Seismic Inversion and AVO Applied to Lithologic Prediction

Seismic Inversion and AVO appliedto Lithologic Prediction Part 1 - Rock Physics and Fluid Replacement Modeling 1-2 Introduction to rock physics • 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 in non-porous rocks, we will take a brief look at Biot- Gassmann theory. • Those interested in a more comprehensive overview should purchase the book “The Rock Physics Handbook” by Gary Mavko. 1-3 Basic Rock Physics • The AVO response is dependent on the behaviour of P- wave velocity (V P ), S-wave velocity (V S ), and density (µ) 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 1-4 Density • Density effects can be modeled fairly simply with the following equation: | | | ) S 1 ( ρ S ρ ) 1 ( ρ ρ w hc w w m sat ÷ + + ÷ = . subscripts water , n hydrocarbo matrix, saturated, w , sat,m,hc , saturation water w S porosity, density, ρ where: = = = = | • This is illustrated in the next graph. Note the linear responses for both a gas and oil sand. 1-5 Density vs Water Saturation - Porosity = 33% Densities: Oil = 0.8 Gas = 0.001 1.7 1.8 1.9 2 2.1 2.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Water Saturation D e n s i t y Oil Gas 1-6 (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. P- and S-waves 1-7 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: µ µ ì 2 V P + = µ µ = s V where: ì , µ = the Lamé parameters and: µ = density. 1-8 Velocity Equations using K and µ • Another common way of writing the velocity equations is with bulk and shear modulus: µ µ 3 4 K V P + = µ µ = s V parameter Lame 2 the modulus shear the ility, compressib s rock' the of inverse the , 3 2 modulus bulk the K nd = = = + = = µ µ ì 1-9 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 ÷ ÷ = ¸ ¸ o 2 S P V V : where | | . | \ | = ¸ • The inverse to the above formula, allowing us to derive V P or V S from o, is given by: 1 2 2 2 ÷ ÷ = o o ¸ 1-10 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 o = 0 – If V P /V S = 1.5, then o = 0.1 (Gas Case) – If V P /V S = 2, then o = 1/3 (Wet Case) – If V P /V S = ·, then o = 0.5 (V S = 0) • A plot of Poisson’s ratio versus velocity ratio is shown on the next slide. 1-11 Vp/Vs vs Poisson's Ratio -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0 1 2 3 4 5 6 7 8 9 10 Vp/Vs P o i s s o n ' s R a t i o Gas Case Wet Case 1-12 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: | A | A | A A ) S 1 ( t S t ) 1 ( t t w hc w w m sat ÷ + + ÷ = V, / 1 t where: = A • Unfortunately, the above equation does not hold for gas sands, and this lead to the development of other equations. 1-13 Velocity vs Sw with Volume Avg. Eq. Por = 33%, Voil = 1300 m/s, Vgas = 300 m/s 1000 1500 2000 2500 3000 3500 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Water Saturation V e l o c i t y ( m / s e c ) Oil Gas 1-14 Velocity in Porous Rocks (cont) • Other empirical equations have been proposed: fl m 2 P V V ) 1 ( V | | + ÷ = • However, the best fit to observation has been obtained with the Biot-Gassmann equations. C 18 . 