Impedance Matrix

30

L1 0.099 + j0.270 0 + j0 0 + j0 0.099 + j0.270 0 + j0

L2 0 + j0 0 + j0 0 + j0 0 + j0 0.032 + j0.06

This matrix of self and mutual impedances completely defines the performance of the

transmission system of Figure 2.10.

Example 2.3

Consider the following problem.

Let

=

2

1

PP

P , [ ]21 PPPtr =

Find PtrBP.

SOLUTION

Performing the PtrB operation, we have

[ ]

2221

121121 BB

BBPP

++

=222121

122211

BPBPBPBP

Performing the (PtrB) P multiplication, we obtain

+++

=

++

22222121

12121111

2

1

222121

212111

PBPPBPPBPPBP

PP

BPBPBPBP

(2.18)

In terms of index notation, we obtain

Pm Bmn Pn = P1 B1n Pn + P2 B2n Pn

by letting m assume the values of 1 and 2. Letting n = 1,2, we obtain

Pm Bmn Pn = P1 B11 P1 + P1 B12 P2 + P2 B21 P1 + P2 B22 P2

= P12B11 +P2

2 B22 +2P1P2B12 (2.19)

which is the same as the result obtained by the arrow rule.

2.4THE CONCEPT OF TRANSFORMATION In our analysis of transmission losses it is necessary that all changes of transformations

in the equivalent circuit of Figure 2.7, as denoted by equation 2.6, be made in such a manner

that all source powers, the load powers, and transmission losses remain invariant. These

transformations may be made through means of transformation matrices which result in

logical and systematic steps in the analysis. Also, the use of transformation matrices leads

31

to orderly computational procedures which are very adaptable to calculation on both manual

and automatic digital computers.

The concept of the transformation matrix C allows a given circuit to be modified to a

new circuit in such a manner that the power input remains invariant. Denote the quantities

pertaining to the original circuit by the subscript old and quantities pertaining to the desired

new circuit by the subscript new. In general, it has been shown by G. Kron that if the set of

currents iold pertaining to the old circuit is related to the new currents inew by a

transformation matrix C such that

iold = C inew (2-10)

and id the power is to remain invariant the new set of voltages is given

enew = Ct*eold (2-21)

and the new set of impedances is given by

Znew = Ct*Zold C (2-22)

The matrix Ct* is obtained by conjugating the elements of the matrix C1.

Let it be required to obtain the new circuit that exists if the old currents are related to the

new currents by the relation

i1 = i1

i2 = K2 i4

i3 = K3 i4 (2-23)

where the current i1 remains unchanged but i2 and i3 are constant proportions K2 and K3

respectively, of the new current 4.

The relation between the two sets of currents as given by equation 2-27 may be denoted

by the following matrix of transformation:

=

4

1

3

2

3

2

1

ii

K0K001

iii

(2-24)

where

=

3

2

00

01

KKC (2-25)

The transpose of C will then be

32

=

320001KK

Ct (2-26)

Since the elements of Ct are real numbers, Ct* = Ct.

The new voltages are given by

==

3

2

1

320001

eee

KKeCe oldtnew (2-27)

=

+ 3322

1

eKeKe

(2-28)

As indicated previously, the new impedances are given by

Znew = Ct* Zold C

The product Zold C is first calculated as

+

+

+

=

33323231

32322221

31321211

3

2

333231

232221

131211

00

01

KZKZZKZKZZKZKZZ

KK

ZZZZZZZZZ

The operation Ct* (Zold C) is calculated as

=

+++

4441

1411

33323231

32322221

31321211

320001

ZZZZ

KZKZZKZKZZKZKZZ

KK

where Z14 = Z12 K2 + Z13 K3

Z41 = K2 Z21 + K3 Z31

Z44 = K2 Z14 K2 + K2 Z33 K3 + K3 Z33 K3 (2-29)

The new set of equation is

=

4

1

4441

1411

4

1

ii

ZZZZ

ee

(2-30)

Example 2.4

This concept is first illustrated in terms of the simple example show in Figure2.14 and

Figure 2.15. As will be noted from the work already presented, this network may be

defined by the set of equations which follow:

33

Figure2.14. Three-source system.

SOLUTION

Figure 2.15. Schematic diagram of illustrative three-source system.

=

3

2

1

333231

232221

131211

3

2

1

iii

ZZZZZZZZZ

eee

=

3

2

1

4.0000.03.03.00.03.05.0

iii

In terms of numbers

=

3

2

2

3

2

K4.00K3.03.0K3.05.0

K0K101

4.00003.03.003.05.0

In terms of number we have

34

( )

=

23

222

2

3

2

2

32 KK3.03.0KK3.05.0

K4.00K3.03.0K3.05.0

KK0001

where the voltages are given by equation 2-30 and the impedances by equation 2-31.

In terms of number we have

+=

+

=

4

123

222

2

3322

1

4

1

4.03.03.0

3.05.0ii

KKKK

eKeKe

ee

The corresponding new circuit is drawn in Figure 2.16.

Figure 2.16. Modified circuit after transformation.

2.5 TRANSFORMATION TO REFERENCE FRAME 2 It is desired to eliminate the individual equivalent load currents as variables, since the

final result should involve only generator powers. As will be recalled, the equivalent load

current at a bus is defined as the sum of the line-charging, synchronous condenser, and load

current as that bus. The first assumption involved in the development of a loss formula is

the following:

It is assumed that each equivalent load current remains a constant complex fraction of

the total equivalent load current.

Define ∑= LjL ii (2-32)

By the above assumption

LjLj iii = (2-33)

For the system given by Figure 2.10 and equation 2-12 we may write

35

LL

LL

GG

GG

GG

iiiiii

iiiiii

22

11

33

22

11

=====

The preceding relation may be written in terms of a matrix of transformation.

=

2

1

12

000000

010000100001

LL

G (2-34)

Thus the currents of reference frame 1 (I1) are related to the currents of reference frame 2

(I2) by a matrix of transformation C12 where

=

2

1

3

2

1

2

1

2

1

3

2

1

000000

010000100001

L

L

G

G

G

L

L

G

G

G

iiiii

ll

iiiii

(2-35)

The symbol C jk is used to indicate the transformation from reference frame or step j to

reference frame or step k.

The new impedance matrix is given as indicated by equation 5-24 by C *t Zold C.

