
Publication History:
This article is based on
"Crain's Seismic Petrophysics" by E. R. (Ross) Crain, P.Eng., first
published in 2003, and updated annually until 2016.
This
webpage version is the copyrighted intellectual
property of the author.
Do not copy or distribute in any form without explicit
permission. 
Modeling the Sonic Log Response From BiotGassmann Equations
An alternate and more rigorous approach is the BiotGassmann
equation set:
The velocity of sound in a rock is related to the elastic properties
of the rock/fluid mixture and its density, according to the
Wood, Biot, and Gassmann equations.
The
composite compressional bulk modulus of fluid in the pores (inverse
of fluid compressibility) is:
1: Kf =
1/Cf = Sw / Cwtr + (1  Sw) / Coil
OR 1a: Kf = 1/Cf = Sw / Cwtr + (1
 Sw) / Cgas
The pore space bulk
modulus (Kp) is derived from the porosity, fluid, and matrix
rock properties:
2: ALPHA = 1  Kb / Km
3: Kp = ALPHA^2 / ((ALPHA  PHIt) / PHIt / Kf )
The
composite rock/fluid compressional bulk modulus is:
4: Kc = Kp + Kb + 4/3 * N
Compressional velocity (Vp) and shear
velocity (Vs) are defined as:
5: Vp = KS4 * (Kc / DENS) ^ 0.5
6: Vs = KS4 * (N / DENS) ^ 0.5
7: Vst = KS4 * (DENSW * (1/N + 1/Kf)) ^ 0.5
Although it is not a precise solution, we often invert equations
5 and 6 to solve for Kb and N from sonic log compressional and
shear travel time values.
Where:
ALPHA = Biot's elastic parameter (fractional)
Cgas = gas compressibility
Coil = oil compressibility
Cwtr = water compressibility
DENS = rock density (kg/m3 or g/cc)
DENSW = density of fluid in the pores (kg/m3 or g/cc)
Kb = compressional bulk modulus of empty rock frame
Kc = compressional bulk modulus of porous rock
Kf = compressional bulk modulus of fluid in the pores
Km = compressional bulk modulus of rock grains
Kp = compressional bulk modulus of pore space
N = shear modulus of empty rock frame
PHIt = total porosity of the rock (fractional)
Sw = water saturation (fractional)
Vp = compressional wave velocity (m/sec or ft/sec)
Vs = shear wave velocity (m/sec or ft/sec)
Vp = Stoneley wave velocity (m/sec or ft/sec)
KS4 = 68.4 for English units
KS4 = 1.00 for Metric units
The BiotGassmann approach looks deceptively simple. However, the major
drawback to this approach is the difficulty in determining the
bulk moduli, particularly those of the empty rock frame (Kb and
N), which cannot be derived from log data. Murphy (1991)
provided equations for sandstone rocks (PHIe < 0.35) that
predict Kb and N from porosity:
8: Kb = 38.18 * (1  3.39 * PHIe + 1.95 *
PHIe^2)
9: N = 42.65 * (1  3.48 * PHIe
+ 2.19 * PHIe^2)
These
can help overcome the lack of empty rockframe data.
RECOMMENDED PARAMETERS:
NOTE: Units of compressibility in this table are PSI^1.
Rock matrix bulk modulus and shear modulus data are listed
HERE.
