dual water Saturation method
This method should be used in conjunction with the bulk volume water (dual water) porosity method described elsewhere. It uses a response equation slightly different than that for the Simandoux equation, and also the value for 1/Fsh is found differently.

      1: COND = Sw / F * Sw * CONDw * (1 - Vsh) (water term)
                   + Sw / F * (1 - Sw) * CONDh * (1 - Vsh) (hydrocarbon term)
                  + Sw / Fsh * CONDwsh (shale term)
                  + (1 - Vsh - PHIe) * Sum (Vi * CONDi) (matrix term)

WHERE:
  CONDh = log reading in 100% hydrocarbon
  CONDi = log reading in 100% of the ith component of matrix rock
  COND = log reading
  CONDwsh = log reading in bound water fraction of shale
  CONDw = log reading in 100% water
  F = formation factor of clean rock (fractional)
  Fsh = formation factor of shale (fractional)
  Sw = water saturation in un-invaded zone (fractional)
  Vi = volume of ith component of matrix rock
  Vsh = volume of shale (fractional)

CONDh and CONDi both equal 0.0 so these two terms disappear. 1/F is equal to (PHIe^M)/A, CONDwsh equals 1/ RWSH, and CONDw is equal to 1/RW@FT. For this equation, 1/Fsh is set equal to
Vsh*(BVWSH^M)/A, where BVWSH equals the apparent porosity of 100% shale.

Instead of solving for Sw by the quadratic solution, the bulk volume water method is solved directly by calculating Ro, the resistivity of the rock as if Sw = Vsh = 0.0, then solving for total water saturation by comparing Ro with RESD. Finally effective water saturation is obtained by comparing total porosity, effective porosity, total water saturation, and bound water volume.

The method is also called the dual water method.

Swd - Water Saturation from Bulk Volume Water
Calculate the apparent water resistivity in shale.
      2: IF PHIe > 0.0
      3: THEN RWSH = (BVWSH ^ M) * RSH / A

Calculate the fraction of water in the shale relative to the total porosity.
      4: D = Vsh * BVWSH / PHIt

Calculate the resistivity of the zone as if it were 100% wet.
      5: Ro = A * RW@FT * RWSH / (PHIt ^ M) / (RWSH + D * (RW@FT - RWSH))

Calculate total water saturation.
      6: Swt = (Ro / RESD) ^ (l / N)

Calculate effective water saturation.
      7: Swd = (PHIt / PHIe) * (Swt - D)
      8: OTHERWISE Swd = 1.0
 

Caution: Various trims are needed in real computer programs to eliminate silly answers ; Swt cannot be allowed to exceed 1.0 and Swb cannot be less than 0.0.

WHERE:
  A = tortuosity exponent (fractional)
  D = fraction of bound water relative to total porosity (fractional)
  BVWSH = bulk volume of water attached to shale, also known as bound water volume (fractional)
  M = cementation exponent (fractional)
  N = saturation exponent (fractional)
  PHIe = effective porosity (fractional)
  PHIt = total porosity (fractional)
  PHIuse = useful porosity (fractional)
  Ro = equivalent resistivity of wet formation (ohm-m)
  RSH = resistivity of shale (ohm-m)
  RWSH = water resistivity of water bound to shale (ohm-m)
  RW@FT = water resistivity at formation temperature (ohm-m)
  Swb = water saturation in effective porosity (fractional)
  Swt = water saturation in total porosity (fractional)
  SWuse = water saturation in useful porosity (fractional)
  Vsh = shale content (fractional)

COMMENTS:
The method is also called the dual water method since there are two water resistivities being considered - the water in the pore space and the water bound to the shale.

The term RWSH is the apparent water resistivity (Rwa) calculated from the resistivity and the apparent porosity of the shale. RWSH is also inverted and referred to as the conductivity of bound water (Cwb) in some technical papers. Do not confuse Cwb with CBW, which is the volume of clay bound water (CBW = Vsh * BVWSH).

