Transcript
S ei e i s m i c IIn n v e r s i o n a n d A V O a p p l i ed ed t o L i tth h o l o g i c P r ed ed i c t i o n
Part 1 - Rock Physics and Fluid Replacement Modeling
In t r o d u c t io io n t o r o c k p h y s i c s • Rock physics is a very large subject, and we will only touch its surface today. • We will concentrate on the effect of fluids on the density, P-wave velocity and S-wave velocity of rocks. • After an overview of velocities velocities in non-porous rocks, we will take a brief look at BiotGassmann theory. • Those interested in a more comprehensive overview should purchase the book “The Rock Physics Handbook” by Gary Mavko. 1-2
In t r o d u c t io io n t o r o c k p h y s i c s • Rock physics is a very large subject, and we will only touch its surface today. • We will concentrate on the effect of fluids on the density, P-wave velocity and S-wave velocity of rocks. • After an overview of velocities velocities in non-porous rocks, we will take a brief look at BiotGassmann theory. • Those interested in a more comprehensive overview should purchase the book “The Rock Physics Handbook” by Gary Mavko. 1-2
Basic Rock Physics • The AVO response is dependent on the behaviour of Pwave velocity (VP), S-wave velocity (VS), and density ( ) in a porous reservoir rock. As shown below, this involves the matrix material, the pores, and the fluids filling the pores:
Rock Matrix
Pores / Fluid
1-3
Density • Density effects can be modeled fairly simply with the following equation:
) ρw Sw ρhc ( 1 Sw ) ρsat ρm( 1 where:
density, porosity, Sw water saturation , ρ
sat,m,hc ,w saturated, matrix, hydrocarbo n , water subscripts . • This is illustrated in the next graph. Note the linear responses for both a gas and oil sand.
1-4
Density vs Water Saturation - Porosity = 33% Densities: Oil = 0.8 Gas = 0.001 2.2 2.1 y 2 t i s n e D 1.9
1.8 1.7 0
0.1 Oil
0.2 Gas
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Water Saturation 1-5
P- and S-w aves
(a) P-wave motion
(b) S-wave motion
Since the direction of particle motion for the P-wave is in the same direction as its wave movement, it will be more affected by the 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. Note that there is also S-wave motion out of the plane shown above.
1-6
P and S-wave Velocities •
Initially, we will consider only isotropic rocks, in which the velocities do not depend on direction of travel.
•
There are two different types of velocities of interest to us: – P-wave, or compressional wave velocity – S-wave, or shear wave velocity.
•
For an interactive tutorial on the two waves, go to: http://einstein.byu.edu/~masong/HTMstuff/WaveTrans.html
•
Here are the equations for velocity derived in their most basic form using the Lamé coefficients:
2 V P where: and:
,
V s = the Lamé parameters = density.
1-7
Velocity Equations using K and • Another common way of writing the velocity equations is with bulk and shear modulus:
V P
4 K 3
V s
2 K the bulk modulus , 3 the inverse of the rock' s compressibility,
the shear modulus the 2 nd Lame parameter 1-8
Poisson’s Ratio • A common way of looking at the ratio of V P to V S is to use Poisson’s ratio, defined as:
2 2 2
where :
V P V S
2
• The inverse to the above formula, allowing us to derive V P or V S from , is given by:
2 2 2 1 1-9
Poisson’s Ratio • There are several values of Poisson’s ratio and V P /V S ratio that should be noted: – If V P /V S = 2, then = 0 – If V P /V S = 1.5, then = 0.1 (Gas Case) – If V P /V S = 2, then = 1/3 (Wet Case) – If V P /V S = , then = 0.5 (V S = 0) • A plot of Poisson’s ratio versus velocity ratio is shown on the next slide. 1-10
Vp/Vs vs Poisson's Ratio 0.5 0.4 o i t a R s ' n o s s i o P
0.3 0.2 0.1 0 -0.1 -0.2 0 Gas Case
1
2
3
Wet Case
4
5
6
7
8
9
10
Vp/Vs
1-11
Velocity in porous rocks • The previous derivation was for velocity in solid isotropic rocks. Velocity effects can be modeled by the bulk average equation as seen below and in the next figure:
t sat t m( 1 ) t w Sw t hc ( 1 Sw ) where: t 1 / V, • Unfortunately, the above equation does not hold for gas sands, and this lead to the development of other equations. 1-12
Velocity vs Sw with Volume Avg. Eq. Por = 33% , Voil = 1300 m/s, Vgas = 300 m/s 3500 ) c e s / m ( y t i c o l e V
3000 2500 2000 1500 1000 0 Oil
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Gas
1
Water Saturation 1-13
Velocity in Porous Rocks (cont) • Other empirical equations have been proposed:
V P ( 1 ) V m V fl 2
Raymer et al.