2 93 . 6 59 . 5 ) s / km ( V P ÷ ÷ = | C 89 . 1 91 . 4 52 . 3 ) s / km ( V S ÷ ÷ = | Raymer et al. Han et al, where: C = Volume Clay Content 1-15 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: Rock Matrix Pores / Fluid Dry rock frame, or skeleton (pores empty) Saturated Rock (pores full) 1-16 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: sat sat sat P 3 4 K V µ µ + = sat sat s V µ µ = Note that µ sat is found using the volume average equation discussed earlier. 1-17 Biot-Gassmann - Shear Modulus • In the Biot-Gassmann equations, the shear modulus does not change for varying saturation at constant porosity : dry sat µ µ = where: µ sat = shear modulus of saturated rock, and: µ dry = shear modulus of dry rock. 1-18 Biot-Gassmann - Saturated Bulk Modulus • The Biot-Gassmann bulk modulus equation is often written as follows: 2 m dry m fl 2 m dry dry sat K K K 1 K ) K K 1 ( K K ÷ ÷ + ÷ + = | | • This equation shows that K sat is dependent on the porosity and fluid content of the rock, as expected. where sat = saturated rock, dry = dry frame, m = rock matrix, fl = fluid, and µ = porosity. 1-19 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 K ( K K K K K K K fl m fl dry m dry sat m sat ÷ + ÷ = ÷ | where sat = saturated rock, dry = dry frame, m = rock matrix, fl = fluid, and µ = porosity. • Note that K sat can then be written: ) K K ( K fluid , K K K dry : where K fluid dry 1 fluid dry K fluid dry K K K fl m fl dry m dry m sat sat m sat ÷ = ÷ = ( ¸ ( ¸ + + + = ¬ + = ÷ | 1-20 Biot-Gassmann - Saturated Bulk Modulus The Saturated Bulk Modulus (K sat ) is affected by: Rock frame bulk modulus (K dry ) Porosity Fluid bulk modulus (K fl ) -Saturation -Temperature -Pore Pressure Effective Pressure Overburden – Pore pressure Mineral bulk modulus 1-21 Biot-Gassmann - Shear Bulk Modulus & Density Saturated Shear Modulus (µ sat ) Is Equal to Rock frame shear modulus (µ dry ) Porosity Effective Pressure Overburden – Pore pressure Saturated Density (µ sat ) depends on Rock matrix density (µ M ) Porosity Fluid density -Saturation -Temperature -Pore Pressure 1-22 The Rock Matrix Bulk Modulus • 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 sandstone = 40 GPa, K limestone = 60 GPa. • 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 10 10 dynes/cm 2 . 1-23 The Fluid Bulk Modulus • The fluid bulk modulus can be modeled using the following equation: hc w w w fl K S 1 K S K 1 ÷ + = • 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 gas = 0.021 GPa, K oil = 0.79 GPa, K w = 2.38 GPa where: K fl = bulk modulus of the fluid, K w = bulk modulus of water, and: K hc = bulk modulus of the hydrocarbon. 1-24 Estimating K dry • For known V P , but unknown V S , K dry can be estimated (Gregory, 1977) by assuming the dry rock Poisson’s ratio o dry . Gregory shows that equation (1) can be rewritten as: • For known V S and V P , K dry can be calculated by first calculating K sat and then using Mavko’s equation. 