Water compressibility:
Cwtr: 4.0
* 10^10 salinity 5000 ppm
3.7
* 10^10
35000 ppm
2.9
* 10^10 ___ 200000 ppm
Oil
compressibility:
Coil: 8.5 * 10^10 depth 2000 ft 610 m
9.5
* 10^10 4000
1220
11.6
* 10^10
8000
2440
13.5
* 10^10 12000 3660
Gas
compressibility:
Cgas: 250 * 10^10 depth 2000 ft 610 m
210
* 10^10 4000
1220
180
* 10^10 8000
2440
100
* 10^10 12000 3660
NUMERICAL EXAMPLES  BiotGassmann and Wyllie Equations
Combining equations 1 through 4, we get:
10: Kc = Kb + ((1  Kb / Km) ^ 2) / (PHIe / Kf + (1  PHI) / Km
 Kb / (Km ^ 2))
1. Hard Rock Example  Gassmann
Given:
Dolomitic limestone with 20% porosity at 10,000 feet
grain density = 2.78 gm/cc
acoustic velocity = 14,000 ft/sec at 100% water saturation = 71.4
usec/ft
bulk density = 2.44 gm/cc
formation water = 200,000 ppm total dissolved solids
reservoir temperature = 240 degrees F
reservoir pressure = 4700 psi
What
percent change in velocity could be expected if 75% of the pore
space were filled with methane gas?
0.
Convert velocity to cm/sec.
Vp = 14000 * 0.3048 * 100 = 426720 cm/sec
1.
Invert the velocity equation and solve for Kc (sometimes referred
to as the space modulus M) for 100% waterbearing condition.
Kc = (Vp ^ 2) * DENS = 426750 *
426720 * 2.44 = 44.5*10^10 dynes/cm2
2.
Determine Km from handbook data.
dolomite = 82 *10^10 dynes/cm2
limestone = 67 * 10^10 dynes/cm2
For
this example, Km was estimated to be 74.5 * 10^10 dynes/cm2 because
the rock was a mixture of limestone and dolomite.
3.
Estimate Kb.
Cb = 3.7*10^6 psi1 from handbook data
Kb = 1 / Cb = 1.86*10^10 dynes/cm2
4.
Determine fluid properties at reservoir conditions.
Water compressibility (Cw) was estimated to be:
Cw = 2.26*10^6 psi1 from handbook data
Kw = 1 / Cw = 3.05 * 10^10 dynes/cm2
Water
density = 1.085 gm/cc.
Gas
density was estimated to be 0.157 gm/cc.
Gas compressibility (Cg) was computed from:
Cg = 161*10^6 psi1 from handbook data
Kg = 1 / Cg = 0.0428 * 10^10 dynes/cm2
Gaswater
combination properties.
DENSf = 0.25 * (1.085) + 0.75 * (0.157) = 0.389 gm/cc
Kf
from the Wood equation.
1 / Kf = Sw / Kw + Sg / Kg
Kf = 0.0568*10^10 dynes/cm2
5.
Solve Gassmann equation for Kb at Sw = 100%.
Kb @ 100% water saturation = 44.5 * 10^10  ((1  1.86*10^10 / 74.5 * 10^10) ^ 2) /
(0.2 / 3.05 * 10^10 + 0.8 / 74.5 * 10^10  1.86 / (74.5 * 10^10)
^ 2
= 32.9 * 10 ^ 10 dynes/cm2
6.
Solve this equation for Kc with combination gas/water conditions.
Kc @ 75% gas saturation = 44.5 * 10^10  ((1  1.86 * 10^10 / 74.5 * 10^10) ^ 2) /
(0.2 / 0.0568*10^10 / 74.5 * 10^10  1.86 / (74.5*10^10) ^ 2)
= 33.2*10^10 dynes/cm2
7.
Solve density equation for gas/water combination.
DENS @ 75% gas saturation = 0.2 * (0.389) + 0.8 * (2.78) = 2.30 gm/cc
8.
Solve for Vp for Sg = 0.75.
Vp @ 75% gas saturation = (33.2 * 10^10) ^ 0.5 / 2.30 = 3.80 * 10^5 cm/sec = 12,500
ft/sec
Utilizing
this methodology and the stated assumptions, there would be an
approximate 10% change in velocity (14,000  12,500 / 14,000)
from a 100% water saturated zone to one that has identical rock
properties but 75% gas saturation in the pore space. As a matter
of interest, if the rock frame compressibility were assumed to
be 5.0 * 10^7 psi1, the computed gas/waterbearing velocity
would be 13,100 ft/sec. This is only 7% slower than the observed
waterbearing velocity and helps demonstrate the sensitivity of
the calculation to the assumption of rock frame modulus.