V P ( km / s ) 5 .59 6 .93 2 .18 C
V S ( km / s ) 3.52 4.91 1.89C Han et al, where: C = Volume Clay Content • However, the best fit to observation has been obtained with the Biot-Gassmann equations. 1-14
Dry versus saturated rock • To understand the Biot-Gassmann equations, let us extend the figure we saw earlier to include the concept of the dry rock frame, or skeleton, where the pores are empty, and the saturated rock, where the pores are full:
Dry rock frame, or skeleton (pores empty) Rock Matrix
Saturated Rock (pores full)
Pores / Fluid
1-15
The Biot-Gassmann Equations • Independently, Gassmann (1951) and Biot (1956), developed the theory of wave propagation in fluid saturated rocks, by deriving expressions for the saturated bulk and shear modulii, and substituting into the regular equations for P- and S-wave velocity:
V P
4 K sat sat 3
sat
sat V s sat
Note that sat is found using the volume average equation discussed earlier. 1-16
Biot-Gassmann - Shear Modulus • In the Biot-Gassmann equations, the shear modulus does not change for varying saturation at constant porosity :
sat dry where: sat = shear modulus of saturated rock, and:
dry = shear modulus of dry rock.
1-17
Biot-Gassmann - Saturated Bulk Modulus • The Biot-Gassmann bulk modulus equation is often written as follows:
K dry 2 ( 1 ) K m K sat K dry 1 K dry 2 K fl K m K m where sat = saturated rock, dry = dry frame, m = rock matrix, fl = fluid, and = porosity.
• This equation shows that K sat is dependent on the porosity and fluid content of the rock, as expected. 1-18
Biot-Gassmann - Saturated Bulk Modulus • Mavko et al, in The Rock Physics Handbook, re-arranged the previous equation to give a more intuitive form:
K sat K m K sat
K dry K m K dry
K fl
( K m K fl )
where sat = saturated rock, dry = dry frame, m = rock matrix, fl = fluid, and = porosity.
• Note that K sat can then be written:
K sat K m K sat
dry fluid K m dry fluid K sat 1 dry fluid K dry
K fl where : dry , fluid K m K dry ( K m K fl )
1-19
Biot-Gassmann - Saturated Bulk Modulus
The Saturated Bulk Modulus (K sat) is affected by: Rock
frame bulk modulus (K dry)
Porosity Fluid
bulk modulus (Kfl)
-Saturation -Temperature -Pore Pressure Effective
Pressure
Overburden – Pore pressure Mineral
bulk modulus
1-20
Biot-Gassmann - Shear Bulk Modulus & Density Saturated Shear Modulus ( Is
sat)
Equal to Rock frame shear modulus (
dry)
Porosity Effective
Pressure
Overburden – Pore pressure Saturated Density ( Rock
sat)
depends on
matrix density (
M)
Porosity Fluid
density
-Saturation -Temperature -Pore Pressure 1-21
The Rock Matrix Bulk Modulus • We will now look at how to get estimates of the various bulk modulus terms in the Biot-Gassmann equations, starting with the bulk modulus of the solid rock matrix. Values will be given in GigaPascals (GPa), which are equivalent to 1010 dynes/cm2. • The bulk modulus of the solid rock matrix , K m is usually taken from published data that involved measurements on drill core samples. Typical values are: K s a n d s t o n e = 40 GPa, K l i m e s t o n e = 60 GPa. 1-22
The Fluid Bulk Modulus • The fluid bulk modulus can be modeled using the following equation:
1 K fl where: and:
Sw K w
1 Sw K hc
K fl = bulk modulus of the fluid, K w = bulk modulus of water, K h c = bulk modulus of the hydrocarbon.