2 m dry m fl 2 m dry dry sat K K K 1 K ) K K 1 ( M M ÷ ÷ + ÷ + = | | ) 1 ( ) 1 ( 3 S : and , SK 3 / 4 K M , 3 / 4 K M : where dry dry dry dry dry sat sat o o µ µ + ÷ = = + = + = 1-25 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 | gives the solution. 0 c b a 2 = + + | | | | . | \ | ÷ | | . | \ | ÷ ÷ = + ÷ | | . | \ | ÷ = ÷ = ÷ = 1 K K K M S c K M S 1 K K S b , 1 S a , K K 1 : where fl m m sat m sat fl m m dry | | | 1-26 Porosity Change • Porosity, dry rock bulk modulus, and the matrix bulk modulus can be related by the following equation: m dry P K 1 K 1 K ÷ = | • For a known porosity and a computed K dry , we can write: 1 m dry known P K 1 K 1 K ÷ ( ( ¸ ( ¸ ÷ = | • 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: 1 m P new new _ dry K 1 K K ÷ ( ¸ ( ¸ ÷ = | 1-27 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, S W = 1.0, µ W = 1.0g/cc, µ gas = 0.001 g/cc, µ m = 2.65 g/cc, µ = 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%. m sat sat m sat K fluid dry 1 fluid dry K fluid dry K K K | | . | \ | + + + = ¬ + = ÷ Hint: Note: The velocities will be in km/sec. 1-28 Exercise 1-1A – Worksheet 1 = + ÷ = ÷ + + ÷ = | | | | | w m w gas w w m sat ρ ) 1 ( ρ ) S 1 ( ρ S ρ ) 1 ( ρ ρ ) 1 ( = = ( ¸ ( ¸ = ( ( ¸ ( ¸ ÷ + = ÷ ÷ w 1 w 1 gas w w w fl K K 1 K S 1 K S K ) 3 ( = ÷ = dry m dry K K K dry ) 2 ( = ÷ = ) K K ( K fluid ) 4 ( fl m fl | = | | . | \ | + + + = m sat K fluid dry 1 fluid dry K ) 5 ( 1-29 Exercise 1-1A – Worksheet 2 = = sat S V ) 6 ( µ µ = + = sat sat P 3 4 K V ) 7 ( µ µ = S P V / V ) 8 ( 2 ) V / V ( 2 2 ) V / V ( ) 9 ( 2 S P 2 S P ÷ ÷ = o 1-30 Exercise 1-1B – Worksheet 1 = + + ÷ = ÷ + + ÷ = | | | | | | gas w m w gas w w m sat ρ 5 . 0 ρ 5 . 0 ) 1 ( ρ ) S 1 ( ρ S ρ ) 1 ( ρ ρ ) 1 ( = ( ( ¸ ( ¸ + = ( ( ¸ ( ¸ ÷ + = ÷ ÷ 1 gas w 1 gas w w w fl K 5 . 0 K 5 . 0 K S 1 K S K ) 3 ( = = 1A as same dry ) 2 ( = ÷ = ) K K ( K fluid ) 4 ( fl m fl | = | | . | \ | + + + = m sat K fluid dry 1 fluid dry K ) 5 ( 1-31 Exercise 1-1B – Worksheet 2 = = sat S V ) 6 ( µ µ = + = sat sat P 3 4 K V ) 7 ( µ µ = S P V / V ) 8 ( 2 ) V / V ( 2 2 ) V / V ( ) 9 ( 2 S P 2 S P ÷ ÷ = o 1-32 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.) 1-33 | = 33% K m = 40 K gas = 0.021 K dry = 3.25 µ = 3.3 GPa 1-34 1-35 0 2 4 EFFECT OF WATER SATURATION P-WAVE VELOCITY (km/sec) POISSON'S RATIO Gas Sand ( Phi = 33% ) 0.5 0.4 0.3 0.2 0.1 0 0 50 75 90 94 96 98 99 100 Another way of displaying the data is on a two parameter plot. Here, Poisson’s ratio is plotted against P-wave velocity. 1-36 Velocity vs Sw - Oil Case Porosity = 33%, Koil = 1.0 MPa 1000 1500 2000 2500 3000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Sw V e l o c i t y ( m / s ) Vs Vp 1-37 Poisson's Ratio vs Water Saturation Oil Case 0 0.1 0.2 0.3 0.4 0.