g/cc, s = 0.13 Vp/Vs = 1.53 Shale: V P = 2250 m/s, V S= 1125 m/s, = 2.0 g/cc, s = 0.33 Vp/Vs = 2.0 Notice that the P-wave velocity drops dramatically in the gas sand, when compared to the wet sand, but the S-wave velocity actually goes up.
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The Vertical Incidence Seismic Raypath
Surface
Seismic Raypath
VP
VS
Shale
Gas Sand
Shale depth
As shown above, the seismic raypath is dependent on three physical parameters: density ( ), P-wave velocity (V P ), and S-wave velocity (V S), which were discussed in the rock physics section. 4-6
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. In fact, as we will shortly see, there are three different traveltimes that we can record: t PP , or P-wave down and P-wave up; t SS , or S-wave down and S-wave up; and t PS , or P-wave down and S-wave up (this is called the converted wave). Assuming that the gas sand in the previous slide is at a depth of 2000 m and has a thickness of 20 m , and using the velocities on the slide before the previous one, work out the following traveltimes: 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 =
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The reflection coefficient • If the ray paths in the previous slide were at normal incidence (i.e. vertical) the reflection coefficients for the P and S-waves are as follows:
R P 0
2 V P 2 1V P 1
1 D V P
D
, 2 V P 2 1V P 1 2 V P 2 V S 2 1V S 1 1 D V S D R S 0 , 2 V S 2 1V S 1 2 V S where :
D V P V P 2 V P 1 , D V S V S 2 V S 1 , D 2 1 , V P
V P 2 V P 1 2
, V S
V S 2 V S 1 2
,
2 1 2
.
Exercise 4-2 Compute the parameters for the wet sand interfaces using the approximate formulae for the reflection coefficients: Top Shale: V P1 = 2250 m/s V S1 = 1125 m/s 1 = 2.0 g/cc Wet Sand: V P2 = 2500 m/s V S2 = 1250 m/s 2 = 2.11 g/cc Base Shale: V P3 = 2250 m/s V S3 = 1125 m/s 3 = 2.0 g/cc
V P
V S
DV P DV S
D
D V P D V S D V P
V S
R P0 R S0
Exercise 4-3 Compute the parameters for the gas sand interfaces using the approximate formulae for the reflection coefficients: Top Shale: V P1 = 2250 m/s V S1 = 1125 m/s 1 = 2.0 g/cc Gas sand: V P2 = 2000 m/s V S2 = 1300 m/s 2 = 1.94 g/cc Base Shale: V P3 = 2250 m/s V S3 = 1125 m/s 3 = 2.0 g/cc
V P
V S
DV P DV S
D
D V P D V S D V P
V S
R P0 R S0
Model Values We also found in an exercise that the P and S-impedances for the three cases were: Z = Pgas
3900 m /s*g/cc
Z = 2555 m /s*g/cc Sgas
Z = Pwet
5275 m /s*g/cc
Z = 2638 m/s*g/cc Swet
Z = 4500 m /s*g/cc Pshale
Z = 2250 m /s*g /cc Sshale
Using the above values, the P and S reflection coefficients for the gas and wet cases, where the shale overlies the sand, are: R = -0.071 Pgas
R = 0.063 Sgas
R = 0.079 Pwet
R = 0.079 Swet
An interesting thing to note about the reflection coefficients is that the gas and wet cases for the P-waves show opposite polarity, whereas the gas and wet cases for the S-waves show the same polarity. 4-11
Synthetic Models
The four figures on the next two slides show synthetic zero-offset models of the four cases we have considered: the P and S-wave responses of both the wet case (Model A) and the gas case (Model B). (Note that the parameter values have been changed slightly) We have used a 25 Hz Ricker wavelet as the seismic wavelet, and that this wavelet has a wavelength that is less than the time thickness of the sand. Thus, we are seeing “tuning” of the top and base responses. The key thing to note is that the P-wave response changes polarity in going from a wet to a gas sand, but the S-wave response remains the same polarity. 4-12
(a) P-wave log, density and synthetic from model A
(b) S-wave log, density and synthetic from model A (note the different traveltimes).
