a technique to calculate complex electromagnetic fields by

Portland State University Portland State University

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Dissertations and Theses Dissertations and Theses

1978

A technique to calculate complex electromagnetic A technique to calculate complex electromagnetic

fields by using the finite element method fields by using the finite element method

Davood Asgharian Portland State University

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I I

............. --... __ ......... ...._..... ......

AN ABSTRACT OF THE THESIS OF Davood Asgharian for the Master of Science

in Applied Science presented May 16, 1978. ·

Title: A Technique to Calculate Complex Electromagnetic Fields by

Using the Finite Element Method.

APPROVED BY MEMBERS OF THE THESIS COMMITTEE:

James L. Hein

A computer program based on Maxwell's equations is developed to

calculate two-dimensional complex potentials by the Finite Element

Method.· This study offers a solution to a complex continuum problem by

allowing a subdivision into a series of simpl~ interrelated problems.

The region of interest is divided into triangular elements. For each

node in the grid, the Finite Element Method is used to set uy art equa-

tion for the potential as a function of those of the surrounding nodes.

All these equations are solved by the Gaussian Elimination Method. For

increased accuracy this method requires a high degree· of division of the

region of interest. This could cause a storage problem on the computer.

To eleviate this.problem a half-banded scheme is used. A comparison is

! " J

I j-

1·

I I I

-&·-·-------

provided between the data obtained from the developed algorithm and an

actual experiment. In this experiment two-types of sunken swimming

pools, reinforced and non-reinforced, were used to hold three different

waters of conductivities 29µ"V""/cm, 1500µc.r/cm and 3000µ"21"/cm. In order

to test the accuracy of the computer.program developed, the results of

another solved problem are also compared to another computer program's

results which was based on c.apacitive and resistive d·ist:i;:j,bution of

potentials. The result of this study s~ows the hazard may exist on the

edges of the swimming pool when the resistivity of the surrounding soil is high.

i·

1·

I

A TECHNIQUE TO CALCULATE COMPLEX ELECTROMAGNETIC

FIELDS BY USING THE FINITE ELEMENT METHOD

by

DAVOOD ASGHARIAN

A thesis submitted in ·partial fulfillment of the requirements-for the degree of

MASTER OF SCIENCE in

APPLIED SCIENCE

Portland State University

1978

,,

I I· ! I l I

TO THE OFFICE OF GRADUATE STUDIES Al"'ID RESEARCH:

The members of the Committee approve the thesis of Davood

Asgharian presented May 16, 1978.

M. Young, Department ineering and Applied

Graduate Studies and Research

~~

James L. Hein

~.d:IM

f>NIA01 AW O.L

a~.tv~rcraa

ACKNOWLEDGMENTS

The author gratefully acknowledges indebtedness to Dr. V. J. Garg

for his suggestion of the problem, criticisms, and suggestions during

the course of this study and his sacrifice of time and energy during

the writing of this thesis.

The author wishes to express his gratitude to Dr. F. Rad for his

advice and reviewing this thesis. I would also like to express my

appreciation to Professor R. Greiling for making many constructive

conunents.

Consultation with Dr. W. Mueller was also a great aid in the

writing of the computer portion of this study.

The author wishes to thank Ms. Donna Mikulic for the excellent

and efficient typing of this thesis.

I am especially indebted to my dear wife for her encouragement,

understanding and unfailing patience.

TABLE OF CONTENTS

ACKNOWLEDGMENTS

LIST OF TABLES

LIST OF FIGURES

CHAPTER

I INTRODUCTION

1.1 REVIEW OF LITERATURE 1.2 STATEMENT OF THE PROBLEM

II FINITE ELEMENT METHOD

2.1 DEFINITION 2.2 FORMULATION OF FINITE·ELEMENT METHOD 2.3 FORMULATION OF POTENTIAL PROBLEMS WITH SPATIAL

FINITE ELEMENT SUBDIVISIONS 2.4 FINITE ELEMENT SOLUTION OF COMPLEX POTENTIAL

ELECTRIC FIELDS

III COMPUTER PROGRAM

3.1 SOLUTION TECHNIQUE FOR THE FINITE ELEMENT METHOD

3.2 RESULTS OF THE COMPUTER SOLUTION 3.3 COMPARISON OF RESULTS WITH OTHER COMPUTER

TECHNIQUE

IV EXPERIMENTAL PROGRAM

V RESULTS

5.1 COMPARISON OF CALCULATED VALUES WITH EXPERIMENTAL RESULTS

5.2 POTENTIAL HAZARD TO THE HUMAN BODY 5.3 THE LIMITATIONS AND ACCURACY OF THE

THEORETICAL TECHNIQUE

VI CONCLUSION

BIBLIOGRAPHY

Page

iv

vii

viii

1

1 3

4

4 4

8

10

18

18 29

35

38

42

42 52

59

60

61

Page

APPENDIX A EULER'S THEOREM OF VARIATIONAL CALCULUS 63

APPENDIX B THE GAUSSIAN METHOD 66 APPENDIX C CALCULATION OF CURRENT DENSITY 72

APPENDIX D LISTING OF PROGR.A11S AND SUBROUTINES 76

APPENDIX E COMPUTER RESULTS 87

APPENDIX F EXPERIMENTAL RESULTS 100

Table

3.1

5.1 - 5.4

5.5

C.l - C.3

E.l - E.9

E.10 - E.12

F.l - F.9

F.10 - F.12

LIST OF TABLES

Comparison of Results of a Specific Problem

Calculated Current Travelling Through the Human Body Standing on the Soil

Calculated Current Travelling Through the Human Body Inside the Swimming Pool

Calculated Current Densities on the Surface of the Earth

Current Densities Calculated in Soil

Current Densities Calculated in Water

Current Densities Measured in Soil

Current· Densities Measured.in Water

Page

37

54- 57

58

73- ·75

88- 96

97- ·99

101-109

110

Figure

2.1

2.2

2.3

2.4

3.1

3.2

3.3

3.4

3.5

3.6

3.7

3.8

3.9

3.10

4.1

4.2

5.1-5 .. 8

5.9

B

LIST OF FIGURES

Triangular Division of the Area

One Triangle

Typical Triangle with Vertices Marked

Area Coordinates

Region 's' is Divided in 18 Triangular Elements

Banded Form of a Symmetrical Matrix

Flow Chart, Subroutine "BANDWIDTH"

Flow Chart, Subroutine "FIND"

The Finite Element Model

Regional Division of the Experimental Model

Flow Chart, Subroutine "GOOD"

Flow Chart, Program "TES"

Resistive and Complex Admittance Networks

Two Materials in Series

Schematic Diagram of the Experimental Model

Schematic Diagram of the Electrical Circuit

Calculated Current Densities

Levels of Current Hazards to the Human Body

Flow Chart, Subroutine "SOLVE"

Page

5

6

11

11

19

22

24

25

31

32

33

34

35

36

40

41

44-51

53

70

~

CHAPTER I

1-l INTRODUCTION

I 1.1 REVIEW OF LITERATURE

A considerable amount of work has been done in the past in cal-

culating the self and mutual impedance of two parallel ground return

wires. The following paragraphs summarize these attempts in chronolo-

gical order.

The first attempt was made by Carson (1). He investigated the

problem of wave propagation along a transmission system composed of

an overhead wire parallel to the surface of the earth. However a

complete solution of determining the actual impedance is impossible

because of the non-homogeneity of the earth. The solution to the

problem, where the actual earth is replaced by a plane homogeneous

semi-infinite solid has promoted considerable theoretical and prac-

tical interest.

In 1951, Lacey and Wasley (2) at the Hydro-Electro Power Com.mis-

sion of Ontario, Canada, developed an equation for the mutual impedance

of two finite length earth-return circuits, either parallel or at an

angle. The equation developed by them is to· be a generalization of

Carson's work.

In 1965, Wedepohl (3) published a paper on wave propagation in

multiconductor overhead lines which would permit the earth-return path

to have a relative permeability other than unity, which was not permissi-

j.

I I

2

ble in the analysis by Carson. In this paper, the new approach is applied

to the case of a two-layer earth, including the effects of displacement

currents. The results were in agreement with those obtained for the

case of a homogeneous earth.

In 1966, Krakowski (4) developed equations for the mutual im-

pedances of overhead lines with the earth as a return path. In this

paper the problem deals with two different lines which cross each other

at an angle, a., different from zero. A particular case of this pro.bl em

is the same as Carson's solution for a.= 0. The general solution of

this problem is considered, assuming that the earth is uniformly con-

ducting and that both overhead conductors are parallel to the surface

of the earth.

In 1973, Nakagawa (5) published a paper in this area. This

solution permits the earth-return path to be considered as three layers

of different resistivities, permitivities and permeabilities. A strati-

f ied earth causes marked differences in the earth impedances and the

resultant wave deformations from the homogeneous case. The depth of a

layer is a significant factor to the value of the stratified-earth

impedance. The displacement currents can influence earth-return

impedances. This is only at very high frequencies and under the con-

ditions of high earth resistivity and low conductor height.

All these papers prove that there are several ways of calculating

the distributed impedance of ground return transmission lines.

Magnusson (6) developed a method of calculating the mutual and

self-impedance of overhead lines with the earth as a return path. He

also calculated the mutual and self-impedance of the line under the

following conditions:

A. A conductor height of 35 feet

B. A line-to-ground sh~rt-circuit current of 2000 ~mperes.

C. A ground conductivity of 0.01 mho per meter

By the calculated value of the mutual and self-impedance of overhead

lines with the earth as a return path and the use of the developed

formula, he calculated the current densities in a typical below grade

swimming pool.

3

The densities change with respect to the distance of the swimming

pool from the vertical plane of the transmission line. The calculated

current densities in the pool were found to be hazardous to the swimmer

in the swimming pool.

1.2 STATEMENT OF THE PROBLEM

The purpose of this investigation is to develop a computer code

based on Maxwell's equations to calculate potentials between points

of interest on the surface of the earth and swimming pool by knowing

at least two boundary conditions, using the Finite Element Method.

In order to check the validity of this study, the results are

compared to experimental values.

. ,

CHAPTER II

FINITE ELEMENT METHOD

2.1 DEFINITION

The Finite Element Method is a numerical technique for obtaining

approximate solutions to a wide variety of engineering problems. The

ability to use elements of various types and sizes and to model a

system of arbitrary geometry, are the main advantages of the Finite

Element Method.