• Equations for estimating the values of brine, gas, and oil bulk modulii are given in Batzle and Wang, 1992, Seismic Properties of Pore Fluids, Geophysics, 57, 1396-1408. Typical values are: K g as = 0.021 GPa, K o il = 0.79 GPa, K w = 2.38 GPa
1-23
Estimating K dry • For known V S and V P , K dry can be calculated by first calculating K sat and then using Mavko’s equation. • For known V P , but unknown V S, K dry can be estimated (Gregory, 1977) by assuming the dry rock Poisson’s ratio dry. Gregory shows that equation (1) can be rewritten as: K dry 2 ( 1 ) K m M sat M dry 1 K dry 2 K fl K m K m where : M sat K sat 4 / 3 , M dry K dry 4 / 3 SK dry , and : S
3( 1 dry ) ( 1 )
1-24
Estimating K dry • After a lot of algebra, the previous equation can be written as the following quadratic equation for a term that involves K dry . Solving for the Biot coefficient b gives the solution.
a b 2 bb c 0 where : b 1
K dry K m
,
a S 1,
K m M sat b S 1 S K m K fl M sat K m c S 1 K m K fl
1-25
Porosity Change • Porosity, dry rock bulk modulus, and the matrix bulk modulus can be related by the following equation:
1 1 K P K dry K m • For a known porosity and a computed K dry, we can write:
1 1 K P known K dry K m
1
• If we assume that K P stays constant for a small change in porosity, we can compute a new K dry for a new porosity:
new 1 K dry _ new K K P m
1
1-26
Exercise 1-1 •
Using the equations on the previous pages, compute the saturated densities, velocities (P and S), V P /V S ratio, and Poisson’s ratio of the following two sandstones:
(A)
= 0.33, SW = 1.0, W = 1.0g/cc, gas= 0.001 g/cc, m= 2.65 g/cc, = 3.31 GPa, K m= 40 GPa, K gas= 0.021 GPa, K W = 2.38 GPa, K dry = 3.25 GPa.
(B) Same as (A), but with S W = 0.5, or 50%. Note: The velocities will be in km/sec. Hint:
K sat K m K sat
dry fluid K m dry fluid K sat 1 dry fluid 1-27
Exercise 1-1A – Worksheet 1 ( 1 ) ρsat ρm( 1 ) ρw Sw ρgas( 1 Sw ) ) ρw ρm( 1 ( 2 ) dry
K dry K m K dry
Sw 1 Sw ( 3 ) K fl K K gas w ( 4 ) fluid
1
K fl
( K m K fl )
1 K w
1
K w
dry fluid ( 5 ) K sat K m 1 dry fluid
1-28
Exercise 1-1A – Worksheet 2 ( 6 ) V S sat ( 7 ) V P
K sat
4 3
sat
( 8 ) V P / V S ( V P / V S )2 2 ( 9 ) 2 ( V P / V S )2 2 1-29
Exercise 1-1B – Worksheet 1 ( 1 ) ρsat ρm( 1 ) ρw Sw ρgas( 1 Sw )
ρm( 1 ) 0 .5 ρw 0 .5 ρgas ( 2 ) dry same as 1A
Sw 1 Sw ( 3 ) K fl K K w gas
1
0 .5 0 .5 K K w gas
1
K fl ( 4 ) fluid ( K m K fl ) dry fluid K m 1 dry fluid
( 5 ) K sat
1-30
Exercise 1-1B – Worksheet 2 ( 6 ) V S sat ( 7 ) V P
4 K sat 3
sat
( 8 ) V P / V S ( V P / V S )2 2 ( 9 ) 2 ( V P / V S )2 2 1-31
Data examples • In the next few slides, we will look at the computed responses for both a gas-saturated sand and an oil-saturated sand using the Biot-Gassmann equation. • We will look at the effect of saturation on both velocity (V P and V S) and Poisson’s Ratio. • Keep in mind that this model assumes that the gas is uniformly distributed in the fluid. Patchy saturation provides a different function. (See Mavko et al: The Rock Physics Handbook.)
