dual water Saturation
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.
= 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)
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)
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
where BVWSH equals the apparent porosity of 100% shale.
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.
method is also called the dual water method.
Swd - Water Saturation from Bulk Volume Water
the apparent water resistivity in shale.
2: IF PHIe > 0.0
3: THEN RWSH = (BVWSH ^ M) * RSH / A
the fraction of water in the shale relative to the total porosity.
4: D = Vsh * BVWSH / PHIt
the resistivity of the zone as if it were 100% wet.
5: Ro = A * RW@FT * RWSH / (PHIt ^ M) / (RWSH + D * (RW@FT - RWSH))
total water saturation.
6: Swt = (Ro / RESD) ^ (l / N)
effective water saturation.
7: Swd = (PHIt / PHIe) * (Swt - D)
8: OTHERWISE Swd = 1.0
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.
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)
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.
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).
RWSH = (PHIt ^ M) * RESD / A (in 100% shale)
OR RWSH = (BVWSH ^ M) *RSH /A
and: Cwb = 1000 / RWSH
PHIt = BXWSH = total porosity in 100% shale
RESD = RSH = resistivity of 100% shale
graphical solution is provided for these formulae.
RSH Range = 1.0 to 20.0 Default = 4.0
for sandstone A = 0.62
M = 2.15
N = 2.00
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
A = 1.65 M = 1.33
A = 1.45 M = 1.70
A = 0.85 M = 2.14
Pliocene sands S.Cal. A = 2.45 M = 1.08
A = 1.97 M = 1.29
A = 1.00 M = 2.05 - PHIe
1. Data for Sand "D"
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
PHINDC = 1.00 - (1.00 - (- 0.13)) * (1.00 - 0.30) / (1.00 - 0.03)
BVWSH = (0.184 * 0.03 - (-0.13) * 0.30) / (0.184 - (-0.13)) =
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
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
= (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
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.
Assume Vsh = 0.33 from GR instead of 0.59 from density neutron
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
= (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
dual water model for saturation can be used for SW even if the
shale volume and porosity come form some other method.