Performing the Zold C operation first, we have

Zold C =

−1−−−1−

−−−−1−

−1−−−1−

−1−−−1−

−1−−−1−

2

1

22232222

211131211

23333233

22232222

21131211

000000

010000100001

ll

ZZZZZZZZZZZZZZZZZZZZZZZZZ

LLLLGLGLGL

LLLLGLGLGL

LGLGGGGGGG

LGLGGGGGGG

LGLGGGGGGG

+++++

−−

−−

−−

−−

−−

−−−

−−−

−−−

−−−

−−−

222112

221111

223113

222112

221111

322212

312111

332313

3212212

312111

lZlZlZlZlZlZlZlZlZlZ

ZZZZZZZZZZZZZZZ

LLLL

LLLL

LGLG

LGLG

LGLG

GLGLGL

GLGLGL

GGGGGG

GGGGG

GGGGGG

(2-36)

36

Performing the C *t (Zold C) operation, we have

+

+

+

+

+

=

−−

−−

−−

−−

−−

−−−

−−−

−−−

−−−

−−−

222112

221111

223113

222112

221111

322212

312111

332313

322212

312111

*2

*1

*

000001000001000001

lZlZlZlZlZlZlZlZlZlZ

ZZZZZZZZZZZZZZZ

ll

CZC

LLLL

LLLL

LGLG

LGLG

LGLG

GLGLGL

GLGLGL

GGGGGG

GGGGG

GGGGGG

oldtr

=

−−−

−−−

−−−

ω321

3332313

2322212

1312111

bbbaZZZaZZZaZZZ

GGGGGG

GGGGGG

GGGGGG

(2-37)

where

222*2112

*2221

*1111

*1

23231*13

22*221

*12

12*211

*11

2231133

2221122

2211111

lZllZllZllZllZZlb

ZlZlbZlZlb

lZlZalZlZalZlZa

LLLLLLLL

GLGL

GLGL

GLGL

LGLG

LGLG

LGLG

−−−−

−−

−−

−−

−−

−−

−−

+++=

+=

+=

+=

+=+=+=

ω

(2-38)

From equation 2-23 it is seen than our new voltages of reference frame 2 given by Ctr*eold as

indicated by

*2

*1000

001000001000001

ll

=

−−−−−

RL

RL

RG

RG

RG

EEEEEEEEEE

2

1

3

2

1

−−−−

RL

RG

RG

RG

EEEEEEEE

3

2

1

(2-39)

The calculation for RL EE − is indicated in detail below.

( ) ( ) ( ) RLLRLRL EllElElEEEEl *2

*12

*21

*121

*1 +−+=−+− Ctr

* eold = enew

Define 2*21

*11 LL ElElE +=

Since ( ) 1*1

*1 =+ ll

we have ( ) ( ) RLRLRL EEEElEEl −=+ 2*21

*1

From equations 2-36, 2-37 and 2-39 we have

37

=

−−−−

RL

RG

RG

RG

EEEEEEEE

3

2

1

−−−

−−−

−−−

ω321

3332313

2322212

1312111

bbbaZZZaZZZaZZZ

GGGGGG

GGGGGG

GGGGGG

L

G

G

G

iiii

3

2

1

it will be noted that the effect of each load current has been replaced by a single total load

current.

The preceding steps accomplished by the transformation matrix C12 may be thought of in

terms of a number of algebraic steps. Consider the reference frame1 equations for this

example as repeated below:

−−−−−

RL

RL

RG

RG

RG

EEEEEEEEEE

2

1

3

2

1

=

−−−−−

−−−−−

−−−−−

−−−−−

−−−−−

2212322211

2111312111

2313332313

2211322212

2111312111

LLLLGLGLGL

LLLGGLGLGL

LGLGGGGGGG

LGLGGGGGGG

LGLGGGGGGG

ZZZZZZZZZZZZZZZZZZZZZZZZZ

2

1

3

2

1

L

L

G

G

G

iiiii

(2-40)

As before, let

iL1 = l1 iL

iL2 = l2 iL

Substituting equation 2-41 into equation 2-40., we obtain

−−−−−

RL

RL

RG

RG

RG

EEEEEEEEEE

2

1

3

2

1

=

+++++

−−−−−

−−−−−

−−−−−

−−−−−

−−−−−

222112322211

221111312111

223113332313

222111322212

221111312111

lZlZZZZlZlZZZZlZlZZZZlZlZZZZlZlZZZZ

LLLLGLGLGL

LLLGGLGLGL

LGLGGGGGGG

LGLGGGGGGG

LGLGGGGGGG

2

1

3

2

1

L

L

G

G

G

iiiii

(2-42)

The impedances in equation 2-42 correspond to Zold C.

Define a hypothetical load voltage EL such that the power loss contributed by (EL1-ER)

iL1 + (EL2 – ER)iL2 remains invariant.

Thus

(EL1-ER) iL1 = (EL1-ER) iL1 * + (El2-ER) iL2 *

(EL1-ER) iL1*

iL + (EL2-ER) i2*

iL (2-43)

Dividing equation 2.43 by iL*

(EL1-ER) = (EL1-ER) iL1* + (El2-ER) i2

* (2-44)

38

Performing the operation indicated by equation 2-46 upon 2-44, we obtain equation 2-42 as

before.

221111 lZlZa LGLG −− +=

= ( )( ) ( )( )"2

'22121

"1

'11111 jlljXLRjlljXLR lGGlGG ++−+++− −−

= "111

'221

'111

'221

"111

'221

'111 ( lLRlLXlLXjlLXlLXlLRlLR GGGGGGG −+−+−+−−−−−+−

)"221 lLRG −+

))(())(( 1212

"2

'21111

"1

'1

12*211

*11

GLGLGLGL

GLGL

jXRJlljXRJllZlZlb

−−−−

−−

+−++−=

+=

= 12"212

"112

'211

'1 GLGLGLGL XlXlRlRl −−−− +++

)( 12"211

"212

'211

'1 GLGLGLGL RlRlXlXlj −−−− −−++

Consider next the more general case in which the number of sources = m,n and the

number of loads = j, k and for which reference frame 1 equations are given by equation

2-6 which is repeated below:

=

−−

RL

RGm

EEEE

1

−−

−−

LkLjGnLj

LkGmGnGm

ZZZZ

Lk

Gn

ii

The matrix of transformation C12 is given by

Lk

Gn

ii

=

kl001

LK

Gn

ii

(2-47)

12C

=

λ001

The matrix Lk is a column matrix with the number of elements equals to k, the number of

load currents. This follows by inspection of the matrix, since the number of columns

correspond to l. and the number of rows to Lk. Since there is only hypothetical load current,

then is only one column. The transpose of C12 is given by

39

lj

(C 12 )tr =

lj001

The matrix is a row matrix with the same elements as the column matrix lk

lk but with the numbers written in a row instead of a column. The resultant voltages,

impedances, and currents are given by

−−

RL

RGm

EEEE

=

−

wbaZ

n

mGnGm

L

Gm

ii

(2-48)

where kLkGmm lZa −=

GnLjn Zljb −= *

kLkLj lZljw −= * (2-49)

LjLj EZljEl *=

By means of the above transformation the circuit of Figure 2-7 has been changed to

the circuit given in Figure 2-17. The load point L does not exit in the actual network, and

consequently it is referred to as a hypothetical load point

Figure 2-17. Reference frame 2.