Thus: RWSH = (PHIt ^ M) * RESD / A (in 100% shale)
OR RWSH = (BVWSH ^ M) *RSH /A
and: Cwb = 1000 / RWSH

WHERE:
  PHIt = BXWSH = total porosity in 100% shale
  RESD = RSH = resistivity of 100% shale

No graphical solution is provided for these formulae.

RECOMMENDED PARAMETERS:
RSH Range = 1.0 to 20.0 Default = 4.0
for sandstone A = 0.62  M = 2.15  N = 2.00
for carbonates A = 1.0  M = 2.00  N = 2.00

Asquith (1980 page 67) quoted other authors, giving values for A and M, with N = 2.0, showing the wide range of possible values:

Average sands sands    A = 1.45  M = 1.54
Shaly sands                  A = 1.65  M = 1.33
Calcareous sands         A = 1.45  M = 1.70
Carbonates                   A = 0.85  M = 2.14
Pliocene sands S.Cal.  A = 2.45  M = 1.08
Miocene LA/TX             A = 1.97  M = 1.29
Clean granular             A = 1.00  M = 2.05 - PHIe
 

NUMERICAL EXAMPLE:
1. Data for Sand "D"
PHID = 0.12
PHIN = 0.28
DENSMA = 2650 Kg/m3 (no offset)
PHIDSH = 0.03
PHINSH = 0.30
Vsh = (0.28 - 0.12 ) / (0.30 – 0.03) = 0.59
Select PHIDDC = - 0.13 by calculating dry clay porosity from dry clay density,
PHINDC = 1.00 - (1.00 - (- 0.13)) * (1.00 - 0.30) / (1.00 - 0.03) = 0.184
BVWSH = (0.184 * 0.03 - (-0.13) * 0.30) / (0.184 - (-0.13)) = 0.142
PHIt = (0.184 * 0.12 - (-0.13) * 0.28) / (0.184 - (-0.13)) = 0.186
PHIbvw = 0.186 - 0.59 * 0.142 = 0.103 = PHIe

2. Select Resistivity data for Sand “D”
RSH = 4.0 ohm-m
RESD = 1.0 ohm-m
RW@FT = 0.015 ohm-m
A = 0.62
M = 2.15
N = 2.00

RWSH = (0.142 ^ 2.15) * 4.0 / 0.62 = 0.097
D = 0.59 * 0.142 / 0.186 = 0.450
Ro = 0.62 * 0.015 * 0.097 / (0.186 ^ 2.15) / (0.097 + 0.450 * (0.015 - 0.097)) = 0.558
Swt = (0.558 / 1.0) ^ (1 / 2) = 0.75
Swb = (0.186 / 0.103) * (0.75 - 0.450) = 0.54

The Archie model gave SW = 1.03 and Simandoux gave SW = 0.81 (with Vsh = 0.33) for this sand. When RW is higher, Simandoux and Dual Water models give similar results. When RSH is higher (RSH > 10), both models approach the Archie solution.

3. Assume Vsh = 0.33 from GR instead of 0.59 from density neutron separation
BVWSH = (0.30 + 0.03) / 2 = 0.165 - simplified approximation, close to detailed method
PHIt = (0.28 + 0.12) / 2 = 0.200 - simplified approximation, close to detailed method
PHIe = 0.200 – 0.33 * 0.165 = 0.145 - higher than before because Vsh is lower

RWSH = (0.165 ^ 2.15) * 4.0 / 0.62 = 0.134
D = 0.33 * 0.165 / 0.200 = 0.272
Ro = 0.62 * 0.015 * 0.134 / (0.200 ^ 2.15) / (0.134+ 0.272 * (0.015 - 0.134)) = 0.390
Swt = (0.390 / 1.0) ^ (1 / 2) = 0.62 - a little lower than before because Vsh is lower
Swb = (0.200 / 0.165) * (0.62 - 0.272) = 0.42 - a little lower than before because Vsh is lower

The dual water model for saturation can be used for SW even if the shale volume and porosity come form some other method.
 

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