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 ]