As previously noted, the mutual impedances between the generators and loads are equal.

As noted by equation 2-49 and Figure 2-17. the component of the voltage drop EGm = ER

due to load current iL is given by am iL. Similarly, the component of the voltage drop EL =

40

ER due to current iGm is given by bn iGn. The impedance w is the self impedance existing

between the hypothetical load point and the reference bus.

2.6 TRANSFORMATION TO REFERENCE FRAME 3 As will be noted in equations 2-40 and 2-49 and also figure 2-17. The individual

load currents have been eliminated as variables and replaced by the total load current iL.

The next step in our analysis involves elimination the total load current iL as a variable. We

may accomplish this by the relationship the summation of the source currents must be equal

and opposite to the summation of the load currents. Thus

Ln

Gn ii −=∑ (2-50)

For the system of Figure 5-10 and equation 5-42 we may write

iG1 = iG1

iG2 = iG2

iG3 = iG3

iL = - (iG1 + iG2 + iG3) (2-51)

The above relation may be written in terms of a matrix of transformation as indicated

below:

L

G

G

G

iiii

3

2

1

=

−−− 111000010001

3

2

1

G

G

G

iii

(2-52)

Thus the currents of reference frame 2(I2) are related to the currents of reference frame 3(I3)

by a matrix of transformation C 23 where

23C =

−−− 111100010001

(2-53)

The new voltages of reference frame 3 are given by

41

−−−

110010101001

−−−−

RL

RG

RG

RG

EEEEEEEE

3

2

1

=

−

−

−

ELG

ELG

ELG

EEE

3

2

1

(2-54)

Ct* eold = enew

The new impedance matrix, as indicated by equation 5-24, is given by Ct* Zold C.

*

Performing the C *t Zold operation first. we have

=oldt ZC *

−−−

110010101001

x

−−−

−−−

−−−

wbbbaZZZaZZZaZZZ

GGGGGG

GGGGGG

GGGGGG

321

1312111

1312111

1312111

x

−−−−−−−−−−−−

−−−

−−−

−−−

wabZbZbZwabZbZbZwabZbZbZ

GGGGGG

GGGGGG

GGGGGG

3333223113

2332222112

1331221111

(2-55)

Performing the (C *t Zold) C operation, we have

=CZC oldt*

−−−−−−−−−−−−

−−−

−−−

−−−

wabZbZbZwabZbZbZ

wabZbZbZ

GGGGGG

GGGGGG

GGGGGG

3333223113

2332222112

1331221111

x

−−− 111100010001

=

+−−+−−+−−+−−+−−+−−

+−−+−−+−−

−−−

−−−

−−−

wabZwabZwabZwabZwabZwabZ

wabZwabZwabZ

GGGGGG

GGGGGG

GGGGGG

333332233113

233222222112

133112211111

(2-56)

From equation 2-54. 2-56 and 5-58, the reference frame 3currents, impedances, and

voltages are given by

42

−−−

LG

LG

LG

EEEEEE

3

2

1

=

+−−+−−+−−+−−+−−+−−

+−−+−−+−−

−−−

−−−

−−−

wabZwabZwabZwabZwabZwabZ

wabZwabZwabZ

GGGGGG

GGGGGG

GGGGGG

333332233113

233222222112

133112211111

x

3

2

1

G

G

G

iii

(2-57)

It will be noted equation 2-59 that only the generator currents appear as variables.

Also

−−−

LG

LG

LG

EEEEEE

3

2

1

=

−−−

−−−

−−−

332313

322212

312111

ZZZZZZZZZ

3

2

1

G

G

G

iii

(2-58)

where Zmn = ZGm-Gn - am - bn + w (2-59)

The result of performing the operation indicated by equation 2-22, 2-23, and 2-24

with transformation C 23 may be visualized by a number of algebraic steps. Consider the

reference frame 2 equation for this example as shown in equation 2-24:

−−−−

RL

RG

RG

RG

EEEEEEEE

3

2

1

=

−−−

−−−

−−−

wbbbaZZZaZZZaZZZ

GGGGGG

GGGGGG

GGGGGG

321

1312111

1312111

1312111

L

G

G

G

iiii

3

2

1

Subtraction EL – ER = b1iG1 + b2 iG2 + b3 iG3 + w iL

From each of the previous equations in 2-42 we have

−−−

LG

LG

LG

EEEEEE

3

2

1

=

−−−−−−−−−−−−

−−−

−−−

−−−

wabZbZbZwabZbZbZ

wabZbZbZ

GGGGGG

GGGGGG

GGGGGG

3333223113

2332222112

1331221111

L

G

G

G

iiii

3

2

1

(2-60)

The voltages in equation 2-62 correspond to C *t eold , as indicated by equation 2-56. Also,

the impedances indicated by equation 2-56 correspond to C *t Zold , as indicated by equation

2-57

Subtracting iL = - (iG1 + iG2 + iG3)

Into equation 2-62. we obtain equation 2-59 as before.

From equations 2-59. 2-60, and 2-61 we note, as in reference 2, that the mutual

impeeances are not symmetrical.

43

Thus 1212121212

2121212121

−−−−

−−−−

+=+−−=+=+−−=

jXRwbaZZjXRwbaZZ

GG

GG )622()612(

−−

From equations 2-10., 2-47. and 2-48.

)( "111

"111

'211

'1112121 lXlXlRlRRR LGLGLGLGGG −−−−−− −−+−=

'"222

"121

'222

'121 )( wlXlXlRlR GLGLGLGL ++++− −−−− (2 – 63)

)( "221

"111

'221

'1112121 lRlRlXlXXX LGLGLGLGGG −−−−−− −−+−=

""222

"121

'222

'121 )( wlRlRlXlX GLGLGLGL +−−+− −−−− (2 – 64)

)( "222

"112

'222

'1121212 lXlXlRlRRR LGLGLGLGGG −−−−−− −−+−=

'12

"211

"112

1211

'1 )( wXlXlRlRl GLGLGLGL ++++− −−−− (2 – 65)

)( "222

"112

'222

'1121212 lRlRlXlXXX LGLGLGLGGG −−−−−− −−+−=

"12

"211

"112

'211

'1 )( wRlRlXlXl GLGLGLGL +−−+− −−−− (2 – 66)

where "jwww i += (2 – 67)

It will be not that

)(2 """" 2122211121211111221 GLGLLGLg XXXXRR−−−−−− −−+=− (2 – 68)

)(2 "222

"121

"221

"1211221 lRlRlRlRXX GLGLlGLG −−−−−− ++−−=−

The asymmetry in the real part of Zm-n results from terms involving the products of

imaginary load currents and mutual reactances between generators nad loads. The

asymmetry in the imaginary part of Zm-n results from terms involving the products of

imaginary load currents and mutual resistances between generators and load.