Other approximate methods, for example the Finite Difference

Method, lacks these advantages. Using these approximate methods, a

specific numerical result may be obtained for a specific problem, but

a general computer solution applicable to all cases is not possible.

The Finite Element Method offers a way to solve a complex con-

tinuum problem by subdividing the continuum into a series of simpler

interrelated problems. It gives a consistent technique for modeling

the system as an assemblage of discrete parts or finite elements.

2.2 FORMULATION OF FINITE ELEMENT METHOD

It is desirable to obtain results in a general form applicable to

any situation. For th.is purpose a division of the region into triangu-

lar shape elements is used as shown in Fig. 2.1.

The problem is to calculate the values of ~e) (i.e., voltage)

(e) at each node, (N = 1, 2, ••. , n) by knowing values of RN at some node

5

y

5 4

3

2

x

Figure 2.1 Triangular division of the area.

as boundary conditions.

The integer numbers of 1, 2, ••• , n represent the number of the

particular node and value of H at node 5 which is written as H5

• The

integer numbers written inside parenthesis, for example, (3) represents

the element's number.

Each element has three nodes and each node has its own coordinate

values. For example, element (1) has nodes 1,2,7 and coordinate·

values of (x1

, y1), (x2' y 2)' (x7 , .Y7), and element (5) has nodes 6,7,5

and coordinate values of (x6

, y6),

(x:Y-j Y7)' (x5, Y5). Fig. 2.2 shows a typical triangie from the whole area of Fig. 2.1.

The assumption is that the value of h (i.e., voltage) at any point

inside the triangle is a linear function of Hat the triangle's three

nodes, or simply:

6

HR. h(e) = [N(e) N(e) N(e)] I Hm f ~ [NJ[H] 2-1 R. m m

H m ...

yl R,

m

n

x

Figure 2.2 One triangle element.

Therefore, for the area of Fig. 2.1, the values of h in each

element are:

h(l) = N(l) H + N(l) H + N(l) H 1 1 2 2 7 7 2-2

h(Z) = N(2) H + N(Z) H + N(2) H 2 2 3 3 7 7 2-3

h(J) =·N(J) H + N(J) H + N(J) H 3 3 4 4 7 7 2-4

h(4) = N(4) H + N(4) H + N(4) H 4 4 5 5 ] 7 2-5

h(S) = N(S) H + N(S) H + N(S) H 5 5 6 6 7 7 2-6

h(6) = N(6) H + N(6) H. + N(6) H . 6 6 1 1 7 7 2-7

Where [NJ is called a shape function and will be seen later to

play a paramount role in the Finite Element Method. The shape function

is a function of area coordinates:

7

N(e)= l/2A (e) fa (e)° + b (e) X + c (e) Y] n n n n 2-8

Where A = area of the triangle:

a = x y - x y· n .Q, m m R..

b = y - y n .Q, m

c = x - x n m R..

For example N7

for element (4) is:

N(4) = l/2A(4) [a(4) + b(4) X + c(4) Y] 7 7 7 7

Where: a7 = X4Y5 - x5Y4

b7 = Y4 - Y5

C7 = XS - X4

and so on.

The total h in this area is equal to the summation of hS in the

elements.

E h(e) h=L 2-9

e=l

Where E is the number of the last node. Eq. 2-9 could be written in

matrix form as well as in summation form.

h(l) N(l) 1

N(l) 2

0 0 0 0 N(l) 7 Hl

h(2) 0 N(2) N(2) 0 0 0 N(2) H2 2 3 7

h (3) 0 0 N(3) N(3) 0 0 N(3) H3 I 2-10 = 3 4 7 h(4) 0 0 0 N(4) N(4) 0 N(4)

H4 4 5 7 h(5) 0 0 0 0 N(5) N(5) N(5)

HS 5 6 7 h(6) N(6) 0 0 0 0 N(6) Nj6)1 I H6 1 6

2.3 FORMULATION OF POTENTIAL PROBLEMS WITH SPATIAL FINITE ELEMENT SUBDIVISIONS

8

The current density JT consists of both conduction and displace-

ment components, respectively:

JT = CTE + (d/dt)D 2-11

where

D = jtWEE 2-12

After subseitution of Eq. 2-12 into Eq. 2-11 one may obtain this result:

JT = (cr + jwE)E

Equation 2-13 by Kirchoff's law must satisfy the continuity

equation.

or

but

where

v • J = 0 T

V • (a + jwE)E = 0

E = -VV = 0

v . (cr + jwE)VV = 0

vv = [(a/ax)Va + (a/ay)Va + (a/az)Va ] x y z

Substitute Eq. 2-18 back in Eq. 2-16:

2-13

2-14

2-15

2-16

2-17

2-18

v • (cr + jwE)[(a/ax)Va + (a/ay)Va + (a/az)Va] = o 2-19 x y z

v · A= (a/ax)A.+ Ca/ay)A + Ca/az)A 2-20

Therefore the resultant equation is:

(a/ax)(cr + jwE)(a/ax)V + (a/ay)(cr + jwE)(a/ay)V +

(a/az)(cr + jwE)(a/az)V = o 2-21

In order to solve Eq. 2-21 one may need to know Euler's theorem of

variational calculus, as outlined in Appendix A. By the help of

variational calculus, a function I(V) could be found where oI(V) = 0

everywhere.

( 2 , 2 I(V) = l/2Jn [(a+ jwg)(av/ax) +(a+ jwE)(av/ay) +

2 (a + jwE).(aV/az) ]dx dy dz

but 3

v(e) = .L.: N v = [N] [V] (e) i=l i i

The derivative of I(V) with respect to the Vi is equal to z.ero.

ar(v)(e) /av.= o i

=J0

{r<cr + jwe)(av<e>/ax)(a/av1)(av<e>/ax)J +

[(a+ jwE)(av<e)/ay)(a/avi) av(e) /ay)] +

9

2-22

2-23

[(a+ jwg)(av(e)/az)(a/avi)(av(e)/az)l} dx dy dz 2-24

But from Eq. 2-23 it is obvious that the derivative of V(~) with respect

to x is: 3

av(e)/ax =~ (aNi/ax)Vi = [aN/ax][V](e) i=l

2-25

(a/avi)(av<e)/ax) = (a/avi)[(aN1 /ax)Vi1 = aN1/ax 2-26

where

av(e)/avi =Ni 2-27

The result of the substitution of Eq. 2-25, 2-26 and 2-27 back in Eq.

2-24 is:

ar(v) (e) /avi = o = LJca + jwe:)[aN/axJ[vJ (aNi/ax) +

(a+ jwE)[aN/ay][V](aN./ay) + 1

10

(cr + jwE) [aN/az] [V] [aN/az] }ax dy dz 2-28

Where:

Equation 2-28 could be written in general form as:

[K][V] = [O]

K. . = ( ((a + jwE) (aNi/ax) (aN ./ax) + (cr + jwE) (aNi/ay) i,J Jnl' J

(aNj/ay) + (cr + jwE)(aNi/az)(aNj/az)} dx dy dz

2.4 FINITE ELEMENT SOLUTION OF COMPLEX POTENTIAL ELECTRIC FIELDS

The region of the problem can be subdivided into triangles in

2-29

2-30

any desired manner, insuring only that all different material interfaces

coincide with triangle sides. Figure 2.3 shows a typical region divided

into triangles.

It is assumed that there is a linear variation of potential within

each triangular element with respect to the nodal potentials.

A convenient set of coordinates 11

,12

,13

for a triangle 2,m,n,

Fig. 2.4, is defined by the following linear relation between these and

the Cartesian system:

y

( k: "In

'( \ ~m

Figure 2.3 Typical triangle with vertices marked.

\ \

\ \

(xn,yn)

\ \ \ \ ..

x

L1=1 ~ \ \PCL1 ,L2 ,L3)\ ~ .!l . \ \. \ m

~ ' < ,~,( (x.!l,y.!l) ' \ ' ' xm,ym)

Figure 2.4 Area coordinates.

11

12

x = 11xi + 1 2xm + 1 3xn

y = 11Y~ .~ 12Ym + 13yn

1 = 11 + 12 + 13 '

To every set, 11

, 1 2 , 13

(which are not independent, but are

related by the third equation) corresponds a unique set of Cartesian

coordinates. At point 1, 1 1 = 1 and 1 2 = 1 3 = O, etc. A linear re

lation between the area coordinates and Cartesian coordinates implies

2-31

2-32

2-33

that contours of 11

are equally placed straight lines parallel to side

2-3 on which 11

= 0 etc. It is easy to see that ~n alternative defini

tion of the coordinate 11

of a point P is by a ratio of the area of the

shaded triangle to that of the total triangle.

1 = area Pmn 1 area R-mn

One may write Equations 2-31 through 2-33 in matrix form and solve it

for 11 , 1 2 , 13

.

xi

y R,

1

x m

ym

1

x n

yn

1

11

12

13

x

= y

1

2-34

l l l

UA UIA ?5 t\

8£-Z u Ul ox x x vz £1 ?5 Ul Ul 0

( ?5 AUIX -UIA ?5 X) ( x -x)a + ( a -a)x + 1 1 1

.6. UIA 0.6.

Ul ?5 x x x

l l l

UA UIA o.6.

u Ul ox l£-Z x x vz

z1 u ?5 0 u (UA?fX -oAUX) = = ( x -x).6. + ( .6. -.6.)x + l l l

UA .6. 0 £.

u 0 x x x

l l l

u.6. UIA oa

9£-Z u Ul ?5 x x x vz = 11

(Ul.6.Ux -·UaUIX) = Ul ux).6. + (u£. Ul ( x --.t\)x + .T 1 1

UA UIA ..:\

u Ul x x x

£1

14

Where:

2A = 2*(area of the triangle) = (x y - x y ) + (x Yn - XnY ) + mn nm nx, x,,m

(xRXm - xm Y2) 2-39

The area coordinates are the shape functions: N1

= L1

,N2

= L2

and

N3 = L3.

The potential inside the triangular element is a linear function

of the nodal's potentials:

v(e) = L V + L V + L V 1 R, 2 m 3 n 2-40

After substituting Equations 2-36, 2-37 and 2-38 into Equation 2-40

one obtains:

v(e) = l/2A [[(x y - x y) + x(y - y) + y(x - x )]Vn + mn nm m n n m x,,

[(x y - x y ) + x(y - y ) + y(x - x )]V + ni in n R, R, nm

[(x y - x y ) + x(y - y ) + y(x - x )]V ] R,m mi R, m m R, n 2-41

In order to solve Equation 2-28 the .shape functions must be known.

