64) In a time domain(T-D) induced polarization experiment
with a steady voltage of 10 mV during the current flow interval, the voltage
drop at after the current cut-off is given by v(t)=4.0 e-0.3t. The
Chargeability after current cut-off between t1 =1s and t2 =
4s is ____mS
(Thanks to Chandrasekhar, ANU)
Solution:
The
chargeability for the time domain IP measurement is defined as
M=\frac{1}{V_{c}}\int_{t_{1}}^{t_{2}}V(t) dt
Where
V(t) and Vc have the same units and the chargeability M is in
milliseconds.
Given
that Vc is steady voltage= 10mV
V(t)
is voltage decay = 4.0 e-0.3t
Then
the Chargeability M=\frac{1}{V_{c}}\int_{t_{1}}^{t_{2}}V(t) dt
M=\frac{1}{10}\int_{1}^{4}4
e^{-0.3t} dt
M=\frac{4}{10}\int_{1}^{4}
e^{-0.3t} dt
M=\frac{2}{5}\left\{\frac{1}{-0.3}\left[e^{-0.3t}\right]_1^4
\right\}
M=\frac{-2}{5\times
0.3}\left[e^{-0.3\times4}-e^{-0.3\times1}\right]
M=\frac{-2}{5\times
0.3}\left[0.3-0.74\right] dt
M=\frac{-2\times
0.44}{5\times 0.3}
M=\frac{0.88}{1.5}
M= 0.586 seconds
M= 586 milli seconds
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