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(a) P-wave log, density and synthetic from model B
(b) S-wave log, density and synthetic from model B
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P- and S-wave recording
(a)
(b)
(c)
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. This recording approach, using multi-component geophones, was used over a gas 4-15 sand in Alberta to look for the presence of a gas sand. (Ensley, 1984)
P and SH-waves – Gas Sand Example
(a)
(b)
The above diagram shows recorded and processed seismic sections of (a) P, or compressional, waves, and (b) SH, or horizontal shear-waves, over the Myrnham gas field in Alberta. As predicted by the theory, the P-waves respond to the gas sand whereas the S-waves do not, allowing us to predict the presence of the gas. Note the different time arrivals in the two sections. The arrows indicate the same events and the ellipses outline the anomaly. (Ensley, 1984) 4-16
P and SH-waves – Coal Example
(a)
(b)
The above diagram shows recorded and processed seismic sections of (a) P, or compressional, waves, and (b) SH, or horizontal shear-waves, over a false ”bright spot” due to a coal near the gas field in the previous slide. Note that the P-waves and the S-waves both respond to the coal, allowing us to predict that the “bright-spot” is not due to the presence of gas. Again, the arrows show equivalent events, and the ellipses show the zone of interest. (Ensley, 1984) 4-17
Converted S-waves • The previous example used full S-wave recording, in which S-waves were generated at the surface of the earth using an S-wave vibrator, and the reflections were 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 Pwaves which contain some influence from the Swaves.
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Mode Conversion of an Incident P-wave Incident P-wave
Reflected S-wave Reflected P-wave = RP
r q i
q r
V P1 , V S1 , 1 V P2 , V S2 , 2
q t t
Transmitted P-wave Transmitted S-wave
Consider the interface between two geologic horizons of differing P and 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.
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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, and will be discussed in subsequent parts of the course.
• When we record the converted waves, we need to be very 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 from P-wave reflections at different offsets. 4-20
Converted wave analysis • Before looking at a converted wave interpretation, we will discuss the steps involved in converted wave analysis, 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 have different 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 the wells. – (2) Use seismic pick analysis.
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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 the subsurface, it is hard to correlate between the two sections due to time 4-22 and amplitude differences.
Converted display assuming Vp/Vs =2
(a)
(b)
This slide again shows (a) PP data and (b) PS data. However, now the 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. 4-23
P wave log correlation
We have now correlated the P -wave log at the log intersection on the PP data. Notice the good tie on the right, where the blue trace is the synthetic, and the red trace is the seismic trace. 4-24
PS log correlation
We have now correlated the P and S-wave logs at the log intersection on the PS data. Again, notice the good tie on the right, where the blue trace is the synthetic, and the red trace is the seismic trace. 4-25
PP and PS extracted wavelets
(a)
(b)
(c)
(d)
The wavelets on the previous synthetics were extracted from the seismic data and are shown on the left, where (a) shows the wavelet extracted from the PP section, (b) shows the amplitude spectrum of the PP wavelet, (c) shows the wavelet extracted from the PS section, and (d) shows the amplitude spectrum of the PS wavelet. Notice the difference in frequency content. 4-26
Synthetic to seismic correlation
P S -w a v e o f f s e t s y n t h e t i c
P P -w a v e o f f s e t s y n t h e t i c
The display above shows the offset synthetics computed from the well logs and using the wavelets shown in the previous slide. We will be discussing offset synthetics in the next section, but for now simply 4-27 notice that the PS-wave synthetic has zero amplitude at zero offset.
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.
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P - PS seismic and synthetic ties with well log derived velocities
(a)
(b)
This slide again shows (a) PP data and (b) PS data. However, now the PS data 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 don’t match laterally. 4-29
PP and PS horizon picks
(a)
(b)
This slide shows a more extensive section of the (a) PP data and (b) PS data. To correct for laterally varying velocities, we have picked the major events on both sections, using the picks from the logs. 4-30
Horizon matching
(a)
(b)
This slide again shows the (a) PP data and (b) PS data. Now, the horizons have been matched by computing a laterally varying Vp/Vs ratio. 4-31
Vp/Vs ratio from horizon match
This slide shows the laterally varying Vp/Vs ratio that was computed using the horizon picks in the previous slide. 4-32
Vp/Vs Ratio maps
By applying this technique to all of the lines in the 3D volume, a map of Vp/Vs ratios can be computed. The maps above show the change in Vp/Vs ratio between different pairs of events shown in the previous slides. 4-33
Converted-wave case study
Let us now see how the previous analysis can be applied in a field example. For our case study, we will go back to the Blackfoot example considered in the last part of the course. Recall that this case study involved the delineation of a Lower Cretaceous channel sand system. We will start by re-displaying several of the slides from the previous section, including the PP section. We will then look at the PS converted wave data to see what can be added to the interpretation. 4-34
Blackfoot case study
A repeat from the previous section of the schematic stratigraphy of the Blackfoot area, showing three different incised valleys. The relative age is also indicated, where 30 is oldest and 40 is youngest. (Dufour et al.)