The reference frame 2 equation for the general case are given by equation 2-51:

=

−−

RL

RGm

EEEE

−

wbaZ

n

mGnGm

L

Gn

ii

(2 – 69)

the matrix of transformation C 23 is given by

23C =

ntI

(2 – 70)

where tn = - 1 for all value of n.

By application of equation 2-22, 5-23, and 2-24, we obtain

[ ] [ ]GnnmGnGmLGm iwbaZEE +−−=− −

= [ ][ ]Gnnm iZ − (2 – 71)

44

The circuit of reference frame 2, given by Figure 2-17, has been modified as

indicated by equations 2-59 and 2-73 to that given in Figure 2-18.

Figure 2-18. Reference frame3.

As noted 2-70 and 2-71. the mutual impedances are not symmetrical. Consequently, it is

not possible to represent this equivalent circuit on the network analyzer through the use of

static circuit elements. The losses in the equivalent circuit of Figure 2-18 correspond to the

losses in the transmission line of the original circuit.

2.7 CALCULATION OF LOSS The real losses in the equivalent circuit of Figure 2-18 and equation 2-73 may be

calculated as follows:

3*3 EIL ℜ=Ρ (2 – 72)

33*3 IZIL ℜ=Ρ (2 – 73)

where E3 I3, and Z3 denote reference frame 3 quantities and the symbol denotes the real part

of I 3*3 E .

Let us define the real and imaginary components of iGn by idn and iqn, respectively.

Thus qndnGn jiii += (2 – 74)

For the system of equation 2-60 we have

−−−

LG

LG

LG

EEEEEE

3

2

1

=

−−−

−−−

−−−

332313

322212

312111

ZZZZZZZZZ

+++

33

22

11

dd

dd

dd

jiijiijii

(2 – 75)

where Zm-n = ZGm-Gn – am – bn + w

Then

45

33 IZ =

−−+++

−−++

−−+++

−−++

−−+++

−−++

−−−−−−−

−−−−−−−

−−−−−−−

−−−−−−−

−−−−−−−

−−−−−−−

333223113333223113

333223113333223113

332222112332222112

332222112332222112

331221111331221111

331221111331221111

(

((

((

(

dddqqq

qqqddd

dddqqq

qqqddd

dddqqq

qqqddd

iXiXiXiRiRiRjiXiXiXiRiRiR



22332

313113

332332

212112

2

31331

121221

1333332332

22222

31331

121221

11111333332332

2

222231331

121221

11111333

)2

(2)2

(2)2

(2)2

(2

)2

(2)2

(2)2

(2

)2

(2)2

(2)2

(2

)2

(2)2

(2 =IZ*I

qdqdqdqd

qdqdqqqqqq

qqqqqqdddd

dddddddd

iXXiiXXiiXXiiXXi

iXXiiXXiiRiiRRiiRi

iRRiiRRiiRiiRiiRRi

iRiiRRiiRRiiRi

−−−−−−−−

−−−−−

−−−

−−−−−−

−−

−−−−−

−

+−

−−

+−

−−

−−

−−+

+++

++

++++

++

++

++

+

PROBLEMS

Problem. 2.1

Figure is a simplified one-line impedance diagram of the Indiana Division of the

American Gas and Electric Service Corporation.

Choose the Muncie bus as the reference bus and calculate the open-circuit self and

mutual impedances as indicated in the following:

46

Problem. 2.2

The generator and load currents and bus voltages for a given operating condition of

the system are

EG1 = 1.04 + j 0.15 iG1 = 1.183 +j0.070

EG2 = 1.00 + j 0.049 iG2 = 0.922 +j0.070

EG3 = 0.997 - j 0.005 iG3 = 00+j0.14

EL1 = 0.03 + j 0.114 iL1 = -0.732 – j 0.098

EL2 = 0.927 - j 0.085 iL2 = -1.373 + j 0.098

47

The impedance matrix with generator G2 as reference .

From the above data determine the arithmetic value of the frame 2 impedances and voltages.

It is suggested that these quantities be calculated by using the transformation matrices

equations .

Problem2.3

The generator and load currents and bus voltages for a given operating condition of the

system are

EG1 = 1.04 + j 0.15 iG1 = 1.183 +j0.070

EG2 = 1.00 + j 0.049 iG2 = 0.922 +j0.070

EG3 = 0.997 - j 0.005 iG3 = 00+j0.14

EL1 = 0.03 + j 0.114 iL1 = -0.732 – j 0.098

EL2 = 0.927 - j 0.085 iL2 = -1.373 + j 0.098

The impedance matrix with generator G2 as reference .

Determine the arithmetic value of reference frame3 impedance and voltages. It is

suggested that the frame3 quantities be calculated by using the transformation matrices.

Problem 2.4

48

Consider the transmission on systems given in Figure. Find the numerical value of the

reference frame 3 impedances.

Problem 2.5

(a) Find the numerical resultant impedance frame for the given transmission system

taking load as reference.

(b) Find the new voltages and impedances frame taking source 2 and 3 as composite

source 4.

Figure for problem 2.5 Problem2.6

49

If G1 and G2 are transformed to equivalent source, draw the new equivalent network

with required impedances and source voltage .Take L as reference bus .

Problem2.7 (a)Take G2 as reference and find the Zbus .

(b)If iL1 and iL2 are the 40% and 60% of total load current, find the impedance matrix.

P roblem2.8

If G1 and G3 are transformed to equivalent source, draw the new equivalent network

with required impedances and source voltage .Take L as reference bus .

Problem29

50

L1 L2

(a) Take G3 as reference and find the Zbus.

(b) If each load current is 50% of total load curren ,find the impedance matrix.

Problem2.10

(a) Take L as reference and find the Zbus.

(b) If G2 and G3 are transformed to equivalent source G4, draw the new equivalent

network with required impedances and source voltage .

Problem2.11

51

(a) Take L as reference and find the Zbus.

(b) If G2 and G3 are transformed to equivalent source G4and also each source current is

50% of iG4 , find Znew and Enew.

