environmental and exploration geophysics ipages.geo.wvu.edu/~wilson/geol454/lect3/lec3.pdf · tom...

Tom Wilson, Department of Geology and Geography

Environmental and Exploration Geophysics I

tom.h.wilson

[email protected]

Department of Geology and Geography

West Virginia University

Morgantown, WV

Terrain Conductivity

Phone - 293-6431

In – class problem


The first two problems

are based on 1 and 2 as

described.

Problem 3 is based on

the text problem 8.5

diagramed in the

handout.

Show computations


2 0.680

points

show calculations

20.545

3.67

dz

s

Show ratio and

calculation

Show your

calculations

Without calculations

Save all assignments for

reference and grade

cross check

Archie’s law, a and m


The easiest way to figure out the value of the constant a is

to take the log transform of F. The intercept or

logF(log=0) =a, the intercept.

mF a

logF(log=0)=-

0.0969

So a=10-0.0969=0.8

a generally varies between 0.5 and 1.5; m, between 1.7 and 4.1

Main objective for the day is to develop an understanding

of the computational approach discussed in the text on

pages 514-518. Dig into the calculation details

For more detail see McNeill’s technical note on EM Conductivity

at Low Induction Number.

Objective for the day

• Basic definitions

• Units conversions [ohm/m <>mmhos/m]

• Discuss computational procedures used to compute a.

• We will introduce ideas presented in McNeill’s Technical Note to

provide broader context and background on the computational

approach.

• Relative and cumulative response functions will be introduced.

• Class problems will be limited to vertical dipole computations

outlined by Berger et al.

• We’ll go into much more detail than Berger with the objective of

giving you some understanding of the underlying basis for the

computation.

• Two and 3 layer problems

• In class problem

• Problem assignments from the book

Topic list


1) The function R is incorrectly referred to as the relative response

function; it is the cumulative response function.

2) There is an error in the math (e.g. page 517). Basically you

cannot incorporate the instrument height into the solution the way

they do. We will always assume that instrument is on the ground.

So make a note ….

Z should not include instrument height above the ground!

Assume that the coils are on the ground surface.

Z= (depth)/(intercoil spacing)

Depth is depth beneath surface.

Any equation they use that includes depth plus 1 meter (instrument

height) is incorrect.

Recall those text Errors!

( ) / ( ) 8.22z Depth Instrument height Intercoil Spacing

8.22 ( ) / ( ) 8.22should read z Depth Intercoil Spacing

For the problems you have to do we will always

assume the instrument is on the ground


8.22 ( ) / ( ) 8.22should read z Depth Intercoil Spacing

Some elementary physics ideas to re-familiarize yourself with …

• Right hand rule

• Coulomb law (inverse square relationship)

• Lorentz relation (force on a charge moving through a region

containing both an electric and magnetic fields.

• Basic vector notions associated with interactions between a

moving charge and ambient magnetic field.

• Magnetic field lines associated with a bar magnet and current

flowing through a coil

• Faraday’s law of induction

• Lenz’s law

• Ohm’s law

*Visit basic ideas link under lecture topic 3 of class web page.

Basic Review

Basic definitions

Ohm’s Law iRV V is potential difference

i current

R resistance

Resistance, defined as opposition to direct current

flow, is not a fundamental physical attribute of

materials since it varies depending on the

conductor geometry.

A

lR

The geometrical influences are

evident in this relationship

where is the resistivity,

l the conductor length, and

A the cross-sectional area of the

conductor

The resistivity represents a fundamental physical

property of the conductor, and this or its inverse

(the conductivity) are the parameters we wish to

measure.

l

RA

RA

l

In general -

Resistivity is the property of a material which resists current

flow.

A

l

or

Units-

The unit of resistance is the

ohm

l

RA

Balancing units in the definitional

formula for resistivity, we see that

resistivity has units of ohm-meters or

-m.

Conductance is the reciprocal of

resistance and has units of ohm-1. Thus

conductivity ( ) has units of ohm-1/m or

mho/meterRA

l

• The reciprocal of a resistivity of 1 -m corresponds

to a conductivity of 1 mho/meter

• 1 mho/meter = 1000 millimhos/meter

• 1 millimho/meter =0.001 mho/meters

• The reciprocal of 0.001 mho/meters is 1000 -m

Working back and forth between units of conductivity and resistivity

(1/)

1 -m 1(-m)-1

or

1 mho/meter or

1000 millimhos/meter

1000 -m (1/1000) mho/meters

1 millimho/meter

Conductance (1/R) is often measured in

units of Seimens (S) which are equivalent to

mhos.

