Calculation of true thickness of the bed

79) The Bulk resistivity of a carbonate formation having 10% porosity which is 75% saturated with hydrocarbons is 500Ωm. The bulk resistivity of the formation when the porosity is doubled and 100% saturated with water is ___________Ωm (round off to 1 decimal place).(Assume the tortuosity, cementation factor and saturation exponent to be 1, 2, and 2 respectively). (Thanks to Neeraja AU)

Solution: 

Water saturation S_w from Archies formula is

$S_w=\sqrt{\frac{R_o}{R_t}}$---(1)

R_o - Resistivity of the fully saturated rock

S_w – Water saturation of the uninvaded zone

R_w – Resistivity of the formation water

R_t –True Resistivity of the formation

Formation Factor

$F=\frac{R_o}{R_w}$

$FR_w=R_o$----(2)

Substitute eq.(2) in eq.(1)

$S_w=\sqrt{\frac{FR_w}{R_t}}$---(3)

Formation factor, $F=\frac{a}{\phi ^m}$---(4)

a = tortuosity factor = 1

ϕ – porosity

eq.(4) substitute in eq.(3)

$S_w=\sqrt{\frac{a}{\phi^m}\frac{R_w}{R_t}}$----(5)

Given that

R_t =500Ωm

Porosity = 10% = 0.1

Hydrocarbon Saturation, S_h = 75% = 0.75

S_w+S_h=1

S_w = 1- S_h = 1 – 0.75 = 0.25

$S_w=\sqrt{\frac{a}{\phi^2}\frac{R_w}{R_t}}$

$0.25=\sqrt{\frac{1}{(0.1)^2}\times\frac{R_w}{500}}$

$(0.25)^2={\frac{1}{(0.1)^2}\times\frac{R_w}{500}}$

$\frac{R_w}{500}.\frac{1}{(0.1)^2}=(0.25)^2$

$R_w=(0.25)^2\times500\times(0.1)^2$

R_w = 0.3125 Ωm

Therefore,

R_t = ?

when ϕ = 20% = 0.2 (porosity doubled)

S_w = 100% = 1

$S_w=\sqrt{\frac{a}{\phi^2}\frac{R_w}{R_t}}$

$1=\sqrt{\frac{1}{(0.2)^2}.\frac{0.3125}{R_t}}$

$\frac{0.3125}{R_t}.\frac{1}{(0.2)^2}=1$

$R_t(0.2)^2=0.3125$

$R_t= \frac{0.3125}{(0.2)^2}$

R_t= 7.81 Ωm