Download presentation

Presentation is loading. Please wait.

Published byAbigale Marson Modified over 2 years ago

1
Environmental and Exploration Geophysics I tom.h.wilson tom.wilson@mail.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Resistivity II

2
For next class complete in-class problems and hand in next period (Thursday – Sept. 24 th ). 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. I’ll summarize the approach to these problems in class today. They will be due the following Tuesday (Sept. 29 th ). Redo and cross check with lec8.pdf (9/16/2010)

3
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? A + - B d1d1 d4d4 d2d2 d3d3 V

4
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.

5
In part A) we solve for V as follows A + - V B d1d1 d4d4 d2d2 d3d3 and In part B) what happens to d 2 and d 4 ?

6
… d 2 and d 4 go to . We really can’t 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 d 2 and d 4 go to ? Part B)

7
A + - V B d1d1 d4d4 d2d2 d3d3 d 1 = 60 m and d 3 = 120 m. V is now only 0.1 volts.

8
1B

9
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.

10
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).

11
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

13
Homework problem 5.1a (See Burger et al. p. 341) What kind of an array is this? What are d 1, d 2, d 3 and d 4 ? Find the potential difference between points 1 and 2. 2m sourcesink 20m 12m 4m P1P1 P2P2 Surface =200 -m Depth

14
The critical point here is that you accurately represent the different distances between the current and potential electrodes in the array. Use basic equations for the potential difference.

15
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!

17
tan increases with increasing angle What’s your guess? 2 > 1 2 < 1

18
11 22 2 varies as 1 and 1 varies as 2 2 > 1

20
2 < 1

21
2 > 1

22
Incorporating resistivity contrasts into the computation of potential differences. Let’s consider the in-class problem handed out to you last lecture. 0 5m 1 =200 -m 1 =50 -m C1C1 P1P1 5.3. Calculate the potential at P 1 due to the current at C 1 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.

23
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 P A and . Given that d 1 = 50m, d 2 = 100m, 1 = 30 -m, and 2 = 350 -m.

24
Current reflection and transmission d1d1 d 2 = a+b 1 =30 -m 2 =350 -m Image point a b PAPA PBPB PCPC One potential electrode Source Electrode Sink

25
d1d1 d 2 = a+b 1 =30 -m 2 =350 -m Image point a b PAPA PAPA PAPA Reflection point Some current will be transmitted across this interface and a certain amount of current (k) will be reflected back into medium 1. ? At P A

26
Use of the image point makes it easy to estimate the length along the reflection path Path length is distance from image point to P A.

27
Potential measured at A k is the proportion of current reflected back into medium 1. k is also known as the reflection coefficient.

28
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.

29
Potential measured at point C

32
Potentials a hair to the left or right of the interface should be approximately equal.

33
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, 0 5m 1 =200 -m 1 =50 -m C1C1 P1P1

34
Looking ahead In the remainder of today’s class – last chance for questions about the Terrain Conductivity lab. Put in my box Wednesday, 30 th.

35
Uncontaminated fresh water acquifer Not how the apparent conductivities drop with increased depth of penetration 20 30 40 50 Fresh water aquifer Near-surface clay Basal silty clay

38
Constructing a cross section view of your model results

Similar presentations

OK

Direct current injection is required. Can be difficult to get current into the ground so that little or no current reaches the zone of interest and hence.

Direct current injection is required. Can be difficult to get current into the ground so that little or no current reaches the zone of interest and hence.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on tsunami in india Free download ppt on effective communication skills Nokia acquired by microsoft ppt online Ppt on c language fundamentals File type ppt on cybercrime virus Ppt on condition monitoring solutions Ppt on minimum wages act delhi By appt only movie john English ppt on tenses Ppt on right to education act