1 S = 1 mho


Given =1 m

1 1= 1mho/m = 1000 mmhos/m

1 m

1 mho 1 11 mmho/m= where a mmho =

1000 m 1000 m 1000

If you are given that = 20mmhos/m, then

1 1 1 1000 = 50

120 20 201000

m m mm

mmhos

In general when given a resistance the

equivalent conductivity in millimhos/meter

is obtained by taking the inverse of the

resistivity and multiplying by 1000. The

same applies to the computation of

resistivity when given the conductivity.

100 -m = _____ millimhos/meter

20 millimhos/meter = _____ -m

Consider the following …

For most of today’s lecture we are going to discuss the

computation of a for simple models of the type shown below.

This example comes from McNeil’s Technical Note.

Note that Z= depth/intercoil spacing

This is the equation we to use to

compute a and need to

understand

Note that this equation looks like a weighted

sum of individual layer conductivities. It is.

v

h

0.0

0.5

1.0

1.5

2.0

Rela

tive R

esp

onse

0.0 0.5 1.0 1.5 2.0 2.5 3.0

z=d/s

Note that the relative response

function for the horizontal dipole h is

much more sensitive to near-surface

conductivity variations and that its

response or sensitivity drops off rapidly

with depth

Vertical dipole interaction has no

sensitivity to surface conductivity,

reaches peak sensitivity at z ~0.5,

and is more sensitive to conductivity

at greater depths than is the

horizontal dipole.

2/12

2/32

)14(

42)(

)14(

4)(

z

zz

z

zz

H

V

We have to learn how to calculate the R’s. R is

a function of (see McNeill)

2/12

2/32

)14(

42)(

)14(

4)(

z

zz

z

zz

H

V

R represents an area under the curve

The contribution of this thin layer to the overall ground

conductivity is proportional to zz. This becomes 0 as

z goes to 0. So thin layers don’t contribute much to a

Constant

conductivity

Lets see how it works

The contribution of a layer to the overall ground conductivity

is proportional to the area under the relative response

function over the range of depths (Z2 -Z1) spanned by that

layer.

Z1

Z2

The contribution of the layer is weighted by the

area under the curve

As you might expect, the contribution to ground

conductivity of a layer of constant conductivity that

extends significant distances beneath the surface (i.e.

homogenous half-space) is proportional to the total area

under the relative response function.

Total area under the curve is …

So in general the contribution of several layers to the

overall ground conductivity will be in proportion to the

areas under the relative response function spanned by

each layer.

a is the sum of weighted conductivities

You all will recognize these area diagrams as

integrals. The contribution of a given layer to the

overall ground conductivity at the surface above it

is proportional to the integral of the relative

response function over the range of depths

spanned by the layer.

Area is an integral

McNeill introduces another function, R(z) - the cumulative response

function - which he uses to compute the ground conductivity from a

given distribution of conductivity layers beneath the surface.

RV(z)

0.0 0.5 1.0 1.5 2.0 2.5 3.0

z

0.0

0.2

0.4

0.6

0.8

1.0

Cum

ula

tive R

esp

onse

-R

(z)

RH(z)

zzzR

zzR

H

V

2)14()(

)14(

1)(

2/12

2/12

For next time continue your reading of McNeill

and develop a general appreciation of the

relative and cumulative response functions.

R is an integral of from some z to

Each point on the RV(z) curve represents the area under

the V(z) curve from z to .

The following diagrams are intended to help you

visualize the relationship between R(z) and (z).

Here’s how we get the R’s

Z2

To get an area we have to take the difference in

the two integrals

Express area as the difference in R’s

To get the area between Z1 and Z2 …

How would you express this integral as a difference

of cumulative response functions?

How would you get this area?

)()0()( 101 zRRdzz VV

zV

Just the same as before – a difference between

two integrals (areas)

RV(z)

0.0 0.5 1.0 1.5 2.0 2.5 3.0

z

0.0

0.2

0.4

0.6

0.8

1.0

Cum

ula

tive R

esp

onse

-R

(z)

RH(z)

Note that R(0) = 1, hence

)(1)( 101 zRdzz V

zV

Total area under curve is 1

According to our earlier reasoning - the contribution of a single conductivity layer to the

measured ground (or terrain) conductivity is proportional to the area under the relative

response function. The apparent conductivity measured by the terrain conductivity

meter at the surface is the sum total of the contributions from all layers. We know that

each of the areas under the relative response curve can be expressed as a difference

between cumulative response functions.