CHAPTER 3

TRANSMISSION LOSSES AS A FUNTION

OF VOLTAGE PHASE ANGLE

*************************

3.1 INTRODUCTION

For simple circuit confugyatrions, incremetnsl losses and changes in total losses

may be igorously and simply expressed in terms of functions of voltage phase angles,

52

diriving point and transfer impedances, and voltage magnitudes. In certain limiting cases

voltage magnitudes and driving –point and transfer impedances cancel out, leaving an

expression involving X/R ratios and differences in voltage phase angles.

3.2 TWO –MACHINE SYSTEM WITHOUT INTERMEDIATE LOADS

The system under considerartion is indicated in figure 3.1 . it is assummed that

1. Voltage magnitudes remain constant.

2. Reactive power flows in such a manner as to maintain constant voltage.

Figure 3.1 . Two –machines system without intermediate loads.

The power –angle equations may be written as 1,2

)α(θsinZ

VVαsinZVP 1212

12

2111

11

21

1 −+= (3-1)

)α(θsinZ

VVαsinZVP 2121

21

1222

22

22

1 −+= (3-2)

where P1= power at sources 1

P2= power at sources 2

Z11, Z12, Z21, Z22, = absolute values of values drivingpoint and transfer impedances

11

11111 X

Rtanα −=

12

1211221 X

Rtanαα −==

22

22122 X

Rtanα −=

V1= absolute value of voltage at sources 1

V2= absolute value of voltage at sources 2

θ 12= θ1- θ2

53

θ1 = angle of voltage 1

θ2 = angle of voltage 2

If the line –charging of the transmission line is lumped with the var requirements of the

machine and if there are no intermediate loads or generatores, then

Z11= Z12= Z21= Z22

α11= α 12= α 21= α 22

it is intended ot calculate the change inlosses involved when the generation is swing

between sources a 1 and 2 by increasing ther output of source 1 and decreasing the

output of source 2.

3.3CHANGE IN TOTAL LOSSES

The transmission losses are given by

PL = P1+ P2

)sin(sin 121212

211

11

21 1 αθα −+=

ZVV

ZV

+ )αsin(θZ

VVαsin

ZV

212121

122

22

22 2 −+= (3-3)

Assumed that the system has changed toa new condition in which the angle between

V1and V2 increases to θ12 . then

21 PPPL ′+′=′

)sin(sin 121212

211

11

21 1 αθα −′+=

ZVV

ZV

+ )αθsin(Z

VVαsin

ZV

212121

122

22

22 2 −′+= (3.4)

The change in total losses in given by

)]sin()[sin(

)]sin()[sin(

2121212121

21

1212121212

21111

αθαθ

αθαθ

−−−′

+−−−′=−′=∆

ZVV

ZVVPPP

Recalling that

sin α-β = sinα cos β- sin β cos α

54

121212121212121212

211 sincoscossinsincoscos[sin αθαθαθαθ +−′−′×=∆

ZVVP

= ]sin)cos(cos2[ 12121212

12 αθθ ′−×Z

VV (3.5)

3.4 CALCULATION OF INCREMENTAL LOSS

The incremental loss which we are considering is given by dividing the change in loss

by the change in generation of a given source when swinging generation between that

source and one other source. For the system of figure 3.1,

=∆∆

=1

2.1.1

1

2.1.1

PP

dPdP change in transmission loss divided by change in generation at

source 1 when swinging generation between source 1 and

source 2.

Also.

=∆∆

=2

2.1.1

2

2.1.1

PP

dPdP change in transmission loss divided by change in generation at

source 2 when swinging generation between source 1 and source 2.

Also.

From equations 3-1 and 3-2 it is seen that for very small changes from a given operation

condition.

121212121212

211 )(cos θψθαθ ∆=∆−=∆

ZVVP (3.6)

212121212121

122 )(cos θψθαθ ∆=∆−=∆

ZVVP (3.7)

here )αcos(θZ

VVψ 121212

2112 −=

)αcos(θZ

VVψ 212121

1221 −=

the change in loss is then given by

21L.1.2 ∆P∆PP∆ +=

= 21211212 θψθψ ∆+∆ (3.10)

= 122112 )( θψψ ∆−

55

12

2112

1

12

1

2.1. )(ψ

ψψ −=

∆∆

=P

PdP

dP LL (3.11)

The above expression may be further simplified.

12

2112 )(ψψψ − =

)(cos)cos()(cos

1212

21211212

αθαθαθ

−−−−

= 12121212

1212121212121212

sinsincoscossinsincoscossinsincoscos

αθαθαθαθαθαθ

++−+

= 12121212

1212

sinsincoscossinsin2

αθαθαθ

+

Dividing numerator and denominator by 1212 coscos αθ , we obtain

1212

1212

12

2112

tantan1tantan2)(

αθαθ

ψψψ

+=

−

Thus

121212

12

1

2.1

tan/tan2.

θθ+

=RXdP

dPL since 12

1212tan

XR

=α (3.12)

or 1212

12

1

2.1

tantan2.

θθ

+=

KdPdPL

Similarly,121212

12

2

2.1

tan/tan2.

θθ−

−=

RXdPdPL (3.13)

Typical results for dPL.1.2 /dP1 as a function of θ12 are plotted in figure 3.2. The above

equations, 3-12 and 3-13, correspond to those given by Brownlee if

X12 /R12 is replaced by a quantity k equal to the X/R ratio of the line between source 1

and source 2.

It is suggests that the effect of an intermediate sources may be approximated by the

expression.

212

12

1

2.1.

))2/(tan()2tan(4

θθ

+=

∆ KK

PdPL

if K is taken as the X/R ratio of the impedance between source 1 and source 2 with all

other sources and loads open-circuited.

56

Figure 3.2 Plot of dPL12 /dP1 as function of θ12 for system of Figure 3.1.

3.5 SYSTEM WITH INTERMEDIATE LOAD GENERATION

In this section we shall derive a rigorous expression for dPL.1.2 / dP1 for the case in

which there is an intermediate load or sources as indicated in Figure 3.3. It is

assumed that the reactive characteristics of this intermediate source or load are such

as to maintain constant voltage. We shall designate by the subscript 3 the quantities

relating to this intermediate point. The network connecting these three points is not

restricted in any manner as to its configuration.

Figure 3.3 Schematic representation of system with intermediate load or generation.