When they are determined they can be substituted in Equation 2-42.

[K] [V] = [O] . 2-42

Matrix K is calculated for a two dimensional problem.

Kij = J [(cr + jwE)(aNi/ax)(aNj/ax) + '(cr + jwE)(aNi/ay) n

(aN./ay)] dx dy 2-43 J

"l

15

For each element (a + j WE) may be taken outside the integration sign.

Therefore:

2 2 Kl,l = [(dN1/dx) + (dN1/dy) ]dx dy

(x -x ) 2 ( 2 2 n m Y -y ) + (x -x ) + 2 ] dx dy = m n n m

4*A 4*A 2-44

Kl,2 (y -y )(y -yn) (x -x )(xn-x)

= [ m n n x.. + n m x.. m ]dx dy = 4*A2 4*A2

(y -y )(y -yn) + (x -x )(xn-x) m n n x.. nm x.. n 4*A 2-45

Kl,3 (y -y )(yfl-y) + (x -x )(x -xn) mn m nmmx..

4*A 2-46

K2,l = Kl,2 2-47

2 2 (y n -y Jl) + (x Jl -xn)

K2,2 = 4*A 2-48

K2,3 (yn-yJl)(yJl-ym) (xJl-xn)(xm-xJl)

= [ + ]dx dy = 4*A2 4*A2

(y -y )(y -y) + (x -x )(x -x) n Jl Jl m Jl n m 2

4*A 2-49

K3 1 = Kl 3 ' '

2-50

16

K3,2 = K2,3 2-51

K3,3

2 2 (yi-ym) + (xm-xi)

4*A 2-52

Substituting Equations 2-44 thru 2-51 into Equation 2-42 and writing the

result in matrix form:

2 (y

m-y

n)

(yn

-y i)

+

(y

-y

)

+

m

n 2

(x -x

)

(x -x

)

(xi -

x

) n

m

n m

n

(ym

-yn

)(y

n-y

i) +

2

P/4*

A*

I (y

n-y

i)

+

(x -x

)

(x2-x

)

2 (x

2-x

n)

n m

n

(Y

-Y

)(Y

2-Y

) +

m

n

m

(Yn

-Y 2

) (Y

2-Ym

) +

(X

-X

) (X

-X

i)

n m

m

(X

2-Xn

) (X

m -X

i)

(y m

-y n)

(y i -y

m)

+

I I

v (x

-x

)

(x -x

i)

i n

m

m

(yn

-y g_}

(y

i-y

m)

+

I I

v (x

2-xn

) (x

m -x

2)

m

(Y R

,-Ym

)2 +

I I

v (X

m-X

R,)

2 n

I I

I =

I

I I

0 0 0

I 2-

53 1-

4 -...

.J

CHAPTER III

COMPUTER PROGRAM

3.1 SOLUTION TECHNIQUE FOR THE FINITE ELEMENT METHOD

A computer program is written to.solve Eq. 2-53 f~r the region of

interest which consists of n-type of materials and at least two boundary

conditions. This equation in the short form is given by:

[K] [V] = [O] 3-1

Matrix [K] is the coefficient matrix and consists of all the pro

perties of the materials in the region. Each element in the region

could have a different property from the others. Matrix [K] is calcu

lated for each element with its own properties and then transferred to

the final coefficient matrix [F]. One example is given below.

Region S, Fig. 3.1, is divided into 18 triangular elements and

each element has been numbered from 1 to 18.

Also all nodes are numbered in a fashion to create a sparse [F]

matrix to reduce the band-width of the [F] matrix. To do so, the side

which has less nodes than the other is determined. Then the nodes are

numbered from one end to the other and returned to the original side,

as shown in Fig. 3.1. This method insures the smallest possible band

width for the [F] matrix.

19

y

13 14 15 16

9 L -,....c: =-tc ., 12

5 • Naz ==:.: ., 8

l• JM 3>' < - 4 2 j

x Figure 3.1 Region S is divided in 18 triangular elements.

20

The arbitrary element Z has nodes 2,m,n and coordinates of (xN

YN ), (xNm, YNm), (xNn, YN~) and material property of P. By using

Eq. 2-53 we can solve for matrix K:

KR,' R. K Jl,m K

R. ,n

K = I Km,R. K K m,m m,n 3-2

K K K n, R. n,m n,n

By transformation, Kt Jl goes to the [F] matrix in row Jl and column '

Jl and then added to the previous value of F2 , 2• Similarly, KR.,m goes

into the row Jl and column m of the matrix [F] and then added to the pre-

vious values of F0 , and so on. N,m

After completing the matrix [F], Equation 3-1 becomes:

[F][V] = [O] 3-3

where it has the dimension of (No. of nodes by No. of nodes) and K is

a 3 by 3 matrix. Since Equation 3-3 is equal to zero, it requires

the boundary conditions for solution. The boundary conditions are used

to create values on the other side of the equation.

For instance, region S in Fig. 3.1 has two boundaries, one at each

end. Nodes 1,5,9 and 13 from one end and nodes 4,8,12 and 16 from the

. . other end are the boundary nodals and have known values of voltage •

Therefore we can leave these nodes out of our calculations. For example:

element (7) has nodes 5,6,9 where nodes 5,9 have known values and node 6

is an unknown.

The matrix notation for this element after calculating the K

matrix is:

21

0

=

~,5 K5,6

K6,5 K6,6

KS,9

K6,9

K9,9

vs

v6

v9

0

0

3-4

K9,5 K9,6

Therefore there is just one equation and one unknown and it is

easy to transfer the known values to the other side of the equation.

The result is:

[K6,6][V6] = [-K6,5 *VS - K6,9 * V9] = [B6]

Now this equation is transferred to the [F] matrix:

[F][V] = [B]

For the small size of matrix [F] we can find the inverse of the

[F] matrix and multiply it with the [B] matrix to find the values of

the nodes.

3-5

3-6

All finite element solutions require a high subdivision of the

region for the utmost accuracy. This makes matrix [F] so large that it

becomes useless to solve by the invertion of the [F] matrix.

Due to the nature of the problem, provided that the nodes are

numbered in a careful manner, the non-zero terms in matrix [F] will be

concentrated in a narrow band situated adjacent to the leading diagonal.

This fact, combined with the symmetrical nature of matrix [F] indicates

that only a relatively small portion of the matrix is of real interest.

If advantage is taken of these observations, demands on the computer

storage may be considerably reduced. Moreover, if the solution procedure

is so arranged that many of the operations involving the zero terms are

eliminated, the speed of the solution can be increased. Methods which

22

take advantage of the banded nature of matrix [F] are often called

'banded methods'.

Methods which off er potentially greater economies are the so-called

'half-banded schemes'. The upper half of the diagonal band of the matrix

is stored as a rectangular matrix as shown in Figure 3.2.

._B_.

r SIZE

+--" B--+

UPPER HALF BAND

" "

,,~ ,, 1c"'" ZERO

MATRIX F MATRIX A

Figure 3.2 Banded form of a synnnetrical matrix.

The upper half band part of matrix IF] is stored in matrix {A]

which is much smaller than matrix (F]. Matrix [A] has a number of columns

equal to the bandwidth and rows equal to the number of nodes. Each row

of matrix [F] is transferred to matrix [A].

To calculate the band-width of a finite element problem, one must

know the number of all elements and their node numbers, because bandwidth

is equal to the largest difference between two nodes in one element;

that is compared to the rest of elements + 1.

Figure 3.3 is a flow chart of the computer program which finds the

bandwidth of matrix [F] or any other symmetrical matrix. Figure 3.4

is a flow chart which determines the coefficient matrix and transfers

the upper half part of matrix [F] to matrix [AJ.

Equation 3-6 takes the form:

23

[A]fVJ = fB]

It is impossible to find the inverse of [A] because it is no longer a

square matrix. Theref~re, the Gaussian Elimination Technique is used

3-7

to solve Equation 3-7. Another step to save memory space is to eliminate

matrix [V] from the equation. To do so, the problem between [A] and

[B] is solved and the result is stored in matrix [B]. Mat~ix [B] has

the same dimension as matrix [V].

For more understanding of the Gaussian Elimination Technique an

example is solved in Appendix B along with the flow chart.

Appendix D includes a listing of the main program as well as all

subroutines discussed in this chapter.

YES

NO

SET ~ANDWIDTH•O

K = 1

READ NOD.ES AND LAST NODE BEFORE BOUNDARY NODE

CALCULATE THE (DIFFERENCES BETWEEN EACH TWO NODES) + 1

FIND THE LARGEST LET IT BE Bwl

BANDWIDTH = Bl

RETURN

NO K = K+l

Figure 3. 3 Flow chart, subrout_ine "BANDWIDTH".

24

25

START

M-1, NUMBIB OF NODES IN F.ACH ELEMENT

I•1, NUMBER OF NODES IN EACH ELEMENT

YES

YES

NO

MM • N(M) MM• N(M)

I.CULATE A(MM,1) CALCULATE A(MM,1) CALCULATE A(MM,1) ..

Figure 3.4 Flow chart, subroutine "FIND".

NO

YES

NO

NO

CALCUUTE B(~'N)

WHERB h1l = N(M)

Figure 3.4 (Continued)

YES

MM = N(I)-N(M) + l

NN = N(M)

CALCULATE A(NN,MM)

26

f YES

NO

NO

CALCULATE B(NN)

WHERE NN = N(M)

y.r;s

HM • N(I)-N(M) + l NN -= N(M)

CALCULATE A(NN,MM)


27

CALCULATE B(~N)

WHERE NN ::: N(M)

NO

NEXT I

NEXT H

RETURN

Y~S

MM • N(I)-.S(M) + l

NN • N(M)

CALCULAT£ A(:.'iN,MM)


28

29

3.2 RESULTS OF THE COMPUTER SOLUTION

The problem was to calculate current densities everywhere in the

region S. Region S was a large area of soil with a sunken swimming pool

in the center of the region. The region was divided into 76p ~lements

with three types of materials and two boundary conditions. Figure 3·.5

shows the subdivided region of 'S'.

For large problems such as this involving many elements, it is

useful to possess a routine which generates the complete set of data for

the finite element program.

Region 'S' was subdivided into five regions. Region one was below

the swinnning pool, region two and four were the swimming pool ends and

the soil; region three was the swimming pool and surrounding soil; and

region five was above the swimming pool. Figure 3.6 shows these five

regions.