In summary

The weighting terms are just the areas under the

relative response function associated with each layer

The weighting

functions we just

talked about

Areas = R

The weighting functions

The solution in visual form

Compare this result to that of McNeill’s (see page 8 TN6).

In mathematical form

)()()()(1 2321211 zRzRzRzRa

In this relationship a is the apparent conductivity

measured by the conductivity meter. The dependence of

the apparent conductivity on intercoil spacing is imbedded

in the values of z. Z for a 10 meter intercoil spacing will be

different from z for the 20 meter intercoil spacing.

The above equation is written in general form and applies

to either the horizontal or vertical dipole configuration.

General form for a three layer problemuse the right equation, calculate z’s and R’s and do the arithmetic

1 1 2 1 2 3 21a R R R R Often simplified to this

1 1 2 1 2 3 2 0-z z -z za area area area

2 3/2

2 1/2

2 1/2

2 1/2

4 ( )

(4 1)

4 ( ) 2

(4 1)

1 ( ) ( )

(4 1)

( ) ( ) (4 1) 2

V

H

V Vz

H Hz

zz

z

zz

z

z dz R zz

z dz R z z z

These are the relationships we use to

compute specific values of R for given zs.

and highlights that the assumption of low induction number yields simple

algebraic expressions for the relative and cumulative response functions.

McNeill explores the origins of these

expressions in more detail

Problem 8.4: did you try it?

What did you get for a?


15mmhos/m

air=0 not 15, so to simplify our work –

assume instrument measurements

are made on ground surface

100mmhos/m

3m15mmhos/m

1m Inst Heightair=0

not 15

100mmhos/m

3m

15mmhos/m

( ) / ( ) 8.22z Depth Instrument height Intercoil Spacing

( ) / ( )z Depth Intercoil Spacing

What did you get for the problem at the bottom

of page 517 with meter on the surface


• Assume coils are on the surface at z=0

Note that z’s refer to (total depth)/(intercoil spacing)

NOT thickness/IS

This layer has no Z associated with it

S=3.67m

1 =10mS/m

2 =1mS/m

3 =10mS/m

z1

z2

t1=0.5m

t2=0.5m, 1m

The terrain conductivity meter measures an

apparent conductivity a


12 2

1 1( ) (4 1)vR z z

Remember that z =depth/intercoil spacing, so z1=0.5/3.67=0.136

Z is dimensionless

12 2

2 2( ) (4 1)vR z z

Work over this (use handout sheet) and turn in today before leaving

Problem on page 517


Text indicates solutions of 8.92 and 8.11 mS/m. Solutions differ due to

round off. They round off z1 and R1 to one decimal place while

remaining terms are calculated out to two decimal places.

Complete and turn in before leaving


12 2

1 1( ) (4 1)vR z z

Remember that z =depth/intercoil spacing, so z1=0.5/3.67=0.136

Z is dimensionless

12 2

2 2( ) (4 1)vR z z

Calculate all terms out to two decimal places.

A terrain conductivity survey is planned using

the EM31 meter (3.66 m (or 12 foot) intercoil

spacing) – vertical dipole only. Our hypothetical

survey was conducted over a mine spoil to

locate pyrite rich AMD source aeras within the

spoil. Scattered borehole data suggest that

these zones are approximately 10 feet thick and

several meters in width.

Borehole resistivity logs indicate that areas of

the spoil surrounding these zones have low

conductivity averaging about 4mmhos/m. The

bedrock or pavement at the base of the spoil

also has a conductivity of approximately

4mmhos/m. Depth to the pavement in the area

of the proposed survey is approximately 60 feet.

Conductivity of the AMD transport channels is

estimated to be approximately 100mmhos/m.

Mine spoil surface

AMD contamination zone

Pit Floor

~10ft

In-class/take-home problem

4mmhos/m

4mmhos/m

4mmhos/m

100mmhos/m~20ft

~30ft

Answer the following and turn in before leaving


For next time

•Hand in problem from page 517 on handout before

leaving today.

•Look over the AMD problem handed out today. Answer

questions 1-4 on back and turn in before leaving.

•Be prepared to ask questions and clarify issues with

problems next time

•Look over the problems from the book: 8.5, 8.6 and

8.7. Be prepared to hand in a week from today.

• See following slides for additional discussion of air

layer.

