Environmental%20and%20Exploration%20Geophysics%20I

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Environmental and Exploration Geophysics I
Resistivity II
tom.h.wilson tom.wilson_at_mail.wvu.edu
Department of Geology and Geography West Virginia
University Morgantown, WV
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Redo and cross check with lec8.pdf (9/16/2010)
For next class complete in-class problems and
hand in next period (Thursday Sept. 24th). No
class next Tuesday (September 22nd). Read over
problems 5.1 through 5.3 in the text and be
prepared to ask questions about them next
Thursday. Ill summarize the approach to these
problems in class today. They will be due the
following Tuesday (Sept. 29th).
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Review of basic ideas presented last time
1. Assume a homogeneous medium of resistivity 120
ohm-m. Using a Wenner electrode system with a 60m
spacing, Assume a current of 0.628 amperes. A.
What is the measured potential difference? B.
What will be the potential difference if we place
the sink (negative-current electrode) at infinity?
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We know in general that
For the Wenner array the geometrical factor G is
2?a and the general relationship of apparent
resistivity to measured potential difference is
In this problem we are interested in determining
the potential difference when the subsurface
resistivity distribution is given.
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In part A) we solve for V as follows
and
In part B) what happens to d2 and d4?
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Part B)
d2 and d4 go to ?. We really cant think of
this as a simple Wenner array any longer. We have
to return to the starting equation from which
these array-specific generalizations are made.
What happens when d2 and d4 go to ??
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-

?
A
B
V
d1
d2
d4
d3
d1 60 m and d3 120 m. V is now only 0.1 volts.
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1B
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There are many types of arrays as shown at left.
You should have general familiarity with the
method of computing the geometrical factors at
least for the Wenner and Schlumberger arrays. The
resistivity lab you will be undertaking models
Schlumberger data and many of the surveys
conducted by Dr. Rauch and his students usually
employ the Wenner array.
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Note that when conducting a sounding using the
Wenner array all 4 electrodes must be moved as
the spacing is increased and maintained constant.
The location of the center point of the array
remains constant (despite appearances above).
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Conducting a sounding using the Schlumberger
array is less labor intensive. Only the outer two
current electrodes need to be moved as the
spacing is adjusted to achieve greater
penetration depth. Periodically the potential
electrodes have to be moved when the current
electrodes are so far apart that potential
differences are hard to measure - but much less
often that for the Wenner survey
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Homework problem 5.1a (See Burger et al. p. 341)
20m
source
sink
Surface
?
?
4m
12m
Depth
2m
?200?-m
?
?
P1
P2
Find the potential difference between points 1
and 2.
What kind of an array is this? What are d1, d2,
d3 and d4 ?
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Use basic equations for the potential difference.
The critical point here is that you accurately
represent the different distances between the
current and potential electrodes in the array.
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5.2. Current refraction rules
Given these resistivity contrasts - how will
current be deflected as it crosses the interface
between layers? Measure the incidence angle and
compute the angle of refraction.
Actually calculate the angles!
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Whats your guess?
?2 gt ?1
tan ? increases with increasing angle
?2 lt ?1
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?1
?
?2 gt ?1
?2
?2 varies as ?1 and ?1 varies as ?2
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?
?2 gt ?1
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?
?2 lt ?1
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?
?2 gt ?1
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Incorporating resistivity contrasts into the
computation of potential differences.
5.3. Calculate the potential at P1 due to the
current at C1 of 0.6 amperes. The material in
this section view extends to infinity in all
directions. The bold line represents an interface
between mediums with resistivities of ?1 and ?2.
Lets consider the in-class problem handed out to
you last lecture.
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In-Class/Takehome Problem 2 In the following
diagram - Suppose that the potential difference
is measured with an electrode system for which
one of the current electrodes and one of the
potential electrodes are at infinity. Assume a
current of 0.5 amperes, and compute the potential
difference between the electrodes at PA and ?.
Given that d1 50m, d2 100m, ?1 30?-m, and
?2 350?-m.
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Current reflection and transmission
Sink
?
Source Electrode
?130?-m
d1
One potential electrode
PC
a
PA
PB
b
?2350?-m
d2 ab
Image point
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At PA
?130?-m
Some current will be transmitted across this
interface and a certain amount of current (k)
will be reflected back into medium 1.
d1
PA
a
?
PA
PA
?2350?-m
b
d2 ab
Reflection point
Image point
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Use of the image point makes it easy to estimate
the length along the reflection path
Path length is distance from image point to PA.
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Potential measured at A
k is the proportion of current reflected back
into medium 1. k is also known as the reflection
coefficient.
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Potential measured at point B
1-k is the transmission coefficient or proportion
of current incident on the interface that is
transmitted into medium 2.
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Potential measured at point C
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Potentials a hair to the left or right of the
interface should be approximately equal.
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Incorporating resistivity contrasts into the
computation of potential differences. 3.
Calculate the potential at P1 due to the current
at C1 of 0.6 amperes. The material in this
section view extends to infinity in all
directions. The bold line represents an interface
between mediums with resistivities of ?1 and ?2.
Locate an image electrode and incorporate
reflection process,
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Looking ahead
In the remainder of todays class last chance
for questions about the Terrain Conductivity lab.
Put in my box Wednesday, 30th.
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Uncontaminated fresh water acquifer
Not how the apparent conductivities drop with
increased depth of penetration
Near-surface clay
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40
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Fresh water aquifer
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Basal silty clay
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More on equivalence
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Constructing a cross section view of your model
results