The reason for dividing regi~n 'S' into five regions was to make

data preparation easier. Regions 2,3,4 were divided in a different

fashion than 1 and 5. Region 1,5 and 2,4 are identical in values of x

and y with some constant. Also the results of each region can be stored

in a different matrix and recalled when needed. All nodes on each

boundary are given the same number for simplification purposes.

A subroutine was written to find the coordinates of all nodes.

Figure 3.7 shows a flow chart of such a subroutine.

Figure 3.5 shows that nodes 1,10,19,28 and 37 have the same value

of x and nodes 1,2,3,4,5,6,7,8 and 9 have the same values of y. Therefore

coordinates of nodes are calculated and stored in a matrix for later use.

30

Another data file is generated which consists· of all elements with

their nodal numbers. Figure 3.8 shows .a flow chart ot this program

(called "TES") which can read the element's number and their nodal numbers

from the file and find the corresponding coordinate values and store them I

in a separate file, which ~acks the information about the first and last

row of the.region 'S'. This information could be added to the file

easily.

This data file is ready to be given to the main program for cal-

culation of voltages at each node. A program is written to calculate

the current densities in the region in the y-direction.· Results of

computer program, in tabular form are given in Appendix E.

A comparison of the computer results with experimental results is

given in Chapter V.

32

R E G I 0 N 0 N E

L- - - -- - - - - - - - - - - - ... - - - - - ..... - - - - - --

R £,G I 0 N T W 0 L - - - - - - - - - - - - - - - - - - • - -t - - - - - - - - - - - - - - - ":- •

l

R 1': G l 0 N T H RIE ~

l---------------1--. __ .,. ______ ,_ ________ _ R E G I 0 N F 0 U IR

L... - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -a...~

y

R £ G I 0 N F I VE

x

Figure 3.6 Regional division of the experimental model.

START

READ ALL X, Y

LE'T NODES OF AREA BE REPRESENTED BY A 11.ATRIX

I = l,N J = l,M

A(I,J) = K+J

YA(I,J) = Y(I)

K = K+M

NEXT J,I

CALCULATE THE X-COORDINATE

I = l,N J = l,M

XA(J,I) :: X(I)

NEXT J ,I

REnJRN

Figure 3. 7 · Flow chart, subroutine "GOOD".

33

START

READ X(I), Y(I) FROM Sl'OOAGE

RF.AD NUMBER OF ELEMENT,

rrs. NODES AND PROPESTY

L • NODE 1 J •NODE 2 K •NODE 3

XN(1) • X(L), YN(1) • Y(L) XN(2) • X(J), YN(2) • Y(J).

XN(3) • X(K}, YN(3) • Y(K)

READ IN DATA FOR ARE.A 2,4 <F THE REnION'S'

STOP

NO

Fi~ure· 3.8 Flow chart, program "TES".

34

35

3.3 COMPARISON OF RESULTS WITH OTHER COMPUTER TECHNIQUES

In order to check the accuracy of the proposed theoretical technique

which is based on Maxwell's equations, solutions to selected problems were

compared to results obtained using another computer program, which calcu-

lates electric fields in configurations with both capacitive and resis-

tive distribution of potentials (Anderson, 1976, Ref. No. 16).

5 5

4 4

1

3 ·3

Figure 3.9 Resistive and complex admittance networks.

Figure 3.9 shows triangular elements and their complex admittance

network. The resultant equation for the complex potential at the center

node is:

v 0

= 1

6 LCG+jB)n n=l

6 ~ (G+jB). V L__ n n

3.8

n=l

To check the accuracy of the program, results of a specific problem

are compared.

36

A square of 100 x 100 mm is divided up into two series connected

halves, one where the capacitive distribution dominates, and one where

the resistive distribution dominates. Permittivities and conductivities

are chosen in such a way that the voltage across each half has the same

magnitude (Fig. 3.10). A very coarse subdivision of only 16 triangular

elements is used.

v = 100 y

v = 0

I e: = 100 y = 0 f = 50 Hz e: = 0.01 _1_ y = 278.031

GOm

Figure 3.10 Two materials in series.

Table 3.1 shows the comparison of results using Andersen's solution

and the proposed solution, to the actual values. As evident from Table

3.1, agreement between the proposed solution and the actual values is very

close.

, I

37

TABLE 3.1

Sta. Actual Values Andersen's Proposed Solution . Solution

y lOO+jO lOo+jO lOO+jO

c 75+j25 74.99+j24.97 74.9999+j24.9695

B 5o+j50 4 9 • 98+j 50. 02 49.9999+j50.0001

A 25+j25 24.99+j24.99 24.9999+j24.9999

x O+jO O+jO o+jO

Also, the accuracy of the proposed theoretical solution was veri-

f ied by comparing the results of the theoretical solution to known actual

values. In all cases very close agreement was observed.

CHAPTER IV

EXPERIMENTAL PROGRAM

In conjunction with the theoretical analysis, an experimental

program was set up. This experiment was based on an average current

density of 0.07 amp per square meter in the uniform ground under the

transmission line. (See Appendix C.)

In order to create a similar situation for the experiment, a large

box with conductors at two ends was chosen to hold the soil and the

swimming pool. Figure 4.1 shows a schematic diagram of the box.

To create a uniform current density throughout the soil a known

voltage calculated from Equation 4-1 was applied across the conductors.

J • Ea = E/p

where . J = current density

p = resistivity of soil

a = conductivity of soil

E = applied voltage

Resistivity of the soil was calculated from Equation 4-2.

R = V/I = pf/s a t/as

4-1

4-2

The experiment was done for three different resistivity values for

the soil, each soil type with two different types of swimming pools,

reinforced and non-reinforced swimming pool; and each swimming pool

39

containing three different types of water.

Three· resistivity values for soil and conductivity values for water

were:

For Soil For Water

a: 1000 ohm-meter 30 micro-mho/cm

b: 55 ohm-meter 1500· micro-mho/cm

c: 10 ohm-meter 3000 micro-mho/cm

To determine the current densities, first potentials at predeter-

mined points were measured and then current densities were calculated

from potential measurements.

current densities _ difference in two potentials - {distance between two potentials)* conductivity of

the material

Figure 4.2 shows a schematic diagram of the circuit used to measure the

potentials at each point.

Reference 18 contains a detailed description of the experimental

program and results. Selected results of this experimental program are

presented in tabula~ form in Appendix F.

, I

4•

Conductor

/. .,?soil_

-•k-/ ~ soil

'--

9' ~ Transverse Cross Section

1.-3.61~ l. •

40

Conductor

3.51

I-< 14.4' · .1 Longitudinal Cross Section

Figure 4.1 Schematic diagram of the experimental model.

, 1

~1 ~~~ ' __ 1 IW

J1 ' CO~Dt:CTOR l

}I r , 7

PROSE 2 --SWIMMUIG POOL

PROBE 1 --.

CO.\u'i,;CTO~ 2

'

l lj~ _ 2 -

I SWITCH

I DIGITAL M£TER l

0 -' I

0 + T I

SWITCH 1 TO MEASURE VOLTAGES IN THE SOIL. SWITCH 2 -TO MEASURE VOLTAG~S IN THE WAT£.~.

Figure 4.2 Schematic diagram of ·the electrical circuit.

41

CHAPTER V

RESULTS

5.1 COMPARISON OF CALCULATED VALU~S WITH EXPERIMENTAL RESULTS

In order to compare the calculated results with the measured

values, current densities of medium case (resistivity of the soil = 55

ohm-meter) are plotted in Figures 5.1 to 5.8.

In these figures, current densities are plotted versus distance.

Each figure represents calculated and measured current densities in

the soil as well as the swinnning pool.

The calculated current densities in these figures show the expected

synnnetry of the system about the center line, Figures 5.1-5.8. This is

one verification of the accuracy of the computer program.

The measured values of current densities do not show the same

exact synnnetry. This could be explai~ed in terms of the accuracy of the

instruments. Also the conductivity of the soil is not uniform everywhere

and the given value of conductivity is only an average measured value.

Another reason for the discrepancy between the theoretical and measured

values is the two dimensional computer modeling, which assumes the

swimming pool walls to be infinitely long in the z-direction (depth).

The calculated values of the current densities between stations

1 to 6 and 12 to 17, Figures 5.1-5.2, in the soil are higher than

measured values. The measured values of current densities inside and

outside the swimming pool between stations 6 to 12 are higher than

43

calculated values. Between stations 6 to 12 the theoretical model

assumes a plate of iron bars of infinite depth. Due to this plate of

high conductivity, the potential gradients along the plate are zero,

resulting in zero current densities along the line 'C'. Furthermore,

the current flowing along the paths 'A' and 'B' are attracted toward the

infinite iron plate resulting in lower values of current densities along

'A' and 'B' as compared· to the experimental case, where only finite

plates of iron bars exist.

Along the line 'D', the calculated current densities must-go through

the infinite iron plate, while in the measured case the current paths

go through the bottom surface bars of the swimming pool. Results in

tabular form are shown in Appendices D and E.

The reason for higher current densities and potential gradients

along the line 'C' between stations 6-8 and 10-12 is the sharp change in

material conductivities at these stations (soil conductivity = 1.8 x 10-l

mho/m; iron conductivity 1.1 x 106 mhofm). Due to high conductivity

of iron bars the current is attracted toward the pool walls and thus

increasing the field (potential gradien~) around the corners.

CURRENT DENSITIES IN SOIL AND POOL WATER (m-Amps/m2 )

SOIL RESISTIVITY= 5 5 ohm-meters POOL TYPE= 181 Reinforced WATER CONDUCT1VJTY= 1500.micro-mho/cm . DMon-rainforced

LINE C Calculated Measured----A

8 c ; ~-~-:---:-=--~r -t f 1 D -J I I- J I

t Station~I 2 Numbar

~

.0 o· C\1

(\J .

E ....... (/)

n -E

. <(

• .!. ·e ~

(j)

~ 1 <n zo ·wo c-

lz w er. ~ . (.) I

I I

~

3 4 5

r-1-..1- __./

. E?,7,8

~ I I

' ' ' ' ' '

9. 10,11,12

' ' , t I I I l I l I I' l I \ I I I \ I \ . I \ I \ I l- .. - __ J

13· 14 15 1617

\~ \.. . ... ' .

' ..... ... _.oi-"'°',

'I '\

o..-._,_ __ .._~--~----...-1~--~--~----""-'-----"~~----.____,.._,

Figure 5.1 Calculated current densities.