For this case, we have

)sin()sin(sin 131313

311212

12

2111

11

21

1 αθαθα −+−+=ZVV

ZVV

ZVP (3-15)

57


322121

22

1211

22

22

2 αθαθα −+−+=Z

VVZ

VVZVP (3-16)


233131

31

1333

33

23

3 αθαθα −+−+=Z

VVZVV

ZVP (3-17)

Then 131312121 θψθψ ∆+∆=∆P (3.18)

232321212 θψθψ ∆+∆=∆P (3.19)

323231313 θψθψ ∆+∆=∆P (3.20)

where )cos( jkjkjk

kjjk Z

VVαθψ −= with j ≠k (3.21)

In swinging generation between sources 1and 2 the power P3 is to ramainconstant. Thus

,

03 =∆P (3.22)

This relation is used to express 13θ∆ and 23θ∆ in terms of 12θ∆ . From equation 3-20 we

obtain

∆θ31 = - ψ32 ∆ θ32

32

31

θθ

∆∆

= 31

32

23

13

ψψ

θθ

=∆∆

=

Recalling that

∆θ13 = ∆θ12 +∆θ23 (3.23)

we obtain 31

32

23

2312

ψψ

θθθ

−=∆

∆+∆

23θ∆ = 3132

3112 ψψ

ψθ+

−∆ (3.24)

By substituting equation 3-24 into equation 3-23, the angle ∆θ13 may be related to

∆θ13as indicated below:

∆θ13= ∆θ12

+ 3132

32

ψψψ

(3.25)

As before

∆P1.1.2 = ∆P1 +∆P2

=Ψ12 ∆θ12 + Ψ13∆θ13 + Ψ21∆θ21+ Ψ23θ23

=(Ψ12- Ψ21) ∆θ12+ Ψ13∆θ13 + Ψ23θ23 (3.26)

58

Substituting equation 3-24 and 3-25 into 3-26

∆PL.1.2= ( ) 123132

2331

3132

32132112 θ

ψψψψ

ψψψψψψ ∆

+

−

+

+− (3.27)

Substituting equation 3-25 into 3-18

∆P1= 123132

321312 θ

ψψψψψ ∆

+

+ (3.28)

Then

( )

3132

321312

3132

2331

3132

32132112

1

2.1.

1

2.1.

ψψψψ

ψ

ψψψψ

ψψψψ

ψψ

++

+

−

+

+−

=∆∆

=P

PdP

dP LL (3.29)

The foregoing expression may be simplified if the three-point system assumes the

particular form shown in figure 3.4 . for the this particular configuration

Z12= Z21=∞

Z13= Z31=impedance of line 1-3

Z23= Z32=impedance of line 2-3

Since Z12= Z21=∞ , Ψ12 =Ψ21=0

Equation 3-29 then reduces to

( )1332

3123

1

2.1.1

**1ψψψψ

−=∆P

dP (3.30)

From equation 3-21 and 3-30,

( )1332

3123

**ψψψψ =

)cos()cos()cos()cos(

13133232

31312323

αθαθαθαθ

−−−−

=

+−

−+

1313

1313

2323

2323

tantan1tantan1

tantan1tantan1

αθαθ

αθαθ

Dividing with tanα23 tanα13

+

−

−

+

−=∆

1313

1323

23

23

1313

1323

23

23

1

2.1.1

tantan

tantan1

θθ

θθ

RX

RX

RX

RX

PdP (3.31)

[ ][ ][ ][ ]13132323

13132323

1

2.1.1

tantantantan

1θθθθ

+−−+

−=∆ KK

KKP

dP

59

Assume that KRX

RX

==13

13

23

23

13θ = 212

32θθ =

Figure 3.4 Simplified three-point system.

Equation 3-31 then becomes

212

12

1212

1212

1

2.1.1

2tan

2tan4

2tan

2tan

2tan

2tan

1

+

=

+

+

−

−

−=θ

θ

θθ

θθ

K

K

KK

KK

dPdP

It is suggested that the expression 3-14 may be used without consideration of

A. Transfer impedance between plants

B. System load distribution, or

C. Generation of other plants

And that equation 3-12 may be used directly when the angle θ12 is less than 15°

The general applicability of equations 3-12 and 3-14 will be investigated

with respect to a three-point system similar to figure 3-4 . in this and the following

comparisons we shall compare the exact value of given by equation 3-31 with that obtained

by the approximations given by equations 3-12 and 3-14 . These formulas are tabulated

below:

12

12

1

2.1.1

tantan2

θθ

+==

∆ KPdP

2122

1122

1

1

2.1.1

)/tan(/tan4θθ

+==

∆ KK

PdP

60

[ ][ ][ ][ ]13132323

13132323

1

2.1.1

tantantantan

1θθθθ

+−−+

−==∆ KK

KKP

dP (3-31)

The effect of an intermediate load is illustrated by considering the system shown in

figure 3-5 and 3-6.

The quantity K13 is defined to be the X/R ratio of line1-3 ; similarly , K23 is the X/R

ratio of line 2-3 . The ratio of K23 to K13 is given by n . The value of k for the transmission

line impedance between 1 and 2 was maintained at a constant value of 2.5 . Thus equation

3-12and 3-14 are used with K= 2.5. the various curves illustrated correspond to various

values of n; thus n= 1 corresponds to both lines having an X/R ratio of 2.5 similarly n=2

corresponds to line1-3 having an X/R ratio of 1.875 and live 2-3 having an X/R ration of

3.75.

In figure 3.5 the angle θ12 was maintained constant at 30° . It will be noted that θ12 is

equal to θ13 -θ23 . With θ12 maintained constant at 30° , θ23 andθ13 were varied as indicated

by the abscissa. We note that if the X/R ratios of both lines are the same the effect of the

intermediate load and its relative angular position is not very great.

Figure 3.5 Effect of intermediate load upon results obtained by various formulas with θ12=

30°.

61

However, if the X/R rations of the line are different, considerable discrepancies may

occur. For example, with n = 2 , dP L.1.2 /dP1 may very from 0.27 to 0.47, instead of being

constant at 0.35 , as indicated by equation 3-14 or 0.375 , as indicated by equation 3-12.

Figure 3.6 illustrates that similar differences occur when the angular difference θ12is

maintained constant at 15° ; thus dPL1.2 /dP1 may vary from 0.13 to 0.25 for n= 2 where the

approximate answer given by equations 3-12 and 3-14 is approximately 1.19 .Here we note

that the correct result is dependent upon the relative angular position of the intermediate

load and the X/R ratio of the transfer impedances Z13 and Z23 , even though the K of the

transmission line between 1 and 2 remains constant at 2.5.

The effect of intermediate plants is illustrated by the simple system shown in figure

3.7.

.

