Appendix I Numerical Illustration of Equations in Chapter III

Appendix 1

NUMERICAL ILLUSTRATION OF EQUATIONS IN CHAPTER I11

1. Summary

The major equations contained in Chapter 111 will be illustrated here by a numerical example shown in Fig. 1.1 (Fig. 3.1 of Chapter 111). The network equations in the orthogonal network will first be solved employing the contours in Fig. 1.2 (Fig. 3.3 of Chapter 111). Some of the equations will also be illustrated using the contours in Fig. 1.3 (Fig. 3.4 of Chapter I l l ) .

FIG.1.1. Network example. 275

276

Appendix 1

2. Example of Node-to-Datum Case

Consider the primitive network represented by Eq. (1.3) below.

bl

bl

b2

62

b3

b3

:

64

64

b5

b5

b6

66

Let the voltage sources in the primitive network have the numerical values

The open-path contours that will be selected are those in Fig. 1.2. Currents Z', 1 2 , and Z3 specified as in Fig. 1.1 (Fig. 3.1 of Chapter 111) are then numerically equal to those in the orthogonal network (Fig. 3.5 of Chapter 111). Let these currents have the numerical values

The transformation matrices associated with the contours are given in Table 1.1.

277

2. Example of Node-to-Datum Case

b6

‘ E X T R E M I T I E S OF’ OPEN PATHS

FIG.1.2. Open- and closed-path current and voltage contours-node-to-datum

From Eq. (3.46), e, bl

2 -5

-2

=

(3.46)

C:e,

b2 b3

b4

case.

65

66

----~

3

From Eq. (3.47),

e, = cpe, bl

b2

b3

b4

(3.47) b5

b6 bl b2 b3 b4 b5 b6

TABLE 1.1 TRANSFORMATION MATRICES ASSOCIATED WITH CONTOURS I N FIG.1.2 General Form

o

Example Case

1

2

3

cl

c2

c3

bl

b2

b3 b4

b5

b6

1

2

c2

c3

c

1 b

2 3

cl c2 c3

o

Ab'=

3

cl

c

b3

E

1

-1

-1

1

-1

-1

-1

b4 b5 - 1 h6 - I

1 b

2

1

-----1 --I - I -----1 1 ------

3 -1

1 -1 I -----c2 I -1 ------

cl

c3

1

-1

-1

1

2. Example of Node-to-Datum Case

279

From Eq. (2.24),

z,, = c,bz,,cps

(2.24)

As an intermediate step let u s obtain

z,, = c g z b b

Z S b

(2.36)

61

b2

b3

b4

b5

b6

1

2

3

cl

c2

c3

=

Z,, is

1 0 -6 --0 2 4 --3 2 -3 ---

2

5

1

and

From Eq. (3.76),

(3.76)

280

Appendix I

The numerical form of Eq. (3.76) is el 0.016390

____

c2 1.344262 -c3 0.245904

___0.278688 -0.147540 0.180327

c2

I I

I

With all currents J s in the orthogonal network determined, we can illustrate the current relationships given in Chapter 111, Table 3.1.

J b = CPsJs 1 1.360652

bl

b2 -0.229514 0.590166 h3

b2

bl

b4

2.590166

b5

0.344262

--

b6 -0.754096

=

03 b4

2

3

c2

el

c3

-

b5 - 1 b6

-

(3.13)

l b= ChsP

Note that the case considered is one where 1'

-

0 bl 0 b2 h2 b3-I =b3 1 b4 b4 b5 65 - 1 06 b6 -1 bl

-

1

2

3

el

c2

= O',

as stated in Eq. (3.14).

c3 1

2 3 cl

c2 c3

! 0

281

2. Example of Node-to-Datum Case

ib = cbSis

(3.21)

Note that the case considered is one where i"

= 0",

as stated in Eq. (3.22).

0 0 0

-

0.016390 1.344262 -0.245904

J s = Asb J b 61 1

b2

1

_ _ 2 1 -I __--

2 3

-~

63 ~

cl 0.016390

cl

c2

c2

c3 0.245904

c3

b6

-

-I

I --1

-1

--1 ---

I

-I

-I

I" bl

=2

b5

3 -I

-

1

I

b4

(2.9)

b2

b3

=

~p~ J~

b4

1

b5

(3.9) b6

-----1 -I -1 ------

3 -1

1

I

282

Appendix 1

‘i = A:bJb h1

b2

b3

b4

b5

(3.10) b6

t

1

1

2

1

(3.16)

cl3 1 c2 c3

0

0

I”= APblb

1

(3.18)

1

32 1 i

=2

1

-1

3 -1

1

--I

0 ------ b2 -

63 - 1

283

2. Example of Node-to-Datum Case

0' hl

h2

(3.19)

= A!bIb

h3

64

h5

h6

h4

I

h5 - 1 h6 - 1

-

(3.24)

-

1

bl

2

h2

-0.229514

3

h3

1.590166

cl

h4

1.590166

c2

h5

1.344262

c3

h6

0.245904

-

(3.26) 61 1

23

0

0B

1 =2

3

b2

h3

64

b5

h6

---------__-

284

Appendix I

~2 1.344262 = c 2 c3 0.245904 c l r

I:

__-__-__-

__-----

d!