2. Soft Rock Example  Gassmann
Given:
Tertiary sandstone with 32% porosity.
grain density = 2.65 gm/cc at 7,000 ft
acoustic velocity = 8,130 ft/sec = 123 usec/ft
bulk density = 2.17 gm/cc at 100% water saturation
formation water = 225,000 ppm TDS
reservoir temperature = 184 degrees F
reservoir pressure = 7,300 psi
estimated net effective stress < 1,000 psi
What
percentage change in velocity could be expected if 75% of the
pore space were filled with methane gas (gas gravity 0.8)?
Kc = 13.3 * 10^10 dynes/cm2
Km taken as 37.9 * 10^10 dynes/cm2
Estimated Cb = 3.5 * 10^6 psi^1 from experience.
Compressibility
values on poorly consolidated formations are difficult to obtain.
It is believed that many operators are attempting to measure and
catalogue such data. However, most of these data are being held
confidential.
Kb = 1 / Cb = 1.97 * 10^10 dynes/cm2
Kw = 3.71 * 10^10 dynes/cm2
Kg = 0.186 * 10^10 dynes/cm2
Kf = 0.244 * 10^10 dynes/cm2
DENSg = 0.32 gm/cc
DENSw = 1.15 gm/cc
DENSf = 0.528 gm/cc
Kb @ 100% water saturation = 4.59 * 10^10 dynes/cm2
Kc @ 75% gas saturation = 5.27 * 10^10 dynes/cm2
DENS @ 75% gas saturation = 1.97 gm/cc
Vp @ 75% gas saturation = 5,370 ft/sec
The
velocity change computed with these assumptions is approximately
34%. However, the computed velocity of 5,370 feet per second,
or 186 microseconds per foot, compares reasonably well to observed
sonic log values in gas zones having rock and reservoir properties
similar to the assumptions made in this calculation.
As
a matter of interest if the bulk compressibility of the rock frame
is assumed to be 3.0 * 10^5 psi1, or an increase of nearly an
order of magnitude, the computed Gasmann velocity for 75% gas
is 5,000 feet per second. This is an additional 4.5% decrease
in velocity from the waterbearing case.
3.
Hard Rock Example  Wyllie
Same rock properties as Case 1:
porosity (PHIe) = 0.20
shale content (Vsh) = 0.00
water saturation (Sw) = 0.25
shale travel time (DELTsh) = 70 usec/ft
matrix travel time (DELTma) = 44 usec/ft
water travel time (DELTw) = 189 usec/ft
gas pseudotravel time (DELTh) = 250 usec/ft
Solving
the equation:
DELT = (1  0.20  0.00) * 44 + 0.00 * 70 + 0.20 * 0.25 * 189
+ 0.20 * 0.75 * 250
= 82.15 usec/ft
Vp = 10^6 / DELT = 12,170 ft/sec
(compared to 12,500 ft/sec for Gasmann equation)
4.
Soft Rock Example  Wyllie
Same rock properties as Case 2:
porosity (PHIe) = 0.32
water saturation (Sw) = 0.25
shale volume (Vsh) = 0.10
shale travel time (DELTsh) = 90 usec/ft
matrix travel time (DELTma) = 55.5 usec/ft
water travel time (DELTw) = 189 usec/ft
gas pseudotravel time {DELTh) = 550 usec/ft
Solving
the equation:
DELT = (1  0.32  0.10) * 55.5 + 0.10 * 90 + 0.32 * 0.25 * 189
+ 0.32 * 0.75 * 550
= 188.3 usec/ft
Vp = 10^6 / 188.3 = 5,310 ft/sec
(compared to 5,370 ft/sec for Gasmann equation)
Note
that the result depends largely on the choice of gas pseudotravel
time, and this is subject to some error in judgment.