1-32
= 33% Km = 40 Kgas = 0.021 Kdry = 3.25
= 3.3 GPa
1-33
1-34
EFFECT OF WATER SATURATION Gas Sand ( Phi = 33% ) 0.5
0.4
100 0.3 99
POISSON'S RATIO
98
0.2
96 94 90 75 50 0
0.1
0 0
2
4
P-WAVE VELOCITY (km/sec)
Another way of displaying the data is on a two parameter plot. Here, Poisson’s ratio is plotted against P-wave velocity.
1-35
Velocity vs Sw - Oil Case Porosity = 33%, Koil = 1.0 MPa 3000 ) 2500 s / m ( y t i 2000 c o l e V
1500 1000 0
0.1 Vs
0.2 Vp
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Sw 1-36
Poisson's Ratio vs Water Saturation Oil Case 0.5 0.4 o i t a R0.3 s ' n o s 0.2 s i o P
0.1 0 0
0.1
0.2
0.3
Poisson's Ratio
0.4
0.5
0.6
0.7
0.8
0.9
1
Sw 1-37
The Mudrock Line The mudrock line is a linear relationship between V P and V S derived by Castagna et al (1985). The equation and original plot are shown below: V P = 1.16 V S + 1360 m/sec = aV S + b
1-38
The Mudrock Line Notice that this is n o t the same as a constant Poisson’s ratio, since this would be written as follows, without an intercept term:
V P
2 2 V S aV S 2 1
This will be illustrated in the next few slides, where a gas sand is shown below the mudrock line, and then lines of constant are superimposed.
1-39
The Mudrock Line 6000 5000 Mudrock Line
4000 3000 Gas Sand
VP (m/s) 2000 1000 0 0
1000
VS(m/s)
2000
3000
4000 1-40
The Mudrock Line 6000 5000 = 1/3 or Vp/Vs = 2
Mudrock Line
4000 3000 Gas Sand
VP (m/s) 2000 1000 0 0
1000
VS(m/s)
2000
3000
4000 1-41
The Mudrock Line 6000 5000 = 1/3 or Vp/Vs = 2
Mudrock Line
4000 3000 Gas Sand
VP (m/s) 2000 = 0.1 or Vp/Vs = 1.5
1000 0 0
1000
VS(m/s)
2000
3000
4000 1-42
Finally, here is a display of the Mudrock line and the dry rock line on a Poisson’s ratio versus P-wave velocity plot.
1-43
Exercise 1-2 - Instructions On the next page is a blank plot of Poisson’s ratio versus P-wave velocity. Using the tables on the subsequent pages, graph the values of Poisson’s ratio against P-wave velocity for the different values of porosity and water Saturation. Note that this is the same example that was used in the notes.
1-44
Exercise 1-2 - Plot 0.5
0.4
0.3
0.2
0.1
0 0
1
2
3
4
5
P-wave Velocity (km/sec)
6 1-45
Exercise 1-2 - Sw = 100% Porosity 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
Sw 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
Density 2.650 2.567 2.485 2.402 2.320 2.237 2.155 2.072 1.990 1.907 1.825 1.742 1.660 1.577 1.495 1.412 1.330 1.247 1.165 1.082 1.000
Vp 5964 4231 3558 3166 2903 2713 2570 2459 2372 2303 2249 2207 2175 2153 2138 2132 2133 2142 2159 2184 2219
Vs 3919 2417 1923 1661 1494 1379 1294 1229 1179 1140 1109 1085 1066 1053 1044 1039 1038 1040 1047 1058 1074
Vp/Vs Pois Ratio 1.522 0.120 1.750 0.258 1.850 0.294 1.906 0.310 1.943 0.320 1.968 0.326 1.986 0.330 2.000 0.333 2.012 0.336 2.021 0.338 2.028 0.339 2.035 0.341 2.040 0.342 2.045 0.343 2.049 0.344 2.053 0.344 2.056 0.345 2.059 0.346 2.061 0.346 2.064 0.347 2.066 0.347
1-46
Exercise 1-2 - Sw = 5% Porosity 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
Sw 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05
Density 2.650 2.520 2.390 2.260 2.130 2.000 1.870 1.740 1.610 1.480 1.350 1.220 1.090 0.960 0.830 0.700 0.570 0.440 0.310 0.180 0.050
Vp 5964 3722 2994 2615 2382 2227 2121 2049 2002 1976 1969 1980 2010 2061 2140 2254 2421 2676 3102 3965 7340
Vs 3919 2439 1961 1712 1560 1458 1389 1341 1311 1294 1289 1296 1316 1349 1401 1475 1585 1752 2030 2596 4805