Figure3.6 . Effect of intermediate load upon results obtained various formula with

θ12 =15°

62

In this case the angular difference θ12 is maintained at zero degrees and θ13 and θ23

are chosen equal to one another. Equations 3-12 and 3-14 applied to this case would

indicate dP1.1.2/dP1 to be identically zero for this condition. Inspection of Figure 3.3

indicates that if the X/R ratios of the two lines are identical dP1.1.2/dP1is zero. However, if

the X/R ratio of the lines are different, we note that the value of dP1.1.2/dP1may differ

considerably form zero. For example, with n = 2 and θ13= θ23 ±30˚, the value of dP1.1.2/dP1

would be either +0.28 or -0.38, as compared to zero given by equation 3-12 and 3-14. Their

differences become much more pronounces as the X/R ratio of the individual lines become

increasingly different. Such differences in X/R ratio occur because of different conductor

sizes in the system and also become of transformer in the system.

Figure 3.7 Effect of intermediate load upon results obtained various formula with

θ12 =0°

Figure 3.8 also studies the effect of intermediate generation. In this case the angle

θ13 is assumed to be 15° greater than θ23 . The results obtained by application of equations

3-12 and 3-14 are as indicated. The correct value of dP1.1.2/dP1 is plotted as a function of θ23

and θ13 as indicated in figure 3.8. It will be noted that these conditions, even with X/R

ratios of the two lines identical , there is a discrepancy between the results obtained by

equation 3-12 and 3-14 and the correct result when the angular differences are large.

63

From the results presented in Figure 3.5 to 3.8 it appears that the use of the

simplified expressions 3-12 and 3-14 yields a good approximation to incremental

losses when the X/R rations of the elements are very similar.

Figure 3.8 Effect of intermediate generation up on results obtained by various formulas with

θ13 = θ23+ 15˚

3.6 APPLICATION OF SIMPLIFIED PHASE-ANGLE FORMULAS

The phase- angle formulas developed by Cahn parallel the derivations given in section 3.2,

except that a new variable equal to the average of the sending and receiving and powers is

introduced. Define this variable as

221

21PPP −

=− (3.32)

From equation 3-1 and 3-2 and for the system of Figure 3.1,

P1-2 = 12

21

21

ZVV [sin(θ12-α12)-sin(θ21- α21)] (3.33)

Then ∆P1-2= 21−′P -P1-2

=12

21

21

ZVV [sin(θ12- 12α′ )-sin(θ21- α21)]- [sin(θ12-α12)-sin(θ21- α21)] Recalling

that sin(α-β)= sin α cos β +sin β cos α

64

)]sin[(sin(cos 1212121212

2121 αθθα −−′=∆ − Z

VVP (3.34)

Using the relation

)(/sin)(/cos2sinSin 21

21 βαβαβα −+=−

equation 3.34 becomes

)]2

sin([])2

[cos(cos2 1212121212

12

2121

θθθθα −′+′=∆ − Z

VVP (3.35)

Using the relation

Cosα - cosβ = -2 sin ½ (α+β)sin ½ (α-β)

Equation 3.35 becomes

)]2

sin([])2

[sin(sin4 1212121212

12

2121

θθθθα −′+′=∆ − Z

VVP (3.36)

Dividing equation 3-36 by 3-35, we obtain

+′

=∆∆

− 2tantan2 1212

1221

θθαPPL (3.37)

211212

2tan22 −∆

+′

−=∆ PK

PLθθ (3.38)

As 21−∆P approaches zero , equation 3-35 becomes

1221

tan22 θKdP

dPL =−

(3.39)

Compare this formula with equations 3-12 and 3-13.

Equation 3-38 is the formula which is used in the comparisons that follow.

The approximations involved in the determination of the Bmn loss formula used in

the comparisons includes.

1. Each load current remains a constant completes fraction of the total load

current.

2. Generator voltage magnitudes remain at constant values.

3. Generator voltage angles remain fixed.

4. Generator Q/P rates remain fixed at their base case values.

65

PROBLEMS

Problem3.1

.(a)Show that the incremental loss dPL12/dP1 = 2tanθ12/(K+ tanθ12) for two machine

system without intermediated generation.

(b)Plot incremental loss as a function of θ for above system when

θ = 0°,10°,20°,30°,40°,50°,60°and Z12 =0.25+j 0.625.

Problem3.2.

(a)Show that the incremental loss dPL12/dP1 = 2tanθ12/(K+ tanθ12) for two machine

system without intermediated generation.

(b)Plot incremental loss as a function of θ for above system when

θ = 0°,°,20°,40°, 60°,80°.

Problem3.3.

(a)Prove that the incremental loss for without intermediate generation is

dPL12/dP1 = 2tanθ12/(K+ tanθ12).

(b) Show that the above equation is nearly the same as approximate equation of

1

12

dPdPL =

212

12

2

24

)tanK(

tanK

θ

θ

+ for θ12 ≤15°.Take K=2.0.

Problem3.4

.(a)Prove that the incremental loss for without intermediate generation is

dPL12/dP1 = 2tanθ12/(K+ tanθ12).

(b) Show that the above equation is nearly the same as approximate equation of

1

12

dPdPL =

212

12

2

24

)tanK(

tanK

θ

θ

+ for θ12≤15°.Take K=2.5.

Problem3.5.

66

With intermediate load ,show that )tanK)(tanK()tanK)(tanK(

dPdPL

13132323

13132323

1

12 1θθθθ

−−+

−= .

If θ12 is maintained as 15°, plot the loss with corresponding values ofθ13 and θ23.

Take Z13 =0.02+j0.05 p.u and Z23 = 0.015+j0.0375 p.u.

Problem3.6.

With intermediate load ,show that )tanK)(tanK()tanK)(tanK(

dPdPL

13132323

13132323

1

12 1θθθθ

−−+

−= .

If θ12 is maintained as 20°, plot the loss with corresponding values ofθ13 and θ23.

Take Z13 =0.02+j0.04p.u and Z23 = 0.015+j0.030 p.u.

Problem3.7.

With intermediate load ,)tanK)(tanK()tanK)(tanK(

dPdPL

13132323

13132323

1

12 1θθθθ

−−+

−= .

(a)If θ12 is maintained as 30°, plot the loss with corresponding values ofθ13 and θ23.


(b)If θ12 is maintained as30°, plot the loss with corresponding values ofθ13 and θ23.


Problem3.8.


dPdPL

13132323

13132323

1

12 1θθθθ

−−+

−= .



(a)If θ12 is maintained as 30°, but K23/K13 is changed to 1.5 find the difference value of

dPL12/dP1.

Problem3.9.


dPdPL

13132323

13132323

1

12 1θθθθ

−−+

−= .



(a)If θ12 is maintained as 30°, but K23/K13 is changed to 2.0 find the difference value of

dPL12/dP1.

Problem3.10.

67


dPdPL

13132323

13132323

1

12 1θθθθ

−−+

−= .