44

t-.

l

I 45

CURRENT DENSITIES IN SOIL AND POOL WATER (rn-Amps/m2 )

SOIL RESISTIVITY= 5 5 ohm-meters POOL TYPE= ~Reinforced WATER CONDUCTIVITY:: 15 00 micro-mho/cm . DMon-reinforced

1 ·LINE D Calculated Measured----A

8 c

D

.f y

Station~I 2 Number

.

0 o· C\J

<§"' • E

' (/).

0. -E

. <( . .!. .E -CJ) w -l-m zo wo· o-

-r

3

Jz w 0:::: I _, _ _... 0:: :::> • (.)

I ,_

4 5

''I • -----x •• l 1 "'/.~,-

"I I

I" + -1 I I- t·I h

6;7,8 9. 10,11,12. 13' 14 15 1617

I I

I I

I··

.,... ____ ..... _____ .

'1\ I \ ~ \

' ' ' - ......... ,

"'

Q.J.._.J......--.L--1-~~..L----LlU===========::W--~.J---......L---l-~.J.....I

l

I

I CURRENT DENSITIES IN SOIL AND POOL WATER (m-Amps/m2 )

SOIL RESISTIVITY= 5 5 ohm-meters POOL TYPE: ~Reinforced WATER CONDUCTIVITY=30QO micro-mho/cm DNon-reinforced

LINE C Calculated Measured----

AT ll I Ill ! HI : : : r-x 8 ' i I I 1:1 I "Ii '

~ ~ II I - ::1 t 1:: - : : : - : i t i JI. . A.

y <t Station~ I 2 3 4 5 6;7,8 9 IO,fl>l2 13 14 15 1617 Number

t<J' E

........ (f)

a.. E <! • ..!. . E

en w t-

0 0 C\J

(f)

zo wo o-

rz w 0::: 0::: :::> (.)

0

-

.

.

.

.

-.---

. r

I I

I I

.

.

h .. \

\ \

\ \ \ . \ \ \ \ \ \ \

' \

' \ \ I '

__./" I ,.

I "'--- ,/ \ I \ - r--. -: ,, \ I " - ... ., . \ ........ ' """"" ..... _,,,,

I -~· \ \

I ~

\ \ I

' . ' I \ .... -' i...-


46

l

!

CURRENT DENSITIES 1N SOIL AND POOL WATER (m-Amps/m2 )

SOJL RESISTIVITY= 5 5 ohm-meters POOL TYPE= ~Reinforced WATER CONDUCTIVITY= 3Q00micro-mho/cm DMon-reinforced

LINE D Calculated Measured ----A ---~~~--~~~~~~------~~--~...,.__,__ ______ ~x

8 l •

c 1 i J ) i li' I

D I ' I I '

.. t . y

Station~ I 2 3 4 Number

~ E

.......

0 o· N

CJ)

~ E .. . <(

• .!. .E .

(/) w t-oo zo

. l.1.1 0 c-l-z ,,,..... w ,.. 0:: 0:: => (.)

5 ~;r,a 9. lqfl)2 13

11------1-- - - -

t-- l I <t··

14 15 1617 ..

'

Q I I I I I Ill I IJI I l I I I


47

CURRENT DENSITIES IN SOIL AND POOL WATER (m-Amp:>/m2 )

SOJL RESISTIVITY= 59.5 ohm-meh~rs POOL TYPE: 0 Reinforced WATER CONDUCTJVITY = 16 OQ_micro-mho/cm . rzl Non-rainforced

LINE C Calculated Measured -- --A

~ y Station-:-- I 2 3 Number

-

o_ 0 C\J

c:\J E

........ CJ).

a.; . -E

. <( • ..!.. .E --CJ) w -I-en zo WQ o-I- -z w er: 0:: :::> (..)

0

r--

4 5

.

Ill

. 6,7,8

\ \ \ \

t 9.

I

;:1 l l 1 ,V!-~x

..

r 10,11,12

'Ill I I I I I I I

-1 I J·- i t-t.

13 14 15 1617

F·igure 5. 5 Calculated current densities.

48

, i I


SOtL RESISTIVITY =59.5 ohm-meters POOL TYPE= D Reinforced

WATER CONDUGT!VITY =1600 micro-mho/cm . r2I Non-reinforced

LINE D Calculated Measured----

AT I : : ::: I l:: : : : r-x ~: ' .. , l !·· .

D I I • A - "I i 1:, - I ~' =ft t

Station~ I 2 3 4 5 6;7,8 Numbar

&" E

......... (/)

a.. E <( . .!. . E (f) w 1-

0 0 C\J

(f)

zo wo o-1-z w 0:: 0:::: :::> u

0

.

.

.

.

. "I;"

.

.

.

i...--

I I I

)! ___/ / - i..

/ '- - --./

<t 9. !O,fli12 13 14 15 1617

•

~ -_,,,. \ - \ -- \ \ \ \ \

~ ~ -\... /\ .... /

' '-/'

'

Figure 5.6 Calculat~d current densities.

49

SOIL RESISTIVITY =59. 5 ohm-meters POOL TYPE= 0 Reinforced

~'.'ATER CONDUCTl'/ITY =3..QQQ.micro-mho/cm . f81 Non-reinforced

LINE C Calculated Measured----1

t-•• t JJ

" t __ ~x AT I l l Ill I ,, ' '

~ I : : : : : ;! I ~ 1 l ; : J . I i •

nl f f 1 - , 1 , - • 1 ·r o--

t . . y

Station~ I 2 3 4 Number

t<i'" E

......... U> a. E

. <l: • ..!. .E

(/) w .. _

0 0 C\.I

en zo WQ c-

I l I I

-

-

-

.

5 ~,7,8 I I

\ \ \ \ \ \ . \ \

.

lz -~

•/

./ \ \

w 0:: n: ::> (.)

0

-

-

-

.

-. .... r....._ -,_

I

/

\ _,, r _...,,.. - \ \ \ \ l

9. 10,tl,12 13· 14 15 1617 . I I : I

j

I ' I I I ' I ' I I\-' I --- --... ..... I -1-- r ', I

I I I I I

- r-- ..J

-l I L .


50

CURRENT DENSITIES 1N SOIL AND POOL WATER (m-Amps/m2)

SOIL RESISTIVITY=59 5 ohm-meters POOL TYPE= 0 Reinforced WATER CONDUCTIVITY= 3000 micro-mho/cm l8l Non-reinforced

LINE D Calculated Measured----A

--..-~~~~~~~-,.-~--+~~-,..,~----:~~~-:-___,r:-7~x

B I : • l l 1 •• I I!' l 1 : I l c I · ; 1 1 1:· ..

; I L! [I 0-+-+-

t Station-)'> I Numbar

t\J E

' (I) . a. E

. <(

• ..!..

"E~ -en w ..__

0 0 ~

(f)

zo wo o-..... z w· 0:: 0:: ::::> (.)

0

2

--

LJ I I I I I

I <t

3 4 5 s;?,a 9. 1~11)2 13 14 15 1617

...... /

/ JO .;I'

.... ·/ ,_,,..,,.,.

I

\ \ \ \ \

' \ \ .

I / l

\

IJ '\\ ~ /

/ \'----"' ' -... ~ / - ~

'-. ------./

...... -.. ......,,.


<t

51

52

5.2 POTENTIAL HAZARD TO THE HUMAN BODY

It is known that the real measure of shock intensity lies in the

amount of current (amperes) forced through the body, and not the .voltage

(17). Figure 5.9 shows levels of current hazards to the human body.

To define how hazardous the observed current densities are to

humans, currents through the human body are calculated. The human body's

resistance is in the neighborhood of 1000 ohm (17).

the human body if one is standing in the vicinity of the swimming pool.

Similar calculations are done for a person who is inside the swimming

pool and results are shown in Table 5.5.

In comparing the calculated currents traveling through the human

body with Fig. 5.9, one concludes that hazard may exist on the edge of

the swinuning pool where the resistivity of the surrounding soil is very

high. However, this analysis does not include the presence of a human

body in the model. Also, the effects of short duration currents (1-10

cycles) on the human body need further investigation.

.en LL.I a:: w a.. ~ <(

SEVERE BURNS .BREATHING STOPS

53

0. 2 t--r--,.--r-~------.----.____.,-__..,._..,...-.,... _____ _,,_-.---.

0.1 t-

EXTREME BREATrlING t- DIFFIOJLTIES

BREATrlING UPSET LABORED SEVERE SHOCK MUSCULAR PARALYSIS CAi'INOT LET GO

0.01 PAINFUL.

MILD SENSATION

TI-fRESHOLD OF SENSATION

.OOIL---~~~~~~--~--~~~~~~~

Figure 5.9 Levels of current hazards to the human body.

l

TABLE 5.1

CALCULATED CURRENT TRAVELING THROUGH THE HUMAN: BODY STANDING ON THE SOIL

LINE A STANDING ON THE SOIL

SOIL RESISTIVITY=ll60 OHM-METER WATER CONDUCTIVITY=l500 MICRO-MOH/CM

POOL TYPE: REINFORCED

FOOT TO FOOT RESISTANCE OF HUMAN BODY = 1000 OHM

FOOT TO FOOT DISTANCE= 50·CM.

54

BETWEEN VOLTAGE GRADIENT VOLTAGE ACROSS BODY CURRENT THROUGH BODY

ST/I ST/I VOLTS/METER VOLTS AMP.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -1 2 102.500 51. 250 0.05125 2 3 101.836 50.918 0.05092 3 4 100.472 50.236 0.05024 4 5 96 .182 48.391 0.04839 5 6 84.769 42.384 0.04278 6 7 69.020 34.510 0.03451 7 8 67.368 33.684 0.03368 8 9 53.210 26.605 0.02660 9 10 51. 741 25.870 0.02587

10 11 64.211 32.105 0.03211 11 12 65.882 32.941 0.03294 12 13 80.417 40.208 0.04021 13 14 91. 500 45.750 · ·o.04575 14 15 94.874 47.437 0.04744 15 16 96.447 48.224 0.04822 16 17 96.235 48.118 0.04812

- - - - - - - - - - - - - - - - - - - - ~ - - - - ~ . .

1 l

55

TABLE 5.2

CALCULATED CURRENT TRAVELING THROUGH THE. HUMAN BODY STANDING ON THE SOIL

LINE B STANDING ON THE SOIL

SOIL RESISTIVITY=ll60.00 OHM-METER WATER CONDUCTIVITY=1500 MICRO-MOH/CM


FOOT TO FOOT RESISTANCE 9F HUMAN BODY = 1000 OHM

FOOT TO FOQT DISTANCE = 50 CM.·

BETWEEN VOLTAGE GRADIENT VOLTAGE ACROSS BODY CURREN~ THROUGH BODY

ST# ST# VOLTS/METER VOLlS AMP.