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Blackfoot case study
Another look at the index map from the previous section showing seismic cross-line 95, and two east-west cross-sections. The wells are also indicated.
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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.)
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Blackfoot case study
Seismic cross-line 95 from the PS data. Note that resolution is not as good as the PP data and shows only a single valley. (Dufour et al.)
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Blackfoot case study
(a)
(b)
A comparison of the (a) PP data, and (b) PS data from line 95. The lack of resolution in the PS data is now clear. (Dufour et al.) 4-39
Blackfoot case study
In this case study, seismic amplitude inversion was not performed on the PS data. Instead, the authors extracted information about the V P /V S ratio using the seismic time picks, which can be thought of as a type of inversion. The formula used was: V /V / t PP ) – 1 , where P S = 2( t PS
t PS is the
PS isochron and t PP is the PP isochron.
From our earlier discussion of P and S-waves, we know to expect that the V P /V S ratio should go down when we encounter a gas sand, since V P goes down but V S goes up slightly.
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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 ratio for the gas sand example of slide 4-5, and show that this method gives an accurate answer.
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Blackfoot case study
(a)
(b)
Extracted amplitude slices from the (a) PP data, extracted from the upper valley (40), and (b) PS data, extracted from the Glauconitic channel. The white outlines shown the outline of the valley and the anomalous amplitudes are defined by the red outlines. (Dufour et al.)
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Blackfoot case study
(a)
(b)
Computed V P /V S ratio slices the (a) Mannville-Wabamun interval, and (b) top of Glauconitic-incised valley-Wabamun interval. The white outlines shown the outline of the valley. Notice the good match of the anomalously low V P /V S ratios to the productive wells. (Dufour et al.) 4-43
Conclusions
In this section, we have discussed the use of recorded shear wave sections for the computation of reservoir parameter change. Our first example showed how we could differentiate a gas sand “bright-spot” from a coal “bright-spot” using SH wave generation and multi-component recording. We then discussed the use of converted wave data, where the PS conversion (which is an SV wave) is recorded using multi-component geophones. We showed how to integrate the PP and PS recorded section to produce a Vp/Vs estimate and then showed a case study in which this technique was used to explore for channel sands. 4-44
Exercise 4-1 Answers
To base of shale: t PP1 = 1778 ms
To base of sand: t PP2 =1798 m s
t PS1 = 2667 m s t SS1 = 3556 m s
Isochron:
t = 2692 m s PS 2 t PS2 = 3586 ms
t PP = t PP 2 - t PP 1 = 20 m s t PS = t PS2 - t PS1 = 25 ms t SS = t SS2 - t SS1 = 31 ms
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Exercise 4-2 Answers In the following table, we have computed the parameters for the wet sand interfaces using the approximate formulae for the reflection coefficients: Top Shale: V P1 = 2250 m/s V S1 = 1125 m/s 1 = 2.0 g/cc
V P
V S
DV P DV S
2375 1187.5 2.06 250
125
D 0.11
D V P D V S D V P
V S
R P0 R S0
0.105 0.105 0.05
0.079 0.079
- 0.11 -.105 -.105 -.05
-.079 -.079
Wet Sand: V P2 = 2500 m/s V S2 = 1250 m/s 2 = 2.11 g/cc 2375 1187.5 2.06 - 250 - 125
Base Shale: V P3 = 2250 m/s V S3 = 1125 m/s 3 = 2.0 g/cc
Question: Why do you think R P0 and R S0 are identical? 4-46
Exercise 4-3 Answers In the following table, we have computed the parameters for the gas sand interfaces using the approximate formulae for the reflection coefficients: Top Shale: V P1 = 2250 m/s V S1 = 1125 m/s 1 = 2.0 g/cc Gas sand: V P2 = 2000 m/s V S2 = 1300 m/s 2 = 1.94 g/cc Base Shale: V P3 = 2250 m/s V S3 = 1125 m/s 3 = 2.0 g/cc
V P
V S
DV P DV
D
S
D V P D V S D V P
V S
R P0 R S0
2125 1212.5 1.97 - 250 175
-0.06 -.118 0.144 -.03
-.074 0.057
2125 1212.5 1.97
0.06
0.074 -.057
250
- 175
0.118 -.144 0.03
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?
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