
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.
This
webpage version is the copyrighted intellectual
property of the author.
Do not copy or distribute in any form without explicit
permission. 
Simandoux saturation
The response equation for the resistivity log follows the classical
form also, but it must be linearized, so the result does not often
look similar to the usual response equation. To linearize it,
porosity is replaced by (1  Vsh) * Sw/F, and the equation is
set up in terms of conductivity instead of resistivity.
This
guarantees that the response falls back to the Archie equation
when Vsh = 0.0. Thus:
1:
COND = Sw / F * Sw * CONDw * (1  Vsh) (water term)
+ Sw / F * (1  Sw) * CONDh * (1  Vsh) (hydrocarbon term)
+ Sw / Fsh * CONDsh (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
CONDsh = log reading in 100% 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, CONDsh equals 1/ RSH, and CONDw is equal
to 1/RW@FT. For the Simandoux equation, 1/Fsh is set equal to
Vsh/2. Various authors have proposed other assumptions for the
term 1/Fsh, so several forms of this equation are in use.
This
gives:
2 / RESD = (PHIe ^ M) / (A * RW@FT} * (1  Vsh) * Sw ^ N
+ Vsh
/ (2 * RSH) * Sw
When
N = 2.0, this can be solved for Sw by using the standard quadratic
equation solution.
The
first term is the standard Archie equation for the nonshale portion
of the matrix, and the second term is a correction factor for
the shale fraction. The Simandoux equation is one of the most
popular and successful equations in common use which includes
a shale correction for the saturation calculation.
3:
IF PHIe > 0.0
4: THEN C = (1  Vsh) * A * (RW@FT) / (PHIe ^ M)
5: D = C * Vsh / (2 * RSH)
6: E = C / RESD
7: Sws = ((D ^ 2 + E) ^ 0.5  D) ^ (2 / N)
8: OTHERWISE Sws = 1.0
WHERE:
A = tortuosity exponent (fractional)
C = intermediate term in Simandoux equation
D = intermediate term in Simandoux equation
E = intermediate term in Simandoux equation
M = cementation exponent (fractional)
N = saturation exponent (fractional)
PHIe = effective porosity from any method (fractional)
RESD = deep resistivity log reading (ohmm)
RSH = resistivity of shale (ohmm)
RW@FT = water resistivity at formation temperature (ohmm)
Sws = water saturation from Simandoux method (fractional)
Vsh = shale content (fractional)
COMMENTS:
The Simandoux equation requires a three chart solution, given
below. Because of their complexity,
charts are not recommended.




Nomographs for solving Simandoux equation 
There
are many variations of the Simandoux equation. This version seems
to work well in all areas of the world; in all environments.
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. Since Sands A, B and C are 100% clean, the saturation from
Simandoux will be the same as for the Archie method. For Sand
D:
RSH = 4.0 ohmm
PHIe = 0.11
RESD = 1.0 ohmm
A = 0.62
M = 2.15
N = 2.00
RW@FT = 0.015 ohmm
C = (1  0.33) * 0.62 * 0.015 / (0.11 ^ 2.15) = 0.717
D = 0.717 * 0.33 / (2 * 4.0) = 0.0296
E = 0.717 / 1.0 = 0.717
Sws = ((0.0296 ^ 2 + 0.717) ^ 0.5  0.0296) ^ (2 / 2.0) = 0.81
This
compares with an Sw of 1.03 for the Archie method.