1 2 104.737 52. 3'68 0.05237 2 3 104.492 52.246 0.05225 3 4 104. 961 5·2. 480 0.05248 4 5 105.509 52.755 0.05275 5 6 103.148 51. 5.74 0.05157 6 7 71.176 35. 588' 0.03559· 7 8 .62. 632 . ·3i. 316 . o. 03132 8 9 28.275 14.·137 0.01414 9 10 27.056 13.528 0~01353

10 11 58. 94 7 29.474 0.02947 11 12 67.255 33.627 0.03363 12 13 97.199 48.600 0.04860 13 14 99.493 49.747 0.04975 14 15 98.984 49.492 0.04949 15 16 98.852 49. 426, o-. 04943 16 17 98.209 49.105 0.04910

- - - - - - - - - - - - - - - - - - - - ~ - -~- - - - - - - - -

TABLE 5.3

CALCULATED CURRENT TRAVELING THROUGH THE HUMAN BODY STA..~DING ON THE SOIL

LINE C STANDING ON THE SOIL

SOIL RESISTIVITY=ll60 OHM-METER WATER CONDUCTIVITY=l500 MICRO-MOH/CM



FOOT TO FOOT DISTANCE= 50-.CM

56


ST# ST# VOLTS/METER VOLTS AMP.

- - - - - - - - - - ~ - - - - - - - - - - - - - - - - - - -1 2: 105.066 '52.533 0.05253 2 3 104.951 52.475 0.05240 3 4 105.945 52.972 0.05297 4 5 108.403 54.201 0.05420 5 6 122.431 61. 215 0.06122 6 7 200.980 100.490 0.10049 7 8 0.000 o.ooo 0.00000 8 9 0.000 0.000 0.00000 9 10 0.000 0.000 0.00000

10 11 0.000 0.000 0.00000 11 12 188.235 94.118 0.09412 12 13 115.208 57.604 0.05760 13 14 102.185 51. 093 0.05109 14 15 99.882 49.941 0.04994 15 16 99.276 49.638 0.04964 16 17 98.536 49.268 0.04927

- - - - - - - - - - - - - - - -.- - - - - - - - - - - - - - - -

TABLE 5.4

CALCULATED CURRENT TMVELING THROUGH THE H~~ BODY STANDING ON THE SOIL

LINE D STANDING ON THE SOIL

57

SOIL RESISTIVITY=ll60 OHM-METER WATER CONDUCTIVITY=l500 MICRO-MOH/C~


FOOT TO FOOT RESISTANCE OF HUMAN BODY = .lOQO OHM

FOOT TO FOOT DISTANCE= 50 CM·



- - - - - - - ~ - - - - - - - - - - - - - - - - ~ - - -1 2 105.461 52.730 0.05273 2 3 105.410 52.705 0.05270 3 4 106.929 53.465 0.05346 4 5 110.787 55.394 0.05539 5 6 127.153 63.576 0.06358 6 7 131.961 65.980 0.06598 7 8 IN WATER IN WATER IN WATER 8 9 IN WATER IN WATER IN WATER 9 10 IN WATER IN WATER IN WATER

10 11 IN WATER IN WATER IN WATER 11 12 123.922 61. 961 0.06196 12 13 119.630 59.815 0.05981 13 14 104.394 52.197 0.052~0 14 15 100.795 50.398 0.05040 15 16 99' 717 49.859 0.04986 16 17 98.863 49.431 0.04943

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

TABLE 5.5

CALCULATED CURRENT TRAVELING THROUGH THE HUMAN BODY INSIDE THE SWIMMING POOL

58

SOIL RESISTIVITY=59.SO OHM-METER WATER CONDUCTIVITY=3000 MICRO-MOH/CM

POOL TYPE: NON-REINFORCED


FOOT TO FOOT DISTANCE = SO CM



- - - - - - - - - - - - - - - - - ~ - - - - - - -1 2 0.875 0.438 0.00044 2 3 0.800 0.400 0 .. 00040 3 4 o. 775 0.388 0.00039 4 5 0.787 0.394 0.00039 5 6 0.788 0.394 0.00039 6 7 0.788 0.394 0.00039 7 8 0.800 0.400 0.00040

- - - - - - - - ~ - - - - - - - - - - - - -· - - ~ -

~ I

. 59

5.3 THE LIMITATIONS AND ACCURACY OF THE THEORETICAL TECHNIQUE·

The teehnique used in the proposed so·lut:i.on is c~ll'ed The Finite

Element Method. This is a powerful numerical technique to solve problems

which require a high degree of accuracy. The solution of problems solved

using this technique are comparatively more accurate than those solved

by other numerical methods such as the Finite Difference Method.

However, there are some limitations as described below:

(1) This program as it exists now can only handle two-dimensional

problems. However, with furth~r development, it would be

possible to .. solve three-dimensional complex electro-magnetic

and electro-static field problems.

(2) This program uses only the triangular division of the region

of interest. Rectangular or other shapes can be accommodated

if the program is modified.

(3) The computer storage is another limitation. This limitation

was improved by using the half-banded method.

CHAPTER VI

CONCLUSION

In this study a computer code based on Maxwell's Equations was

developed to use the Finite Element Method to calculate complex voltage

gradients and current densities on the surface of any desired region.

In order to evaluate the accuracy of this program, the solution to

a selected problem was compared to the solution using another computer

technique. In addition, solution to several problems were compared to

actual known values. In all cases close agreement between the theoretical

solution and actual values was observed.

Also in order to check the validity of the program, theoretical

results were compared to results obtained from experimental tests, and the

comparison showed close agreement.

1

l I !

BIBLIOGRAPHY

1. John R. Carson; "Wave Propagation in Overhead Wires with Ground Return", The Bell System Technical Journal, 1926, 5, pp. 539-554.

2. L. J. Lacey; "The Mutual Impedance of Earth-Return Circuits", IEEE, Vol. 621, 392.2, 1952.

3. L. M. Wedepohl and R. G. Wasley; "Wave Propagation in Multiconductor Overhead Lines", Proc. IEE, 1966, 113, (4).

4. M. Krakowski; "Mutual Impedance of Crossing Earth-Return Circuits", Proceedings of the IEE, Feb. 76, Vol. 114, No. 2.

5. Mr. Nakagawa; "Earth Return Impedance of Overhead Lines Above a 3-Layer Earth", Proceedings of IEE, Dec. 1973, Vol. 120, Number 12.

6. P. Magnusson; "Wave Propagation Over Parallel Wires: The Proximity Effect", Vol. Xll, Apr. 1921.

7. H. Bateman; "Partial Differential Equations of Mathematical Physics", Cambridge University Press, 1959.

8. E. D. Sunde; "Earth Conduction Effects in Transmission Systems", Van Nostrand, 1949.

9. B. O. Pierce; "A Short Table of Integrals", Boston: Ginn and Company, 1929.

10. N. W. McLachlan, "Bessel Function for Engineers", 2nd Ed. Oxford University Press, 1955.

11. Wiley; "The Finite Element Method for Engineers", New York, 1975.

12. 0. C. Zienkiewicz, "The Finite Element Method in Engineering Science", McGraw-Hill, 1971.

13. P. Silvester; "Finite Element Solution of Saturable Magnetic Field Problems", Transaction on Power Apparatus and Systems, Vol. PAS-89, No. 7, Oct. 1970.

14. O. W. Andersen; "LaPlacian Electrostatic Field Calculations by Finite Elements with Automatic Grid Generations", Transaction on Power Apparatus and Systems, IEEE, Nov. 1972.

15. O. C. Zienkiewicz and Y. K. Cheung; "The Finite Element Method Instructural and Continuum Me~hanics", McGraw-Hill, 1967.

1

l I I

16.

17.

18.

62

O. W. Andersen; "Finite Element· Solution of Complex Potential Electric Fields", IEEE Transactions on Power, Vol. PAS-96, No. 4, July/ August 1977.

~. W. Kimbark, "Direct Currerit Transmission", Vol I, Wiley-Interscience, 1948.

V. K. Garg, F. Rad, T. Killian; "Study of Distribution of Ground Fault Currents in Below Grade Swimming Pools Located Near Transmission Lines", Portland State University, 1978.

l

1

APPENDIX A

EULER'S THEOREM OF VARIATIONAL CALCULUS

The transition from a variational statement to an equivalent

governing differential equation is relatively simple and will be demon-

strated here. The reverse process, however, is more involved and any

generalized processes restrictive for the very reason that frequently

on variational principle can be established.

Let us take a problem which is to be minimized.

g = ( f(x,y,z,H,H ,H ,H )dv +.( (qH + pH2/2)ds )v x y z }c

A-1

I h . . f . b. f . H aH d · n t 1s equation 1s an ar 1trary unction, =-;:;--,etc., an c 1s a x ox

portion of the boundary surf ace on which prescribed values of H are not

imposed. On remainder H = HB.

Considering an arbitrary small variation of the unknown function

and its derivitives

as

S af af af af

og = <aH oH + IB oHx + ail oH + ail oH2

)dv + v x y y z

f c (qOH + pHOH)ds

oH x

o(aH) = ax ~ (oH), etc. ax

A-2

1

l I I

64

Equat•ion A-2 can be written as;

Og = ~v [ 1!_ oH + li_ _l c oH) + li_ _l (oH) + lL -1. c oH) J dv aH aH dx aH ay aH az x y z

+5 v (qOH + pHOH)ds = 0 A-3

In the above we have equated 8x to zero, as at the minimum (or·

stationary point) the 'variation' becomes zero.

Now putting dv = dxdydz and integrating the second term of Equation

A-3 by parts with respect to x

S af a i af J a af - - (cSH)dv = - oHL ds - - (-) oHdv aH ax aH x ax aH v x s x v ·x

In which L is the direction cosine of the normal to the outer surface x

with the x axis. Performing similar operation on the other terms of

Equation A-3 and substituting, it becomes;

Og = Sv OH af a af a af a af dv + aH - ax (ail) - ay <aH ) - ~ (ail)

x y z

fc OH + H + L .Ei_ + L . .il..- + L .£!__ ds A-4 q P x aH y aH z aH

x y z

The second integral is only taken over the boundary C as on the remainder

of surface S we have prescribed values of H and therefore oH = O.