The simple algebraic expressions for RV(z) and RH(z)

make it easy for us to compute the terms in the

problem McNeill gives us. In that problem z1 is given

as 0.5 and z2 as 1 and 1.5. remember z is not d!

z RH RV

.1 .82 .98

.2 .68 .93

.3 .57 .86

.4 .48 .78

.5 .41 .71

.6 .36 .64

.7 .32 .58

.8 .29 .53

.9 .26 .49

1.0 .24 .45

1.1 .22 .41

1.2 .20 .39

1.3 .19 .36

1.4 .17 .34

1.5 .16 .32

1.6 .15 .30

1.7 .14 .28

1.8 .136 .27

1.9 .13 .26

2.0 .12 .24

Assuming a vertical dipole orientation

RV(z=0.5) ~ 0.71

RV(z=1.0) ~ 0.45

RV(z=1.5) ~ 0.32

We can take the equations for R and construct

a table or just compute directly

Note that the problems you will solve in this

class will always use the vertical dipole.

1=20 mmhos/m

2=2 mmhos/m

3=20 mmhos/m

Z1 = 0.5

Z2 = 1 and 1.5

)()()()(1 2321211 zRzRzRzRa

Substituting in the following for the case where Z2=1.

45.02045.07.027.0120 a

45.02025.023.020 a

mmhos/m5.15a

Some details for the McNeill’s numerical solution (page 8)

use correct equation - calculate z’s, R’s &

watch the arithmetic!

1 12 2

2 12 2

2 12 2

1 1 0.71.414

4(0.5) 1

1 1(1) 0.452.24

4(1) 1

1 1(1.5) 0.323.16

4(1.5) 1

R

R

R

R calculations illustrated


12

12

1 2

12

12

2 2

12

12

2 2

1 1 0.724(0.5) 1

1 1(1) 0.4554(1) 1

1 1(1.5) 0.32104(1.5) 1

R

R

R

2 1/2

1( )

(4 1)VR z

z

Remember where the R’s come from

Appendix A: Additional discussion of equations 8.22 to 8.39

and air layer for general reference


This is really a 3-layer problem

where layer 1 is the air layer. Air

generally has very low

conductivity. In this example we

assume 1 = 0.

3 = 50

First: get the depths

Second: compute z’s.

Third: compute R’s.

Fourth: substitute and solve for a.

What is the influence of the 0 conductivity

surface layer


3 = 50

Z1=1/3.67 = 0.272

Z2= 9/3.67=2.45

2 1/2

1( )

(4 1)VR z

z

R1=0.878

R2=0.2

)()()()(1 2321211 zRzRzRzRa

0 .122 10 0.678 50 0.2a

16.78 /a mS m This is a little less than the 18mS/m

reported in equation 8.24

Why is our calculated conductivity less than

that in the text?


3 = 50

Intuitively we might guess the

answer should be less since the

equations in the text introduce a 1

meter thick layer of conductivity

10 mS/m that is not there.

The 1 meter thick layer beneath

the conductivity meter has 0

mS/m conductivity and we

suspect this will decrease a.

We would also expect that as we thicken the air layer or

increase the distance of the meter from the surface, the

apparent conductivity will get smaller and smaller.

Revisions apply to equations 8.22 through 8.39


Equations 8.22 through 8.30 yield incorrect results for computations

of a at points A, B, and C in Figure 8.8. Presented as a two layer

problem, these actually turn out to be 3-layer problems.

Likewise, computation of a for the configuration presented in Figure

8.9 is undertaken as a 3 layer problem, but is actually a 4 layer

problem.

1 1 2 1 2 3 1 2 4 31 ( ) ( ) ( ) ( ) ( ) ( )a R z R z R z R z R z R z

In this problem, 1 through 4 are 0, 10, 1, and 10 mS/m respectively.

2 1 2 3 1 2 4 3( ) ( ) ( ) ( ) ( )a R z R z R z R z R z

This yields

since 1 is 0.

You should get results of

7.14 (t1=0.5m) and 7.39 (t1=1 m).

Again - for next time and before leaving

•Hand in problem from page 517 on handout

before leaving.

•Look over the AMD problem handed out today.

Answer questions 1-4 on back and turn in before

leaving.

•Be prepared to ask questions and clarify issues

with problems next time.

•Look over the problems from the book: 8.5, 8.6

and 8.7. Be prepared to hand in a week from today

environmental and exploration geophysics ipages.geo.wvu.edu/~wilson/geol454/lect3/lec3.pdf · tom...

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