Vp/Vs Pois Ratio 1.522 0.120 1.526 0.123 1.526 0.124 1.527 0.124 1.527 0.125 1.527 0.125 1.527 0.125 1.527 0.125 1.527 0.125 1.527 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125 1.528 0.125
1-47
Exercise 1-2 - Porosity = 5% Porosity 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05
Sw 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
Density 2.518 2.520 2.523 2.525 2.528 2.530 2.533 2.535 2.538 2.540 2.543 2.545 2.547 2.550 2.552 2.555 2.557 2.560 2.562 2.565 2.567
Vp 3723 3722 3720 3719 3718 3717 3716 3715 3714 3714 3713 3713 3714 3715 3717 3720 3726 3737 3758 3815 4231
Vs 2441 2439 2438 2437 2436 2435 2433 2432 2431 2430 2429 2427 2426 2425 2424 2423 2422 2420 2419 2418 2417
Vp/Vs Pois Ratio 1.525 0.123 1.526 0.123 1.526 0.124 1.526 0.124 1.526 0.124 1.527 0.124 1.527 0.124 1.527 0.125 1.528 0.125 1.528 0.126 1.529 0.126 1.530 0.127 1.531 0.128 1.532 0.129 1.533 0.130 1.536 0.132 1.539 0.135 1.544 0.139 1.554 0.146 1.578 0.164 1.750 0.258
1-48
Exercise 1-2 - Porosity = 15% Poros ity 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15 0. 15
Sw 0. 00 0. 05 0. 10 0. 15 0. 20 0. 25 0. 30 0. 35 0. 40 0. 45 0. 50 0. 55 0. 60 0. 65 0. 70 0. 75 0. 80 0. 85 0. 90 0. 95 1. 00
Dens ity 2.253 2.260 2.268 2.275 2.283 2.290 2.298 2.305 2.313 2.320 2.328 2.335 2.343 2.350 2.358 2.365 2.373 2.380 2.387 2.395 2.402
Vp 2618 2614 2611 2607 2603 2600 2596 2593 2590 2587 2584 2581 2579 2578 2578 2579 2582 2590 2608 2661 3166
Vs 1715 1712 1709 1707 1704 1701 1698 1696 1693 1690 1687 1685 1682 1679 1677 1674 1671 1669 1666 1663 1661
Vp/ Vs Pois Ratio 1. 527 0.124 1. 527 0.124 1. 527 0.125 1. 527 0.125 1. 528 0.125 1. 528 0.126 1. 529 0.126 1. 529 0.126 1. 530 0.127 1. 530 0.128 1. 531 0.128 1. 532 0.129 1. 534 0.130 1. 535 0.132 1. 537 0.133 1. 540 0.136 1. 545 0.139 1. 552 0.145 1. 565 0.155 1. 600 0.179 1. 906 0.310
1-49
Exercise 1-2 - Porosity = 33% Poros ity 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33 0. 33
Sw 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
Dens it y 1.776 1.792 1.809 1.825 1.842 1.858 1.875 1.891 1.908 1.924 1.941 1.957 1.974 1.990 2.007 2.023 2.040 2.056 2.073 2.089 2.105
Vp 2084 2074 2065 2056 2048 2039 2031 2023 2015 2007 2000 1993 1986 1980 1975 1972 1970 1972 1983 2024 2500
Vs 1364 1358 1352 1346 1340 1334 1328 1322 1316 1311 1305 1300 1294 1289 1284 1278 1273 1268 1263 1258 1253
Vp/V s Pois Ratio 1.527 0.125 1.527 0.125 1.528 0.125 1.528 0.125 1.528 0.126 1.529 0.126 1.529 0.127 1.530 0.127 1.531 0.128 1.531 0.128 1.532 0.129 1.533 0.130 1.535 0.131 1.537 0.133 1.539 0.135 1.542 0.137 1.547 0.141 1.555 0.147 1.570 0.159 1.609 0.185 1.995 0.332
1-50
Exercise 1-2 - Porosity = 50% Poros it y 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50
Sw 0. 00 0. 05 0. 10 0. 15 0. 20 0. 25 0. 30 0. 35 0. 40 0. 45 0. 50 0. 55 0. 60 0. 65 0. 70 0. 75 0. 80 0. 85 0. 90 0. 95 1. 00
Dens ity 1. 325 1. 350 1. 375 1. 400 1. 425 1. 450 1. 475 1. 500 1. 525 1. 550 1. 575 1. 600 1. 625 1. 650 1. 675 1. 700 1. 725 1. 750 1. 775 1. 800 1. 825
Vp 1987 1969 1951 1934 1918 1902 1886 1871 1857 1843 1829 1816 1804 1792 1782 1773 1765 1762 1767 1800 2249
Vs 1301 1289 1277 1266 1255 1244 1233 1223 1213 1203 1193 1184 1175 1166 1157 1149 1140 1132 1124 1116 1109