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Sw P o i s s o n ' s R a t i o Poisson's Ratio 1-38 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 1-39 The Mudrock Line S S P aV V 1 2 2 2 V = ÷ ÷ = o o This will be illustrated in the next few slides, where a gas sand is shown below the mudrock line, and then lines of constant o are superimposed. Notice that this is not the same as a constant Poisson’s ratio, since this would be written as follows, without an intercept term: 1-40 The Mudrock Line 0 2000 2000 4000 6000 1000 3000 4000 0 1000 3000 5000 V P (m/s) V S (m/s) Mudrock Line Gas Sand 1-41 The Mudrock Line 0 2000 2000 4000 6000 1000 3000 4000 0 1000 3000 5000 V P (m/s) V S (m/s) Mudrock Line Gas Sand o = 1/3 or Vp/Vs = 2 1-42 The Mudrock Line 0 2000 2000 4000 6000 1000 3000 4000 0 1000 3000 5000 V P (m/s) V S (m/s) Mudrock Line Gas Sand o = 1/3 or Vp/Vs = 2 o = 0.1 or Vp/Vs = 1.5 1-43 Finally, here is a display of the Mudrock line and the dry rock line on a Poisson’s ratio versus P-wave velocity plot. 1-44 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. 1-45 Exercise 1-2 - Plot P-wave Velocity (km/sec) 0 0 0.1 0.2 0.3 0.4 0.5 1 2 3 4 5 6 1-46 Exercise 1-2 - Sw = 100% Porosity Sw Density Vp Vs Vp/Vs Pois Ratio 0.00 1.00 2.650 5964 3919 1.522 0.120 0.05 1.00 2.567 4231 2417 1.750 0.258 0.10 1.00 2.485 3558 1923 1.850 0.294 0.15 1.00 2.402 3166 1661 1.906 0.310 0.20 1.00 2.320 2903 1494 1.943 0.320 0.25 1.00 2.237 2713 1379 1.968 0.326 0.30 1.00 2.155 2570 1294 1.986 0.330 0.35 1.00 2.072 2459 1229 2.000 0.333 0.40 1.00 1.990 2372 1179 2.012 0.336 0.45 1.00 1.907 2303 1140 2.021 0.338 0.50 1.00 1.825 2249 1109 2.028 0.339 0.55 1.00 1.742 2207 1085 2.035 0.341 0.60 1.00 1.660 2175 1066 2.040 0.342 0.65 1.00 1.577 2153 1053 2.045 0.343 0.70 1.00 1.495 2138 1044 2.049 0.344 0.75 1.00 1.412 2132 1039 2.053 0.344 0.80 1.00 1.330 2133 1038 2.056 0.345 0.85 1.00 1.247 2142 1040 2.059 0.346 0.90 1.00 1.165 2159 1047 2.061 0.346 0.95 1.00 1.082 2184 1058 2.064 0.347 1.00 1.00 1.000 2219 1074 2.066 0.347 1-47 Exercise 1-2 - Sw = 5% Porosity Sw Density Vp Vs Vp/Vs Pois Ratio 0.00 0.05 2.650 5964 3919 1.522 0.120 0.05 0.05 2.520 3722 2439 1.526 0.123 0.10 0.05 2.390 2994 1961 1.526 0.124 0.15 0.05 2.260 2615 1712 1.527 0.124 0.20 0.05 2.130 2382 1560 1.527 0.125 0.25 0.05 2.000 2227 1458 1.527 0.125 0.30 0.05 1.870 2121 1389 1.527 0.125 0.35 0.05 1.740 2049 1341 1.527 0.125 0.40 0.05 1.610 2002 1311 1.527 0.125 0.45 0.05 1.480 1976 1294 1.527 0.125 0.50 0.05 1.350 1969 1289 1.528 0.125 0.55 0.05 1.220 1980 1296 1.528 0.125 0.60 0.05 1.090 2010 1316 1.528 0.125 0.65 0.05 0.960 2061 1349 1.528 0.125 0.70 0.05 0.830 2140 1401 1.528 0.125 0.75 0.05 0.700 2254 1475 1.528 0.125 0.80 0.05 0.570 2421 1585 1.528 0.125 0.85 0.05 0.440 2676 1752 1.528 0.125 0.90 0.05 0.310 3102 2030 1.528 0.125 0.95 0.05 0.180 3965 2596 1.528 0.125 1.00 0.05 0.050 7340 4805 1.528 0.125 1-48 Exercise 1-2 - Porosity = 5% Porosity Sw Density Vp Vs Vp/Vs Pois Ratio 0.05 0.00 2.518 3723 2441 1.525 0.123 0.05 0.05 2.520 3722 2439 1.526 0.123 0.05 0.10 2.523 3720 2438 1.526 0.124 0.05 0.15 2.525 3719 2437 1.526 0.124 0.05 0.20 2.528 3718 2436 1.526 0.124 0.05 0.25 2.530 3717 2435 1.527 0.124 0.05 0.30 2.533 3716 2433 1.527 0.124 0.05 0.35 2.535 3715 2432 1.527 0.125 0.05 0.40 2.538 3714 2431 1.528 0.125 0.05 0.45 2.540 3714 2430 1.528 0.126 0.05 0.50 2.543 3713 