Electrical and electromagnetic methods used in geophysics.ppt
- 1. GG 450 February 25, 2008 ELECTRICAL Methods Resistivity
- 2. Electrical Methods There are many electrical and electromagnetic methods used in geophysics. These methods are most often used where sharp changes in electrical resistivity (resistance in the ground) are expected - particularly if resistivity decreases with depth.
- 3. Applied current Methods: when a current is supplied by the geophysicist. Currents are either DC or low frequency waves. In the electrical resistivity method, the potential difference (voltage) is measured at various points; in the induced polarization method, the rise and fall time of the electric potential are measured. The electromagnetic method applies an alternating current with a coil and the resulting field magnetic field is measured with another coil.
- 4. Natural Currents: when natural currents in the earth are measured. Movement of charge in the ionosphere and lightning cause telluric currents to be generated in the earth. Variation of the spectra of these current fields and their magnetic counterparts yield information on subsurface resistivity. The self- potential method uses currents generated by electro-chemical reactions (natural batteries) associated with many ore bodies.
- 5. Your text discusses several of these methods in detail. We only have time to talk about one of them - the resistivity method. Some of the others will be discussed a bit when we look at well logging and Ground Penetrating Radar. First, a review of basic electricity:
- 6. Consider the circuit: i V R1 R2 - + battery current meter volt meter resistor A battery acts as an energy supply, pushing electrons around the circuit A resistor resists the flow of current A voltmeter measures the potential difference between two points A current meter measures the current flow at a point
- 7. This is equivalent to the water and pipe system: height difference is potential flow is current tanks with pump is battery friction is resist ance The voltage (potential) of a battery is equivalent to the water level difference between the two tanks.
- 8. Batteries are sold by the potential difference they maintain and by the amount of electricity (charge) they can deliver (size of the tank and strength of the pump). The “pump” is a chemical reaction that pulls electrons from one part of the battery to the other. The electrical current is equivalent to the flow of water. Electrical charge is equivalent to water. Resistance is equivalent to restriction of water flow, like the inverse of permeability. What’s the water equivalent of a dead battery? What’s the water equivalent of a short circuit? What’s the water equivalent of an open
- 9. BASIC EQUATIONS: current= charge/sec past a point: i=dq/dt = coulombs/sec=amperes current density = current/cross sectional area: j=i/A resistance= potential /current = Ohms Ω =volts/ampere (Ohm's LAW) Resistance tells us the total drag on the current, but not the property of the material that is generating the drag.
- 10. We need a measure of the resistance of a material. For a given pipe-shaped material we can define the resistivity as: resistivity = resistance x cross section area/ length: = R A/l. , with units of m. Copper has a very low resistivity (1.7x10-8 Ωm and quartz has a very high resistivity (1x1016 m. Copper is a CONDUCTOR because of its low resistivity and quartz is an INSULATOR. We can expect different geologic materials to
- 11. Exercise: In the NEPTUNE project, a cable 1500 km in length might be installed to service observatories. The cable had a copper conductor with a cross section diameter of 0.4 cm. If they send 10 amps down the cable, what will the voltage drop be from shore to the end of the cable? *(length)* A =1.7x10-8 m x 1500000m/(π (0.004)2 m2 ) = ~507total cable resistance V=iR =10 A*507=5.07 kVolts lost to heating the cable
- 12. As is the case for gravity and magnetics, we will find that electrical potential, measured in Volts, has the same properties as gravity and magnetic potentials, in that it is a scalar, and we can add the effects of different sources of potential to find out where current will flow. Current will flow in a direction normal to equipotential (equal voltage) surfaces. Rather than have current flow only through wires, we will now plug our wires into the earth and see how
- 13. Consider an electrode stuck in the ground with it's matching electrode far away (just like a magnetic monopole). It's potential relative to the distant electrode is measured in Volts. Battery current current equipotential
- 14. If we measure the potential difference between two shells at some a distance D from the electrode, we get € dV = iR = i ρ l A ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= i ρ dr 2πr2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ where dr is the thickness of the shell across which we measure the potential, Recalling that the resistivity of air is so high, no current will flow through it, so we only need have the surface of a hemisphere (2πr2 ).