-

1

1

(3.27) 1.360652 b2 -0.229514 bl b3

1.590166

This completes the current relationships in Table 3.1. Now to the voltages. From Eq. (1.3),

bl

1

b2

0.131138

b2

b3

0.950818

b3

b4

9.672140

b4

-----2

1

-----1 __-----

2

----____

b5 -4.491808

65

b6 -6.491784

66 -2

4

-2

-2

2

__-~__--

From Eq. (1.4),

b bl

21.360652 1 7

b3

0.590166

b4

2.590166

b5

0.344262

b6 -0.754096

=b bl

25 1

0.666667 -03.33334; - 0.333334

b6

2

0.666667

!:

--b4 --b5 ---

9.672140 -4.491808

b6 -6.491784

-~

285

2. Example of Node-to-Datum Case

Eb = Vb - eb 2.868844 0.131138 0.950818 9.672140 -4.491808 - 6.491 784

(2.15) bl

b2

b3

b4

b5

b6

I

1 4.491808

1

bl

2 5.540966

2

b21

3 9.672140 cl 2.999982

-

I 0.131 138 I 2.868844

3 cl 65 1-4.491808 b61-6.491784

I

I (3.53)

bl 1 4.491808

2 5.540966 = 2 3 9.672140

3

b2

b3

b4

b5

b6

Appendix I

286 e , = cfvb bl

b2

b3

b4

(3.54) b5

b6

cl

bl

c2

b2

c3

b3

b4

9.672140

b5 -4.491808 b6 - 6.491 784

(3.33) bl 8.491808

1

2 10.540966

2

1 3

8.672140

-

b2

b3

b4 -1 --1 --

3

cl -0.000018

cl

~2 -0.000006

~2

-

I -

c3

0.000036

1

c3

1-11

1 1 1

-I

(3.37)

E , = C:E, bl

b2

b3

b4

b5

b6

- 1 b2 - 1.868862 63 -2.049182 64

8.672140

b5 -8.491808 b6 -8.491784

287

2. Example of Node-to-Datum Case

(3.39) cl

c2

-----__

b2

----__-1

c3

- 1.868862

b3 -2.049182

1

(3.34)

e, = C; heh

1 -4 -

cl

3

c2 9 c3

4 -

b4

b5

b6

2

-

1 -

b3

1

2 -5 3

b2

bl

-

3 cl

c2 c3

In the inverse relationships, the same numerical vectors will be employed as those obtained previously above.

(2.16) 1

2

3

cl

c2

c3

bl

2.868844

bl

1 4.491 808

b2

0.131138

b2

2 5.540966

b3

0.950818

b3

3 9.672140

b4

9.672140

b4

Cl 2.999982

-

c2 8.999994 c3 4.000036

288

E,

61

1.868544

h2

- 1.868862

63 -2.049182 b4

I

8.672140

b5 -8.491808 b6)-8.491808

= AbsE,

-

1-

Appendix I

(3.41) c2

c3 1

2 10.540966 3

--

II eb = Ace,

bl

h2

-I

2

3 -11

1l-1i-!l

2

I

h2I

4

I

1-1-1' -I-I-I

cl

1-11

1

I

I

c2

b5 66

-1

~2 -0.000006

e;k

I

I

I

I

I

I

1-11

I

I

]

2 -5

c2

9

c3

(2.23)

V , = Z,, J s 1

2

3

cl

c2

c3

1 4.491808

I

1

2 5.540966

2

1

3 9.672140

3

1

cl 2.999982 c2 8.999994 c3 4.000036

c2

0

0

--

(3.48)

1 -4

I I I I I

h51-11

0.000036

-

c3

b3 h4

8.672140

cl -0.000018

c3

1

8.491808

289

3. Example of Not of the Node-to-Datum Type

(2.27)

J" = YSSVS 2

1

1

1

1

2

1

2

1

3

3

-=

cl 0.016390

cl

c2 1.344262

c2

1-1

c3 0.245904

c3

2.6'

I -2.3 I

1 1

1.6 -2.3

3

1.161 -3.6

c2

c3

.-.

1.6 -3.3

1 4.491808

--

-2.3 1.6 2 5.540966 -~ 1.6 -1.3 3 9.612140

1 11 I1 -3.6

1.6 1-3.6

---3.1

cl

7.6 -1.3

10.3 cl 2.999982

1.6 -3.3

--

10.3 -3.3

c2 8.999994

10.6 c3 4.000036

Overbar indicates that number repeats for a total of six decimal places; Le., for 2.6 read 2.666666, for 6.16 read 6.166666.

