Modeling the Sonic Log Response From Biot-Gassmann Equations
An alternate and more rigorous approach is the Biot-Gassmann 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 Biot-Gassmann 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 rock-frame data.

RECOMMENDED PARAMETERS:
NOTE: Units of compressibility in this table are PSI^-1.

Rock matrix bulk modulus and shear modulus data are listed in Section 21.17.

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: 1250 * 10^-10 depth 2000 ft 610 m
_______    ________________510 * 10^-10 _____4000 _1220
_________    ______________180 * 10^-10_____ 8000 _2440
___________    ____________100 * 10^-10 _____12000 3660
 

NUMERICAL EXAMPLES - Biot-Gassmann 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% water-bearing 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 psi-1 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 psi-1 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 psi-1 from handbook data
       Kg = 1 / Cg = 0.0428 * 10^10 dynes/cm2

Gas-water 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 psi-1, the computed gas/water-bearing velocity would be 13,100 ft/sec. This is only 7% slower than the observed water-bearing 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 psi-1, 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 water-bearing 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 pseudo-travel 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 pseudo-travel 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 pseudo-travel time, and this is subject to some error in judgment.
 

 

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