Environmental and Exploration Geophysics I. Resistivity II tom.h.wilson

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

2 Review of basic ideas presented last time. Assume a homogeneous medium of resistivity 20 ohm-m. Using a Wenner electrode system with a 60m spacing, Assume a current of 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 d d 2 d 3 d 4 We know in general that G ρ a = i For the Wenner array the geometrical factor G is 2πa and the general relationship of apparent resistivity to measured potential difference is ρ a 2πa = i In this problem we are interested in determining the potential difference when the subsurface resistivity distribution is given. 2

3 In part A) we solve for as follows iρ = a 2πa (0.628amperes)(20Ω m) = = 0.2volts( ampere Ω) 2π (60m) + - and A B d d 2 d 3 d 4 In part B) what happens to d 2 and d 4? Part B) 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. iρ = + 2π d d2 d3 d4 What happens when d 2 and d 4 go to? 3

4 4 + = d d i π ρ + = d d d d i π ρ = 3 2 d d i π ρ A + - B d d 4 d 2 d 3 d = 60 m and d 3 = 20 m. is now only 0. volts. = 3 2 d d i π ρ Ω = m m m amperes ) )(20 (0.628 π Ω = m m amperes 20 2 ) )(20 (0.628 π ) (. 0 Ω = ampere volts B

5 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. 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). 5

6 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 6

7 Homework problem 5.a (See Burger et al. p. 34) 20m source 4m Surface sink Depth 2m ρ=200ω-m 2m P P 2 Find the potential difference between points and 2. What kind of an array is this? What are d, d 2, d 3 and d 4? Use basic equations for the potential difference. iρ 2 = + 2π d d2 d3 d4 iρ 2 = + 2π d d2 d3 d4 The critical point here is that you accurately represent the different distances between the current and potential electrodes in the array. 7

8 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! 8

9 What s your guess? tan θ increases with increasing angle ρ tan = ρ2 tan θ2 θ ρ 2 > ρ ρ 2 < ρ θ ρ 2 > ρ θ 2 ρ tan = ρ2 tan θ2 θ ρ 2 varies as θ and ρ varies as θ 2 9

10 ρ 2 > ρ ρ 2 < ρ 0

11 ρ 2 > ρ Incorporating resistivity contrasts into the computation of potential differences Calculate the potential at P due to the current at C 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 ρ and ρ2. 0 5m C ρ =50 Ω-m P ρ =200Ω-m Let s consider the in-class problem handed out to you last lecture.

12 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 = 50m, d 2 = 00m, ρ = 30Ω-m, and ρ 2 = 350Ω-m. Current reflection and transmission Source Electrode Sink ρ =30Ω-m One potential electrode d a P C P A d 2 = a+b b ρ 2 =350Ω-m P B Image point 2

13 At P A Some current will be transmitted across this interface and a certain amount of current (k) will be reflected back into medium.? d P A b d 2 = a+b a ρ =30Ω-m P A ρ 2 =350Ω-m P A Reflection point Image point 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. 3

14 Potential measured at A k is the proportion of current reflected back into medium. k is also known as the reflection coefficient. Potential measured at point B -k is the transmission coefficient or proportion of current incident on the interface that is transmitted into medium 2. 4

15 Potential measured at point C 5

16 Potentials a hair to the left or right of the interface should be approximately equal. 6

17 Incorporating resistivity contrasts into the computation of potential differences. 3. Calculate the potential at P due to the current at C 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 ρ and ρ2. 0 5m C ρ =50 Ω-m P ρ =200Ω-m Locate an image electrode and incorporate reflection process, Looking ahead Hand in the in-class problems next Thursday Hand in homework problems 5.-3 the following Tuesday Sept.29 th. Bring questions about them to class next Thursday We ll be begin a resistivity computer lab exercise next week. The data come from a paper which is posted on the web site. In the remainder of today s class last chance for questions about the Terrain Conductivity lab. Put in my box Wednesday, 30 th. 7

18 Uncontaminated fresh water acquifer Not how the apparent conductivities drop with increased depth of penetration Near-surface clay Fresh water aquifer Basal silty clay 8

19 Constructing a cross section view of your model results 9