For Equation A-4 to be true for any arbitrary variation· H first

integral should be equal to zero;

af a af a af aH - ax (ail) - ay <aH )

x y

a af az (aH ) z

0 A-Sa

l

l I 65

Everywhere within the region V, and on the boundary C

L ~+L 2-L+L 2.!_=0 x aH y aH z dH x y z

A-Sb

These two equations, if satisfied by H, minimize g. If the solution

is unique then formulations A-1 and A-5 are equivalent. The above

differential equations are known as the Euler equations of the problem.

APPENDIX B

THE GAUSSIAN METHOD

As an example, the solution of three equations and three unknowns

is described below:

In matrix form:

200

-100

0

200X - lOOY + OZ = -8

-lOOX + 200Y - lOOZ = -8

OX - lOOY + lOOZ = -8

x -8 -100

200

-100

0

-100

100

Y I = I -8

z -8

B-1

B-2

Let us first solve this problem as it is in the form of B-2 then reduce

it to the banded form.

200 -100 0 , -8

-100 200 -100 ' -8

0 -100 100 L-8 L

Divide the first row by the diagonal element of the first row.

1 -0.5 0 1-0. 04 -100 200 -100 ' -8

0 -100 100 I -8

67

Multiply the f ~rst row by the first element of the second row and subtract ·

the first row from the second row.

1

0

0

-0.5

150

-100

0

100

100

-0.04

, -12

-8

This manipulation introduced a zero to the second row~ therefore

these are two unknowns in the second row. Now divide the second row

by the diagonal element of the row.

1

0

0

-0.5

1

-100

0

-2/3

100

-0.04

, I .-0. 08

-8

Multiply the second row by the second element in the third row and subtract

the second row from the third row.

1

0

0

-0.5

1

0

0

-2/3

100/3

, -0.04

-0.08

-16

This introduced two zeros to the third row and the third row

contains just one unknown. The unknown may now be easily calculated.

One may proceed to the second and third rows and calculate all the un

knowns. The answer:

100/3 z = -16 z -0.48

Y - 2/3Z = -0.08

y - 2/3(-0.48) = -0.08 y = -0.4

l

X - 0.5Y = -0.4

x = -0.04 + 0.5(-0.4) = -0.24

Now put the upper half band of the coefficient matrix in a new [D]

matrix and solve the problem.

D

********** 200 -100

200 -100

100 0

B ******

-8

-8

-8

68

For simplicity let us not multiply or divide the first columns by

any numbers. At the end substitute 1 for all these elements. Also,

due to symmetry of [A], D(l,2) =A (2,1) and D(2,2) = A(3,2).

First store D(l,2) in c, because it is the same as A(2,l) and

there is no A(2,l) in our [D] matrix.

Divide the first row by the first element of the first row.

200 -o.s -0.04

200 -100 9 -8

100 0 -8

Then multiply the first row by the stored value of c and subtract

D(l,2) from D(2,l), because these two elements correspond to the same

unknown in matrix [A].

200 -0.5 -0.04

150 -100 ,

-12

100 0 -8

i

Now ·store D(2,2) in c and divide the second row by the first

element of the row.

200

150

100

-0.05

-2/3

0

-0.04

, I -o. 08

-8

69

Multiply the second row by the stored value of c and subtract

D(2,2) from D(3,l), because these two elements correspond to the same

unknowns.

200

150

100/3

-0.5

-2/3 I ,

0

-0.04

-0.08

-16

Divide the third row by the first element.

200

150

100/3

-0.5

-213 I , 0

Now substitute 1 for column one.

1

1

1

-0.5

-2/3 I _, 0

-0.04

-0.08

-0.48

-0.04

-0.08

-0.48

This is the same result as before in banded form. Therefore,

the problem has been solved by the use of a simpl~r method and also

it saved memory space. Fig. 3.1 shows a flow chart of such a program.

S.TART

F'ORWARD REDUCTION OF .MATRIX

BY GAUSSIAN ELIMINATIOX.

. CALCULATE C c = A(N,L) I A(N,1)

MULTIPLY THE ROW BY C

YES

AND SUBTRACT FROM NEXT ROW

FORWARD REDUCTION OF CONSTANTS

N = 1, SIZE L = 2, BA~D

YES

Figure B. Flow chart, subroutine "SOLVE".

~

70

I = N+L-1

CALCULI\ TE THE CORRE.SPO?"DING CONS TANT

B(N) = B(N) I A(N,1)

SOLVE FOR uNKNOWNS BY

BACK - SUBSTITUTION

M = 21 SIZE

.N = SIZ.E+l-N

CALCULATE THE UNKNOWN

NEXT L

REltlRN

Figure B. (Continued)

71

APPENDIX C

CALCULATION OF CURRENT DENSITY

The equation for mutual impedance (G) between two infinitely long

conductors, at heights h1

and h2

meters above the earth, and separated

by a horizontal distance of y1 meters, in the power series form is:

jwµ 4

wµ 212(h1,o·,y1) = 4w

0 [R,n( '2 '2) + 1 - 2Y] + T-

hl + Y1

' (

wµ h 1 - j) __ .,o 1

. '2 '2 J wµ o (hl - y 1 )

64

'2 '2 '2 + '2 wµo(hl - yl) [2n(hl yl ) + 2Y - 5/2] C-1

327f 4

Where Y is Euler's number, 0.577216

' h1

= h1

lwµ0

a1

h2

= h2

lwµ0

a1 Y1 = Y1 lwµo al

and current density equation is:

J = 212<h1,0,y1)*. a * Isc C-2

Based on equations C-1 and C-2, impedances and current densities

on the surf ace of the earth for various values of y and different values

of p are calculated and given in Tables C.l to C.3.

The worst case is for y = 0 and the current density for such

· value of y is 0.07 amp per square meter.

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APPENDIX D

LISTING OF PROGRAMS AND SUBROUTINES

Listing of Programs:

MAIN GOOD TES SEARCH

and Subroutines:

BAND PRO PT FIND SOLVE

M

I I

c

-c

PROGRAM MAIN

COMMON A<40t,20),N(3,760),XN(3,760),yN(3,760),MATC760),IEL<760) COMMON BC401),VC403) HJTEGEF~ ti, NODE, 1...NODE, IEL, IELE COMrLEX A1B,PRQp,p1,p2,p3,v,DCV

P1=<1.8182E-04,3.542E-11) P?=C30.0E-OO.O.OOE+01) P3=<3.0E-03,7.4374E-10)

C IBW - eAND-WIDTH OF THE COEFFICIENT MATRIX C NODE - NUMBER OF NODES C LNODE- NUMI-{ER OF .LAST NODE BEFORE THE NODE WITH BOUNDFrnY COND. C IEL - NAME OF MATRIX WHICH STORES ELEMENTS NUMBER C IELE - NUMBER OF ELEMENTS c c c

R~ADC15,1>NODE~IELErLNODE,BCV

1 FORMAT(I311x,13,1x,13,2c1x,F7.3)) c

WRITEC66,69)P1 WRITEC66,69)P2 WRITEC66169)P3

69 FORMATC2C1X,E14+6)) c

IBW=O

77

DO 20 I=11IELE READC1512>IEL,NC1,J),NC2,I),N(3,J),XN<1,J),YNC!rI>1XNC2,I),YN<2

c c

C1I>1XNC3,I),YNC3,I)1MAT

2 FORMAT<I3,3(1X113),6C1X,F8+3),1Xrl1) c 20 CONTINUE c

DO 22 I=1rIELE c C THIS SUBROUTINE FINDS THE BAND-WIDTH OF THE MATRIX c

r: ~

22 CALL BANDCNrl1LNODE,IBW>

flO 60 !=1,LNODE DO 60 .. J=:1, IBL.J

60 ACI,J)=(O.o,o.o> [!IJ 1 99 KK= 1 , LNODE

199 V.CKK1=CO.o,o.o>

c

V(UWDF+1 )-:::(0.o,o.o> l) < LNODE+2) =BCV DO 190 K=t,IELE

C SUDROUTINE PROPT FINDS THE PROPERT OF MATERIALS AND AREA OF IHE C ELEMENT.

78

Pl"\OGF\AM MA IN < CDNT. >

CALL PROPT<MAT,K,PROP,P3,P1,P2,S,XN,YN> c C THIS SUBl""\OUTINE FINDS THE COEFFICIENT MATRIX AND CONST ANH>. c c c

c CALL FIND<N,K,LNODE,YN,xN,s,pRQP,B1V1A)

190 CONTINUE c c c c C THIS SUBROUTINE CALCULATES THE VOLTAGES AT EACH NODES+ c

c c

CALL SOLVECLNODEvIBw,A,B) c.~L..L ~-;:E~:;u1 ... T (BCV '1 B)

1000 CONTINUE STOP END

r·F~OGr~AM GOOD

DIMENSION X<401>1YC401),XXC17>1YY<17) RE~DC15,2)L,M,NX1NY

2 FOl:;~MATC4!3)

DD :1.0 I::::1,NX 10 READC15,1>XX

DO 20 :£::::1.,NY 20 READC1511>YY :L F 0 F\: MI~ T ( F B • 3 )

,J:::;()

L1:::::1. DD ~50 I ==L, M ,J!'.::.J-} 1 X(!)::::XXL.J) Y < I) ::::yy < L :J. > Ii::::I-L+:L IF<I1.LT.NX> GO TO 30 f':::: I 1 /FLOAT< NX) L1::::IFIX f='::::P·"·Ll IF<P.EO.O.O> GO TO 35 GO TO 3t»

3~5 J=O :36 L.:t::::l...1+1

30 CONTIMl.JE DO ~rn I=-.:L' M

50 WRITEC1613>I1X,Y 3 FORMAT<3X1I312<2X,F8.3))

79

PROGF~AM TES

DIMENSION NC3,760),XNC3r760),yN(3r760)rMAT<760),JEL<760),XC401) DIMDJSION YC401) fl C' 5 I ~= 1 , 4 0 1

5 READC13,1>KKrXCI>,Y 1 FOPMAT(3X,r3,2(2XvF8.J))

flf:! 10 I=17r744 READC1412)IELCI),N(1,I>rN(2,J)rN(3rI>rMAT

2 F 0 F: ri A r ( I 3, 3 ( 1 x' I 3) ' 1 x, I1 ) L=.'-.J(1rI) .J :: f'.' < 2 , I ) f'(=~l(3d)

XNClrI>=X(L) nH 1 !'I) =Y CL) Xt~(2, I >=X(J) y;{ ( 2, I> =Y ( J > XU<3rI)=X<K> Ytl<31 I >=Y<K>

10 f-ONTINUE DO :20 I=1r16

80

20 R£~~(16?3>IELrN(1,I>rN<2rl)rNC3rI>rXNC1rI>rYNC1rl)rXN<2,I>rYN<2 c, I), XtH3, I) ,.YNC3, I> rMAT

3 FORMATCI3r3C1XrI3>r6C1XrF8.3)r1XrI1> . fl 0 •W I= 7 4 5 ,. 7 6 0

40 RE~DC16r3>IEL(l)rNC1rI>1NC2rl)rNC3rl)rXNC1r!)rYNC1rl)rXNC2rI>rYN<2 C1I>YXNC3rI>,YNC3rl)rMATCI>

DO 100 I=l,760 100 WRITEC17r4>IELCI>,NC1r!)rN<2rl>rNC3rI>rXNC1rI>rYNC1rI>rXNC2rI)rYNC

c2,r>,XN<3r!)1YN<3r!),MATCI> 4 FORMATCI3r3C1Xrl3)r6C1XrF8.3>r1Xrll)

STOP END

.. ...