2429 1.529 0.126 0.05 0.55 2.545 3713 2427 1.530 0.127 0.05 0.60 2.547 3714 2426 1.531 0.128 0.05 0.65 2.550 3715 2425 1.532 0.129 0.05 0.70 2.552 3717 2424 1.533 0.130 0.05 0.75 2.555 3720 2423 1.536 0.132 0.05 0.80 2.557 3726 2422 1.539 0.135 0.05 0.85 2.560 3737 2420 1.544 0.139 0.05 0.90 2.562 3758 2419 1.554 0.146 0.05 0.95 2.565 3815 2418 1.578 0.164 0.05 1.00 2.567 4231 2417 1.750 0.258 1-49 Exercise 1-2 - Porosity = 15% Porosity Sw Density Vp Vs Vp/Vs Pois Ratio 0.15 0.00 2.253 2618 1715 1.527 0.124 0.15 0.05 2.260 2614 1712 1.527 0.124 0.15 0.10 2.268 2611 1709 1.527 0.125 0.15 0.15 2.275 2607 1707 1.527 0.125 0.15 0.20 2.283 2603 1704 1.528 0.125 0.15 0.25 2.290 2600 1701 1.528 0.126 0.15 0.30 2.298 2596 1698 1.529 0.126 0.15 0.35 2.305 2593 1696 1.529 0.126 0.15 0.40 2.313 2590 1693 1.530 0.127 0.15 0.45 2.320 2587 1690 1.530 0.128 0.15 0.50 2.328 2584 1687 1.531 0.128 0.15 0.55 2.335 2581 1685 1.532 0.129 0.15 0.60 2.343 2579 1682 1.534 0.130 0.15 0.65 2.350 2578 1679 1.535 0.132 0.15 0.70 2.358 2578 1677 1.537 0.133 0.15 0.75 2.365 2579 1674 1.540 0.136 0.15 0.80 2.373 2582 1671 1.545 0.139 0.15 0.85 2.380 2590 1669 1.552 0.145 0.15 0.90 2.387 2608 1666 1.565 0.155 0.15 0.95 2.395 2661 1663 1.600 0.179 0.15 1.00 2.402 3166 1661 1.906 0.310 1-50 Exercise 1-2 - Porosity = 33% Porosity Sw Density Vp Vs Vp/Vs Pois Ratio 0.33 0.00 1.776 2084 1364 1.527 0.125 0.33 0.05 1.792 2074 1358 1.527 0.125 0.33 0.10 1.809 2065 1352 1.528 0.125 0.33 0.15 1.825 2056 1346 1.528 0.125 0.33 0.20 1.842 2048 1340 1.528 0.126 0.33 0.25 1.858 2039 1334 1.529 0.126 0.33 0.30 1.875 2031 1328 1.529 0.127 0.33 0.35 1.891 2023 1322 1.530 0.127 0.33 0.40 1.908 2015 1316 1.531 0.128 0.33 0.45 1.924 2007 1311 1.531 0.128 0.33 0.50 1.941 2000 1305 1.532 0.129 0.33 0.55 1.957 1993 1300 1.533 0.130 0.33 0.60 1.974 1986 1294 1.535 0.131 0.33 0.65 1.990 1980 1289 1.537 0.133 0.33 0.70 2.007 1975 1284 1.539 0.135 0.33 0.75 2.023 1972 1278 1.542 0.137 0.33 0.80 2.040 1970 1273 1.547 0.141 0.33 0.85 2.056 1972 1268 1.555 0.147 0.33 0.90 2.073 1983 1263 1.570 0.159 0.33 0.95 2.089 2024 1258 1.609 0.185 0.33 1.00 2.105 2500 1253 1.995 0.332 1-51 Exercise 1-2 - Porosity = 50% Porosity Sw Density Vp Vs Vp/Vs Pois Ratio 0.50 0.00 1.325 1987 1301 1.527 0.125 0.50 0.05 1.350 1969 1289 1.528 0.125 0.50 0.10 1.375 1951 1277 1.528 0.125 0.50 0.15 1.400 1934 1266 1.528 0.126 0.50 0.20 1.425 1918 1255 1.529 0.126 0.50 0.25 1.450 1902 1244 1.529 0.126 0.50 0.30 1.475 1886 1233 1.530 0.127 0.50 0.35 1.500 1871 1223 1.530 0.127 0.50 0.40 1.525 1857 1213 1.531 0.128 0.50 0.45 1.550 1843 1203 1.532 0.129 0.50 0.50 1.575 1829 1193 1.533 0.129 0.50 0.55 1.600 1816 1184 1.534 0.130 0.50 0.60 1.625 1804 1175 1.535 0.132 0.50 0.65 1.650 1792 1166 1.537 0.133 0.50 0.70 1.675 1782 1157 1.540 0.135 0.50 0.75 1.700 1773 1149 1.543 0.138 0.50 0.80 1.725 1765 1140 1.548 0.142 0.50 0.85 1.750 1762 1132 1.556 0.148 0.50 0.90 1.775 1767 1124 1.572 0.160 0.50 0.95 1.800 1800 1116 1.612 0.187 0.50 1.00 1.825 2249 1109 2.028 0.339 1-52 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 K m 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. 