3. Example of Not of the Node-to-Datum Type Some of the equations in Tables 3.1 and 3.2 will be illustrated for the contours in Fig. 1.3, the corresponding transformation matrices being given in Table 1.2.

bl

\ E X T R E M I T I E S OF/ OPEN PATHS

FIG.1.3. Open- and closed-path current and voltage contours-not case.

node-to-datum

Let Z', 12, and I 3 in Fig. 1.1 be unity; l 4 correspondingly is minus three. The elements of the open-path current vector I" ( I t , Zz,1 3 ) are no longer numerically equal to those specified in Fig. 1.1 , but can be obtained from those in Fig. 1.1 by equating the sum of the respective currents in Fig. 1.3 at each

290

Appendix I

bus to the values specified; in Chapter VIIl this procedure is shown to be equivalent to establishing a transformation matrix between one set of openpath currents and another. By following this procedure, current vector I" is

Z,, and voltage vector eb will be taken to be the same as those in the previous example. With eb and I" established, the solution can be obtained as outlined for the previous case. The orthogonal network appears in chapter 111(Fig. 3.6.)

J" = AsbJb b6

1 2 31

2

I

-

3

1

----I - 1 -1 -_ ---

cl c2

0.590166

-I --1

b4

2.590166

-1

h6 -0.154096

--I

b2 -0.229514 b3

-1 I ---

-

c3

-1

(3.16)

1

1

2

1

3

2

:Ij

c3

I

I

b31 b4 b6)

-1

I

3. Example of Not of the Node-to-Datum Type

TRANSFORMATION

0

29 1

TABLE 1.2 MATRICES ASSOCIATED WITH

c

Cf, =t Cb, Cbc

CONTOURS I N

FIG.1.3

I

2

3

cl

c2

c3

bl

62

h3

b4

b5

b6

I

2

3

cl

c2

c3

bl

b2

b3

b4

b5

b6

bl 62 b3

b4 h5 bt

1

b

2 3 cl

c2 c3

bl O

C

h2 03

b4 b5 66

b

292

Appendix I

(3.24) b2

bl

b3

b4

05

h6

1

2

62 -0.229514

3

b3

1.590166

b4

1.590166

1

-1

1

-I

65

1.344262

-1

1

h6

0.245904

J h = CbsJs 1

2

3

cl

c2

c3

61

1

b2

2

b3

3

1

I

-2

b4

Cl 0.016390

65

c2 1.344262

b6

c3 0.245904

(3.13)

bll

O

I

1

1

2

1

1 3

2

293

3. Example of Not of the Node-to-DatumType

(3.21) 2

1

bl

1.360652

bl

b2

- 0.229514

b2

b3

1 S90166

b4

1.5901 66

-

3

c2

cl

c3

b3 b4

b5

1.344262

b5

b6

0.245904

b6

(2.15) bl

1

b2

b4 b5

b6

I I bl I

2.868844

1-1

0.131138

l~-----4.491808

2 -4.131174 3

9.672140

cl

2.999982

1

21

1-1I-1I-1-I

1-1 1-1

8.999994 ___-4.000036 c3

I

b21 I

I

E,

I

I

8.491808

2

I .868826

2

3

8.672140

3

-0.00001 8

cl

-ITG~T

I

-1

1 -I-I11 -1 -

c2

cl

b3

1

c2 -0.000006

c2

c3

c3

=

(3.33)

C,bE,

-1

------1 __----__

-1

-1

bl

1.868844

b2

- 1.868862

1 63 ------

I 1 --__--1

b4

1

1

1

1

1

------1

I

b5

-2.049182 8.672140 -8.491808

b6 -8.491784

294

Appendix I

e, = C,be, bl

b2 63

(3.34)

b4 b5

b6 61

1

62

2

3

b3

3 1

3

1

Cl

3

cl

64

c2

9

c2

b5

c3

4

c3

-

4 ___b6 2

-

(2.16)

V, = A: V,

bl

2.868844

b2

0.131138

b3

0.950818

64

9.672140

I I

2

3

cl

c2

bl b2

-1

c3 I

1

4.491808

-1

2

-4.131 I74 9.672140

64

--

-b5 - 1

b5 -4.491 808

66 -6.491784

1

I

cl

2.999982

c2

8.999994

66 -1

4.000036

(3.41)

E , = A:E, 1

2

3

cl

c2

c3

1.868844

8.491808 2

63 -2.049182 b4

~3

8.672140

cl

b5 -8.491808

1.868826 8.672140 -0.00001 8

c2 -0.000006

66 -8.491784

c3 I

0.000036

I

3. Example of Not of the Node-to-Datum Type

295

(3.48)

eb = A:e, 2

1

b3

cl

c2

c3

3

b3

I

b4

4

b5

c2

ILI b6

c3

- =

b4

__

b5 b6

3

__

(2.23)

V, = Z,, J s 2

1

1

4.491808

1

2 -4.131174 3

9.672140

-

3

2

3

cl

c2

c3

- 2 2

__-

2

- 4 4

0

1

-

cl

2.999982

c2

8.999994

c3

4.000036

cl

(2.27)

JS = YSSV, I 1 -=

2

3

c2

Cl

c3

5.6

2 -2.3

2

0.016390 1.344262 0.245904

I

I

I

I

I

I