- 15. We now integrate in from infinity (where potential is zero) to get the potential at a point a distance D from the source: V = dV = iρ 2π D ∞ ∫ dr r2 D ∞ ∫ = iρ 2πD IF the resistivity of the ground is UNIFORM. The current, i, above is the current IN THE WIRE, not the current in the ground, which varies.
- 16. This is the basic equation of resistivity, in that we can add the potentials from many sources to obtain a "potential" map of a surface. By contouring that map, we have equipotential lines, along which no current flows. Current flows in directions perpendicular to equipotential lines. Sound familiar? It should! Magnetic lines of force are perpendicular to magnetic equipotential surfaces, and the pull of gravity is perpendicular to gravity equipotential surfaces.
- 17. TWO ELECTRODES: What if we move the other current electrode in from far away? We can calculate the potential at point P1 by just adding the potentials from both current electrodes - remembering that one is positive, and the other negative: Battery d P1 z x
- 18. € VP 1 = iρ 2πr1 − iρ 2πr2 = iρ 2π 1 r1 − 1 r2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ What is the potential between the electrodes vs. depth? We just change the r values to x-z coordinates: € VP 1 = iρ 2π 1 d 2 + x ( ) 2 + z2 − 1 d 2 − x ( ) 2 + z2 ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪
- 19. 0 20 40 60 80 100 120 140 160 0 25 50 75 100 125 150 -350.0 -300.0 -250.0 -200.0 -150.0 -100.0 -50.0 0.0 potential depth distance from center Potential at depth This function yields the figure below for one side of the array (under one half of the array): and at the surface.
- 20. We can’t measure the potential below the surface in the field, though. We’re stuck at the surface. The surface potential looks like the profile below for a current of 1 A, 100 m between electrodes, and a resistivity of 10km Potential along surface -400.0 -300.0 -200.0 -100.0 0.0 100.0 200.0 300.0 400.0 -200 -150 -100 -50 0 50 100 150 200 Distance Potential
- 21. If we put voltage probes at –65m and +65m along the x axis above, what voltage would we see? Potential along surface -400.0 -300.0 -200.0 -100.0 0.0 100.0 200.0 300.0 400.0 -200 -150 -100 -50 0 50 100 150 200 Distance Potential
- 22. This allows us to contour equipotential lines, but how much current is flowing in what areas? Current flows ALL THROUGH the subsurface, not just directly from one electrode to the other. With some difficulty, it can be shown that the fraction of the total current (if) flowing above a depth z for an electrode separation d is given by: if = 2 π tan−1 2z d In a region of equal resistivity - about 70% of the current flows at depths shallower than the distance
- 23. With this information, we can sketch lines perpendicular to the equipotentials that show where most of the current is flowing. Be careful, though, this only works where the resistivity is constant throughout the model! Note that all current lines are perpendicular to equipotential lines - no current flows between two points where the potential is equal.
- 24. This is the pattern of equipotentials (blue) and current flow (red) expected for a constant-resistivity material.
- 25. The most common form of resistivity measurement uses two current electrodes and two potential electrodes:
- 26. We use the same argument, summing potentials, to obtain the voltage across two electrodes we get: i C 1 P 1 C 2 P 2 r 2 r 4 r 3 r 1 + - + - curren t potentia l The potential difference between P1 and P2 is: VP1 −P2 =VP1 −VP2 VP1 = VC1 +VC2 ( ) P1 , VP2 = VC1 +VC2 ( ) P2 VP1 −P2 = iρ 2πr1 − iρ 2πr2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟− iρ 2πr3 − iρ 2πr4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= iρ 2π 1 r1 − 1 r2 − 1 r3 + 1 r4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟
- 27. Solving for the resistivity,: € ρ = 2πVP 1 −P2 i 1 1 r1 − 1 r2 − 1 r3 + 1 r4 Thus, we can measure the current, voltage, and appropriate distances and solve for resistivity.