1-53 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. 1-54 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) 1-55 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. 1-56 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. 1-57 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. 1-58 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. 1-59 The “Fizz Water” issue:  When multiple pore fluids are present, K fl is usually calculated by a Reuss averaging technique: Kfl vs Sw and Sg 0 0.5 1 1.5 2 2.5 3 0 0.25 0.5 0.75 1 Water saturation (fraction) B u l k m o d u l u s ( G p a ) This averaging technique assumes uniform fluid distribution! -Gas and liquid must be evenly distributed in every pore. g g o o w w fl K S K S K S K 1 + + =  This method heavily biases compressibility of the combined fluid to the most compressible phase. 1-60 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). 1-61 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). 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): g g o o w w fl S K S K S K K + + = g e w g w fl K S ) K K ( K + ÷ = 1-62 The “Fizz Water” issue: Patchy Saturation: Gassmann predicted velocities Unconsolidated sand matrix Porosity = 30% 100% Gas to 100% Brine saturation 1.5 1.7 1.9 2.1 2.3 2.5 0 0.25 0.5 0.75 1 Water Saturation (fraction) V p ( k m / s ) Patchy Voigt Reuss 1-63 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. 1-64 The “Fizz Water” issue 1-65 The “Fizz Water” issue Note the change of Fluid Modulus as a function of pressure. 1-66 The “Fizz Water” issue 1-67 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 Biot- Gassmann equations do not hold. •The ARCO mudrock line is a good empirical tool for the wet sands and shales. 1-68 Exercise 1-1A – Worksheet 1 Answers 11 . 2 ρ ) 1 ( ρ ) S 1 ( ρ S ρ ) 1 ( ρ ρ ) 1 ( w m w gas w w m sat = + ÷ = ÷ + + ÷ = | | | | | 38 . 2 K K 1 K S 1 K S K ) 3 ( w 1 w 1 gas w w w fl = = ( ¸ ( ¸ = ( ( ¸ ( ¸ ÷ + = ÷ ÷ 088 . 0 K K K dry ) 2 ( dry m dry = ÷ = 192 . 0 ) K K ( K fluid ) 4 ( fl m fl = ÷ = | 754 . 8 K fluid dry 1 fluid dry K ) 5 ( m sat = | | . | \ | + + + = 1-69 Exercise 1-1A – Worksheet 2 Answers s / m 1250 V ) 6 ( sat S = = µ µ s / m 2500 3 4 K V ) 7 ( sat sat P = + = µ µ 2 V / V ) 8 ( S P = 333 . 0 2 ) V / V ( 2 2 ) V / V ( ) 9 ( 2 S P 2 S P = ÷ ÷ = o 1-70 Exercise 1-1B – Worksheet 1 Answers 94 . 1 ρ 5 . 0 ρ 5 . 0 ) 1 ( ρ ) S 1 ( ρ S ρ ) 1 ( ρ ρ ) 1 ( gas w m w gas w w m sat = + + ÷ = ÷ + + ÷ = | | | | | | 042 . 0 K 5 . 0 K 5 . 0 K S 1 K S K ) 3 ( 1 gas w 1 gas w w w fl = ( ( ¸ ( ¸ + = ( ( ¸ ( ¸ ÷ + = ÷ ÷ 088 . 0 1A as same dry ) 2 ( = = 0032 . 0 ) K K ( K fluid ) 4 ( fl m fl = ÷ = | 356 . 3 K fluid dry 1 fluid dry K ) 5 ( m sat = | | . | \ | + + + = 1-71 Exercise 1-1B – Worksheet 2 Answers s / m 1305 V ) 6 ( sat S = = µ µ s / m 2000 3 4 K V ) 7 ( sat sat P = + = µ µ 53 . 1 V / V ) 8 ( S P = 13 . 0 2 ) V / V ( 2 2 ) V / V ( ) 9 ( 2 S P 2 S P = ÷ ÷ = o