V p/ Vs P ois Ratio 1. 527 0.125 1. 528 0.125 1. 528 0.125 1. 528 0.126 1. 529 0.126 1. 529 0.126 1. 530 0.127 1. 530 0.127 1. 531 0.128 1. 532 0.129 1. 533 0.129 1. 534 0.130 1. 535 0.132 1. 537 0.133 1. 540 0.135 1. 543 0.138 1. 548 0.142 1. 556 0.148 1. 572 0.160 1. 612 0.187 2. 028 0.339
1-51
Tips for using of Gassmann’s equation Km: Mineral term
“Text book” values have been measured on pure mineral samples (crystals).
Mineral values can be averaged using Reuss averaging to estimate Km for rocks composed of mixed lithologies. Kdry: Rock frame
Represents the incompressibility of the rock frame (including cracks and pores).
Often pressure dependent due to cracks closing with increased effective pressure.
Difficult to obtain accurate values in many cases. Laboratory measurements of representative core plugs under reservoir pressure may be the best source of data. 1-52
Tips for using of Gassmann’s equation: CAUTIONS:
Rocks with large Km and Kdry values (most carbonates) appear insensitive to saturation changes in Gassmann theory.
Gassmann assumed that pore pressure remains constant during wave propagation. This implies fluids are mobile between pores and all stress is carried by K dry.
This assumption is violated at “high frequencies” in highly variable and compressible pore systems.
Carbonates with an abundance of crack-type pores and heterogeneous pore systems are not suitable for standard Gassmann theory. 1-53
Fluid Replacement Modeling (FRM) Estimates
Vp, Vs and density changes that occur when saturation
changes. FRM requires: Top and bottom depth of the reservoir P wave velocity log Porosity and/or density information Shear
wave velocity information (log or estimate)
Saturation
information (consistent with input well logs)
Rock
matrix information (from mineral tables)
Fluid
properties (From B-W fluid calculator)
1-54
Fluid Replacement Modeling (FRM) Input P wave and Density information: FRM
operates on the log data on a sample by sample basis.
Areas
with low porosity, or high shale content should be excluded using gamma ray, density or porosity cut-offs Density
and porosity information are required. This information must be consistent. FRM
can accept: -Density log with saturation data, matrix and fluid densities (porosity is calculated) -Porosity log with saturation data, matrix and fluid densities (density log is calculated) -Density and porosity logs with saturation data and fluid densities (matrix densities are calculated)
FRM
can be sensitive to poor quality or inconsistent log data. 1-55
Fluid Replacement Modeling (FRM) Shear wave information: Shear wave information is required to calculate K dry from the saturated P wave log information. Shear wave information can come from: Dipole Shear wave sonic logs Estimated S-wave velocity logs using the ARCO mudrock line Dry rock Poisson’s ratio (try values from .12 to .25 for sandstones) The Mudrock line underestimates S wave velocities in unconsolidated, highly porous sands. This may result in incorrect estimates of the dry rock Poisson’s ratio and K dry. In that case, suggest: replace the estimated S wave velocities for these sands in a synthetic S wave log with a Vp/Vs of 2.0. 1-56
Fluid Replacement Modeling (FRM)
Water Saturation information: Water saturation for the initial reservoir conditions may be provided as a constant value or as a log. Saturation information must agree with the recorded sonic log and density values.
