GATE-2018
78.
The apparent resistivities obtained at 0.1 Hz and 10 Hz in the
frequency domain I.P. measurement are 100 ohm m and 80 ohm m,
respectively. The Percentage Frequency Effect is_________.
Solution:
f
and F are the frequencies (F>f) and f=0.05 to 0.5 Hz and F=1-10 Hz.
Then $\rho_f$ >$\rho_F$ then, frequency effect is
$FE
= \frac{\rho_f-\rho_F}{\rho_F}---(1)$
The
ratio FE is often multiplied by 100 to express it as a precentage
frequecy effect (PFE)
$PFE
= 100*\frac{\rho_f-\rho_F}{\rho_F}---(2)$
$PFE
= 100*\frac{\rho_{0.1}-\rho_{10}}{\rho_{10}}----(3)$
$\rho_f$
- 100 ohm m
$\rho_F$
- 80 ohm m
Substitute
above values in eq (3)
$PFE
= 100*\frac{100-80}{80}$
$PFE
= 100*\frac{20}{80}$
$PFE
= 100*0.25$
$PFE
= 25$
Reference
: Fundamental of Geophysics, William Lowrie.
Post a Comment
Post a Comment