
Publication History:
This article is based on
Chapter 8 of "The Log Analysis Handbook" by E. R. Crain, P.Eng., published by Pennwell Books 1986 Updated 2004,
2016, 2018.
This
webpage version is the copyrighted intellectual
property of the author.
Do not copy or distribute in any form without explicit
permission. 
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 uninvaded 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.
Water Saturation from
Dual Water Model
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 (ohmm)
RSH = resistivity of shale (ohmm)
RWSH = water resistivity of water bound to shale (ohmm)
RW@FT = water resistivity at formation temperature (ohmm)
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:
Reference:
1. The Theoretical and Experimental Bases
for the Dual Water  Model for Interpretation of Shaly Sands
C. Clavier, G. Coates, J. Dumanoir,
SPWLA, 1977
The method is 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
carbonates A = 1.00
M = 2.00
N = 2.00 (Archie Equation as first published)
for sandstone A = 0.62
M = 2.15
N = 2.00 (Humble Equation)
A = 0.81 M = 2.00 N = 2.00 (Tixier Equation 
simplified version of Humble Equation)
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 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 ohmm
RESD = 1.0 ohmm
RW@FT = 0.015 ohmm
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.