1-57
Fluid Replacement Modeling (FRM) Water Saturation information: The sonic tool measures the fastest travel path from source to receiver. In many cases, the sonic velocity represents the flushed well bore annulus rather than the hydrocarbon saturation formation. Petrophysicists can provide water saturation logs that represent the conditions of the invaded region. Flushed regions often exhibit patchy saturation. 1-58
The “Fizz Water” issue: When multiple pore fluids are present, K fl is usually calculated by a Reuss averaging technique:
1 K fl
Sw K w
So K o
Sg K g
This method heavily biases compressibility of the combined fluid to the most compressible phase. T h i s a v e r ag i n g t ec h n i q u e a s s u m e s uniform fluid distribution! -G a s an d l i q u i d m u s t b e e v en l y d i s t r i b u t e d in every po re.
Kfl vs Sw and Sg
3 ) a p 2.5 G ( s 2 u l u d 1.5 o m 1 k l u 0.5 B 0 0
0.25
0.5
0.75
Water saturation (fraction)
1 1-59
The “Fizz Water” issue Patchy Saturation:
When fluids are not uniformly mixed, effective modulus values can not be estimated from Reuss averaging.
Non-uniform (or patchy) fluid distributions are defined relative to permeability, fluid viscosity and frequency bandwidth (scale dependent: millimeters for logs and meters for seismic).
1-60
The “Fizz Water” issue Patchy Saturation: When patch sizes are large, with respect to the seismic wavelength, Voigt averaging gives the best estimate of K fluid (Domenico, 1976).
K fl K w Sw K oSo K g Sg When patch sizes are of intermediate size, Gassmann substitution should be performed for each patch area and a volume average should be made (Dvorkin et al, 1999). This can be approximated by using a power-law averaging technique (Brie et al, 1995):
e K fl ( K w K g )Sw K g 1-61
The “Fizz Water” issue: Patchy Saturation: Gassmann predicted velocities Unconsolidated sand matrix Porosity = 30% 100% Gas to 100% Brine saturation 2.5 2.3 ) s / m2.1 k ( p1.9 V
Patchy Voigt Reuss
1.7 1.5 0
0.25
0.5
0.75
1
Water Saturation (fraction) 1-62
The “Fizz Water” issue According to a paper by Han and Batzle, The Leading Edge, April, 2002:
the “Fizz Water” effect is greatly dependent on the pressure of the formation.
1-63
The “Fizz Water” issue
1-64
The “Fizz Water” issue
Note the change of Fluid Modulus as a function of pressure.
1-65
The “Fizz Water” issue
1-66
Conclusions • An understanding of rock physics is crucial for the interpretation of AVO anomalies.
•The volume average equation can be used to model density in a water sand, but this equation does not match observations for velocities in a gas sand. •The Biot-Gassmann equations match observations well for unconsolidated gas sands. •When dealing with more complex porous media with patchy saturation, or fracture type porosity (e.g. carbonates), the BiotGassmann equations do not hold. •The ARCO mudrock line is a good empirical tool for the wet sands and shales. 1-67
Exercise 1-1A – Worksheet 1 Answers ( 1 ) ρsat ρm( 1 ) ρw Sw ρgas( 1 Sw ) ) ρw 2 .11 ρm( 1
( 2 ) dry
K dry K m K dry
Sw 1 Sw ( 3 ) K fl K w K gas ( 4 ) fluid
1
K fl
( K m K fl )
0 .088
1 K w
1
K w 2 .38
0 .192
dry fluid K m 8 .754 ( 5 ) K sat 1 dry fluid
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Exercise 1-1A – Worksheet 2 Answers ( 6 ) V S 1250 m / s sat ( 7 ) V P
K sat
4 3 2500 m / s
sat
( 8 ) V P / V S 2 ( 9 )
( V P / V S )2 2 2 ( V P / V S ) 2 2
0 .333
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Exercise 1-1B – Worksheet 1 Answers ( 1 ) ρsat ρm( 1 ) ρw Sw ρgas ( 1 Sw )
ρm( 1 ) 0 .5 ρw 0 .5 ρgas 1.94
( 2 ) dry same as 1A 0 .088 Sw 1 Sw ( 3 ) K fl K w K gas
1
0 .5 0 .5 K w K gas
1
0 .042
K fl ( 4 ) fluid 0 .0032 ( K m K fl )
dry fluid ( 5 ) K sat K m 3.356 1 dry fluid
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