- 28. In the example above we have current electrodes at ±50m, and voltage electrodes at ± 65m, so: r1= 15m r2= 115 r3= 115 r4= 15 and the current is 1.0 A So we can solve for =(2 π*185 /1)*1/(1/15- 1/115-1/115+1/15)= 1025.6 m. Which is pretty close to the model value of 135 m. BUT, this is a boring model; what we really want to know is what to expect as the resistivity changes with depth.
- 29. RESISTIVITY THAT CHANGES WITH DEPTH Consider a single horizontal interface with a constant resistivity above and a different constant resistivity below. Burger presents a formula (5-18) to give the fraction of current that will penetrate into the lower layer, ( programmed in Table 5-3 with results in Figure 5-11). Note that the x axis of this plot is: SEE NOTES
- 30. CURRENT DENSITY AND FLOW LINES If we think about current flow lines crossing the boundary between two resistivities, it's almost like a seismic ray passing between two materials with different velocities - but the formula is different: € tanθ1 tanθ2 = ρ2 ρ1 1.276 tan( θ2 )/tan( θ1 )= ρ1/ρ2 y y θ2 θ1 ρ1 ρ2 0.111 ρ1/ρ2=1.276/0.111
- 31. 1.276 tan( θ2 )/tan( θ1 )= ρ1/ρ2 y y θ2 θ1 ρ1 ρ2 0.111 ρ1/ρ2=1.276/0.111 Note that this is equivalent to z1/z2=r1/r2, where z is the distance along the vertical axis. So, if we make zi proportional to ri then z2 is proportional to r2 , holding y constant, then we will get the current flow direction easily. € tanθ1 tanθ2 = ρ2 ρ1
- 32. Note that he current flow lines get closer together when the current moves into a region of lower reisistivity: implying that the current density increases as we cross to the lower resistivity material. If resistivity increases with depth, then current density decreases. If resistivity in a region is VERY high ( insulator), then few flow lines will cross a boundary with a conductor, and those that do will be directed perpendicular y y z1 equipotential
- 33. If we measure the resistivity when a horizontal resistivity boundary is present with a system like that shown on page 6, what would we get? We can define the APPARENT RESISTIVITY as the resistivity we would get assuming that no boundary or change in resistivity is present. So that apparent resistivity equation is identical to the equation for a material with constant resistivity: ρa = 2πΔVP1−P2 i 1 1 r1 − 1 r2 − 1 r3 + 1 r4 ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟
- 34. So, what does this tell us? Recall that current density is qualitatively measured by the number of flow lines. What is the relationship between potential difference V and current density j ? Since j=i/A, and i=V/R, and =RA/l, j=V/RA, or j=V/l. Thus, current density is proportional to potential within a tube extending along the flow line from the current electrode to the to the potential electrode. Does this mean that if
- 35. Consider measuring the potential between a wire and some point on the outside of insulation around that wire. We would measure a high potential difference between a point on the wire and a point outside the insulation - does that mean that the current across the wire through the insulation will be higher than the current through the wire?? !! What's wrong here? w ire insulat ion The POTENTIAL difference is a function of the BATTERY - not the material. So the higher the voltage of the battery, the higher the current density, but, across an insulator, the current will be very low because the
- 36. Variations in current density near the earth’s surface will be reflected in changes in potential difference, and will result in changes in apparent resistivity. Consider the cases below -
- 37. surface interface ρ1 ρ2 surface interface ρ1 ρ2 ρ1> ρ2 ρ1= ρ2 surface interface ρ1 ρ2 ρ1< ρ2 What happens if we move the layer up and down? If the interface is very deep, RELATIVE TO THE ELECTRODE SPACINGS, the lower layer should have no effect, and our readings shouldn't reflect its presence. How deep is very
- 38. What if we plot apparent resistivity vs. electrode spacing? As spacing increases, we should "feel" deeper and deeper. At some point, if our electrodes are far enough apart, the top layer will have considerably less effect than the bottom! ρ1> ρ2 ρa electrode spacing Top layer ρ bottom layer ρ ρ1< ρ2 This change in apparent resistivity with electrode spacing should give us the information we need to interpret data and determine the depth to an interface and the
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