shaly sand equations
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sistivity
5
0.401.&91.95
.1%
.%4
.5
. '%.1&
%.54%.9'4.445.005.'4'.%&. 0
.19.09
10.011.0111.&91 .%1 .49
0.01 0.10 1.00
1.00
10.00
100.00
1000.00
10000.00
00.050.10
0. 00.40
w
* t
Vsh
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Shaly sand equations: Estimate of Sw from resistivity and shale conten
Equations:
Poupon+ 1954 ,ndonesia + 19&1
Input: Material parameter Type into the yellow fields the parameters
10.00 Ohm m
0.05 Ohm m
&.00 Ohm m
0. 0m = .00 deri!ed parameter: F =
n = .00
Calculation:
-aria$le Poupon ,ndonesia imandouVsh Sw Sw Sw
0.00 0.%54 0.%54 0.%540.05 0.%% 0.%4' 0.%490.10 0.%11 0.%% 0.%450.15 0. 9 0.% 9 0.%400. 0 0. '& 0.% 0.%%'0. 5 0. 45 0.%14 0.%%0.%0 0. 4 0.%0& 0.%0.%5 0. 0 0.%00 0.% 40.40 0.1&9 0. 94 0.% 00.45 0.15& 0. 0.%1'0.50 0.1%4 0. % 0.%1
R t=
R w
=
R sh =
Φ =
R o =
0.00 0.05 0.10 0.150.000
0.050
0.100
0.150
0. 00
0. 50
0.%00
0.%50
0.400
0.450
0.500
) w
.
S w = [ R w⋅( 1 R t
− V sh R sh )⋅
1φ m⋅
11 − V sh ]
1
n S w = {V sh(
1
−Vsh
2⋅
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imandou + 19'%
5
1. 5
0. 0 0. 5 0.%0 0.%5 0.40 0.45 0.50
Poupon,ndonesia
imandou
-/shale
S W = 1
2⋅ RW
φ m⋅
[√4 ⋅ φ
m
R W ⋅ ρ t
+
(V sh R sh )
2
− V sh R sh ]S w = {V sh
(1
−Vsh
2⋅√ R t
R sh+√ Rt
R0 }−
2
n
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(V sh R sh )
2
− V sh R sh ]