'')

··~

PI E:<5. :l 41. 5926 t~<~T7. 0 XMU=4*PIE*<10E-8> 'ZEGMA=O + 1. GAMA::::O. ~57'7216 H1==10. 68 l·J::::H :l *Smn ( W*XMU*ZEGMA) l,Jf~ITE < 1~5, :L) lJfU TE< 15 ~.::;.~ >

81

Pf~OGF~AM S[f\f~CH ...... ______ ...,. ____ _

F OF~ MAT <:~ X , • D I ST AN CE • , 4 X , • f~ EAL F' A Fff • , 5 X , • I MAG 1 NARY • , ~) X ' • MUTUAL • c,5x,·ANGLE•,3x,•cuRRENT 11

)

FORMAT<2x,·FROM CENr•,3x,•IMPEDANCE•,5x,•1MPEDANCE•,26X,•DENSITY·~ DO 99 N==O, 30 Y1:=N*:LO+O Y=Y1*SQRT<W*XMU*ZEGMA> f.t :::: H * * ~~ + Y * >.'< 2 B::::H**2-Y**2 C====Al ... OG ( A/4 > [l::::Al ... OG < 4/A > REAL_1=(1.0/8.0>-CH/(4+24264*PIE>>-<<BIC32*PIE>>*<C+2*GAMA-2+5)) XIMAG1=C<D+1.0-2*GAMA>l<4*PIE>>+CH/C4+24264*PIE>>-<B/64> l=i:EAl...=W*XMUi!<f~EAL 1 XIMAG=W*XMU*XIMAG1 Z=SQRT<REAL**2+XIMAG**2> PHI::==ATAN ( X IMAG/f:;:EAL > PHI=PHI*C180+0/PIE> XJ=Z*lEGMA*20000.0 WRITEC15,3)Y1,REAL,XIMAG,z,pHI,xJ

3 FORMATC2X,F6~1,5X,E9+4,5X,E9.4,5X,E9.4'5X,F5+1,3X,FB+6,2X, cuAMP+ PER SQ. METERU)

?? CONTINUE STOP END

40

4 1:." ,.}

~:) ()

!55

60

SUBf\DUT I NE BAND

SUBl~OUT I NE BAND< N, I, LNDDE, 1 BW) DIMENSION NC3,760) IFCNC1,l)tGT.LNODE> GO TO 60 IFCNC-~,I) .GT.UWDE> GO TD 60 IF(N(3,I>.GT+LNDDE> GO TO 60 ID1=ABS<NC1,I>~N(2,I>> ID2=ABSCNC2,I)-N(3,I>> IB3~ABS<NC1,I>-NC3,!))

IF<IB:l+LT.IB2> GO TO 40 IFCIB1+LT+IB3) GO TO 45 I BlJ :I.=--= J: B 1 + :L GO TO !7;0 IF<IB2+LT+IB3) GO TO 45 ID~J1:=IB2+1

GO TO 50 :r B~J l :::: I B:3+ 1 IF<IBW.LT.IBW1> GO TO 55 GO TO 60 I BW:::: I BW 1 I B l·J 1. ::::0 HETlJl~N END

82

1

SUDF~OUT I NE PF~OPT

SUBROUTINE PROPT<MAT,K,PROP,p3,p1,p2,s,x~,yN) DIMENSION MAT<760),XN<3,760),yN(3,760) COMPLEX PROP,P1~p2,p3 IFCMAT<K>+EO+l) GO TO 10 IFCMATCK>+EQ.2) GO TO 20 PROP=P3 GO TO 30

10 PROP=P1 GO TO 30

20 PROP=P2

83

30 S=2*<CXNC1,K>*YNC2,K>-XNC2rK>*YNC1,K>>+<XN<2rK>*YNC3,K>-XN<3,K>*YN C(2,K>>+<XNC3rK>*YNC1,K>-XNC1rK>*YNC3rK>>>

RETURN END

l

SUBROUTINE FIND

SUBROUTINE FIND<N,K,LNODE,YN1XN,S,PROP,D,V,A> DIMENSION A(401,2Q),V(403),B(401),N(3,760),XN<3,760),yN(3,760) COMPLEX PROPrA,B,x,v DO 99 M=!,3 DO 90 J:::1,3 IFCM.£Q,I) GO TO 80 IF«t'U1,K>-N<J,K)) .GT.O> GO TO 90 IFCM.EQ,1) GO TO 68 IFCM.EQ.2) GO TO 85 IFCI.EQ.1) GO TO 82 00 Tfl 83

68 IFCI.EQ,2) GO TO 81 GO TO 82

85 IFCI.EQ.1) GO TO 81 GO TO 83

81 IFCN(M,K>.LE.LNODE> GO TO 181 GO TO 90

84

181 IF<N<I,K>.LE.LNODE> GO TO 281 X=CCCYNC2,K>-YNC3,K>>*<YN(3,K>-YN<1,K>>+<XNC2,K>-XN<3,K))*(XN<3,K)

C-XN<l,K>>>IS>*PROP MM=N(!,K) Wl=N CM, t\) BCNN>=-X*V<MM>+B<NN> GO TO 90

82 IF(N(M,K>.LE.LNODE> GO TO 182 GO TO 90

182 IF<N<I,K>.LE.LNODE> GO TO 282 X=<<<YNC2,K>-YN(3,K>>*<YNC1,K>-YN~2,K>>+CXNC2,K>-XNC3,K>>*<XN<1•K>

C-XN(2,K)))/S>*PROP MM=N<I,K) NN=N(M,K> B<NN>=-X*V<MM>tBCNN> GO TO 90

83 IFCNCM,K>.LE.LNODE> GO TO 183 GO TO 90

183 IFCN(J,K>.LE.LNODE> GO TO 283 X=<<<YN(3,K>-YNC1,K>>*<YN(1,K>-YNC2,K>>+<XN<3,K>-XNC1,K>>*<XN(1,K>

C-XN<2•K>>>IS>*PROP MM=N<I,K> NN=N < rh K) DCNN>=-X*V<MM)fBCNN> GO TO 90

281 MM=NCI,K>-NCM1K>+l NN:=N < f"1, K > A<NN•MM>=<<<YN(2,K>-YNC3,K>>*<YN(3,K>-YN(1,K>>+<XNC2,K>-XN<3,K))*(

CXNC3,K>-XNC1,K>>>IS>*PROP+A<NN,MM> GO TO 90

282 MM=NCJ,K)-N(M,K>+l NN=N <l'h K) A<NN1MM>=CCCYNC2,K>-YN(3,K>>*<YNC1,K>-YN<2,K>>+<XN<2,K>-XNC3,K))*<

CXNC1,K>-XN<2,K>>>IS>*PROP+ACNN,MM>

85

SUBROUTINE FINDCCONT.>

GO TO 90 283 MM=NCI,K>-NCM,K>+l

tm=N < M, K) . ACNN1MM>=C<CYNC3,K>-YN(1,K>>*<YN(1,K>-YNC2,K>>+<XNC3,K>-XNC1,K>>*<

CXNC1,K>-XNC2,K)))/S>*PROP+A<NN,MM> GO TO 90

80 IF<NCM,K>.GT.LNODE> GO TO 90 IF<M.EQ.1) GO TO 51 IFCM.EQ.2) GO TO 52 MM==N < M, 1--\) A<MM,1>=<<<<YN<1,K>-YNC2,K>>**2>+<XNC1,K>-XN(2,K>>**2>1S>*PROPfACM

CM,,.1) GO TO 90

51 MM=N <M, 10 A<MM11>=<<<<YNC2,K>-YNC3,K>>**2>+<XNC2,K>-XNC3,K>>**2)/S)*PROPtACM

CM'1) GO TO 90

52 MM=NCM,K> A<MM,1>=<<<<YNC3,K>-YNC1,K>>**2>+<XNCJ,K>-XNC1,K>>**2)/S)*PROP+A<M

CM,1) 90 CONTINUE 99 CONTINUE

RETURN END

fiUBr.:OUT I NE SOLVE

SUBROUTINE SOLVECNSIZE,MBAND,A,B> DIMENSION AC401,20),B(401) COMPLEX A,a,c

C FORWARD REDUCTION OF MATRIX<GAUSS ELIMINATION> DO 100 N=11NSIZE DO 200 L=2~MBAND IF<ACN,L).EQ.(O.o,o.O>> GO TO 200 I=N+L-1 C=A(N,L>IACN,1> J=O DO 30 K=L,MBAND J=Jt1

30 A<I,J>=ACI,J>-C*A(N,K> ACNYL>=C

200 CONTINUE 100 CONTINUE

C FORWARD REDUCTION OF CONSTANTS . DO 10 N=l,NSIZE DO 20 L=2,MBAND IF<ACN,L>.EQ.C0+010+0>> GO TO 20 I=N+L-1 BCI>=BCI>-ACN,L>*B<N>

20 CONTINUE . 10 B<N>=B<N>IA(N,1). C SOLVE FOR UNKOWNS BY BACK-SUBSTITUTION

DO 40 M=2,NSIZE N=NSIZE+l-M DO 50 L=21MBAND IFCACN1L).EQ.CO.o,o.o>> GO TO 50 K=N+L-1 B<N>=DCN>-ACN,L>*B<K>

50 CONTINUE 40 CONTINUE

RETURN END

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N

a technique to calculate complex electromagnetic fields by

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