If θ12 is maintained as 30°, plot the loss with corresponding values ofθ13 and θ23 for the

K23/K13=1,1.5,2.0.

69

CHAPTER 4

PRACTICALCALCULATION, EVALUATION, AND APPLICATION

OF ECONOMIC SCHEDULING OF GENERATION

4.1 GENERAL SUMMARY OF METHOD

This section is intendance to present a chronological outline of the required

data and calculating procedure involved in the analysis of the economic scheduling of

system generation;

4.1.1 Preparation of Data

1. Electrical-system Data

(a) Impedance diagram of transmission and sub-transmission facilities whose

losses are dependent upon the manner in which generation is scheduled.

(b) Daily load cycles for typical week’s operation.

(c) Load –duration curve for period of operation to be considered.

(d) Selection of base-case loading period and tabulation of loads, voltages,

and probable generation schedules and interconnection flows for base case.

2. Plant Data

(a) Thermal characteristics of units and in particular the incremental fuel-rate

data on all units

(b) Cost of fuel at various plants in cents per million Btu.

(c) Determination straight-line equations of incremental production costs of

various units.

4.1.2 Determination of Transmission-Loss Formula

1. Resistance measurements on open-circuited transmission network.

2. Base-case load –flow data.

3. Transcribing of base data to punched cards if data steps 1 or 2 are taken from

network analyzer.

4. Calculation of loss-formula coefficients on digital calculator.

70

4.1.3. Evaluation of saving to be obtained by considering Transmission losses in the

scheduling off generation

1. Determine generation schedules by equation incremental production costs.

2. Determine generation schedules by coordination equations.

3. Determine cost of received power for each schedule.

4. Determine difference in cost of received power.

5. There is use of loss-duration curve, determine annual savings.

4.1.4. Practical Application of coordination equations to power-system operation.

1. Pre-calculated generation schedules.

2. Use of special computers built particularly for use of load dispatcher to

calculate schedules as need arises.

3. Economic automation system which automatically and simultaneously

maintain economic allocation of generation, system frequency, and net

interchange.

4.2 .DETERMATION OF IMPORTANCE OF TRANSMISSION –LOSS

CONSIDERATION

It is prudent to evaluate the saving involved when scheduling generation with

the effect of incremental transmission losses included for typical operating conditions

and typical fuel-cost data. It is suggest that this step be taken before undertaking

point 4.1.4 of the outline in order to determine the amount of effect and dollars it is

worthwhile to expend for part 4.1.4.

4.3 SOLUTION OF COORDINATION EQUATIONS

The determination of generation schedules by equal incremental production

costs is discussed in chapter1. The theory involved in determining the coordination

equations to include the effect of both incremental production costs and incremental

transmission losses is covered in chapter 1. Various methods of solving these

coordination equations in order to undertake step 4.1.3 and also to pre-calculate

generation schedules for various operating conditions if desired, in step4.1.4.

The coordination equations may be solved by analogue or digital methods. We

shall first discuss those suitable for analogue computers, such as the network analyzer

and the differential analyzer.

71

4.4 ANALOGUE METHODS

From chapter 4 we may write the following coordination equations for a two-

plant system:

F11P1+λ(2B11P1+2B12P2) = -(f1- λ)

F22P2+λ(2B22P1+2B12P2) = -(f2- λ)

Regrouping coefficients, we obtain

P1(F11+ λ2B11) + P2(λ2B12) = -(f1- λ)

P1(λ2B12) + P2 (F22+ λ2B22) = -(f2- λ)

Defining A11 = F11+ λ2B11

A12 = λ2B12 (6-1)

A21 = λ2B12

A22 = F22+ λ2B22

C1= - (f1- λ)

C2 = -(f2- λ)

We obtain A11 P1+ A12 P2 = C1 (6-2)

A21 P1+ A22 P2 = C2

The use of a network-analyzer mesh-circuit analogue is illustrated Figure 6.1. The Pn

are considered to be currents; the Amn resistances; and the Cn , voltages.

Figure 6.1 Mesh-circuit analogue of coordination equations.

This method has certain practical limitations which restrict is usefulness:

1. Errors introduced by mutual transformers in representing mutual coefficients.

72

2. Limitation in number of mutual transformers. The use of conductive coupling

may eliminate the need for some mutual transformers, and conductive

coupling is very time consuming.

3. Difficulties involved in representing negative mutual coefficients.

4. Difficulty in maintaining constant current output of a network analyzer

generator when the particular plant in question is at maximum or minimum.

5. Time involved in plugging the analyzer.

If a nodal-circuit analogue is used, the Pn are considered to be voltage; the Amn

inductive or capacitive reactances; and the Cn currents. The corresponding circuit

is given in figure 4.2. In this analogy the difficulties discussed under points1,2,

and 3 are usually eliminated.

Figure 4.2 Nodal-circuit analoge of coordination equations.

However, and error is introduced because of the resistance in the reactance

units. Also, driving function requires the setting of constant currents with its

attendant difficulties, since the network-analyzer generators are constant voltage

devices.

The use of an electronic differential analyzer is a better analogue method of

solving the coordination equations than the a-c network analyzer. The electronic

differential analyzer is composed of amplifiers (for adding, inverting, and

integrating) and of resistance and capacitance units. In this computer the analogy

is between d-c voltages and the variables of the system under study. Consider the

set of simultaneous equations.

Amn Pn = Cn (4.3)

73

or

AP = C (4.4)

Rewriting equation 4.3.

Amn Pn -Cn= 0 (4.5)

An auxiliary set of equations with the same steady –state solution is given by

nmmnn CPAPdtd

− (4.6)

Thus for m, n = 1, 2 we have

12121111 CPAPAPdtd

+= (4.7)

22221212 CPAPAPdtd

+=

The method of solution of these equations may also be solved in another manner

by using electronic differential-analyzer elements, as indicated in Figure 4.3.

Figure 4.3 alternative electronic differential analyzer solution of coordination

equations.

For this method the coordination equations are written in the form

∂∂

−=1

1

1

1 1Pp

dPdF λ

74

4.5 DIGITAL METHODS

A number of digital methods which may be undertaken manually or by means

of digital computers are discussed.

1. Determinants

2. Matrix inversion. If it is assumed that our original set of equations is given by

AP = C

Then

A-1 AP = A-1 C

P= A-1 C

Where A-1= inverse of A

The following method of obtaining an inverse is taken from the book tensor

analysis of works by G Kron.

(a) Interchange rows and columns.

(b) Replace each element by its minor. The minor of any given element is

obtained by striking out the row and column in which the element lies and then

calculating the determinant of the remaining matrix.

(c) Multiply every other minor with minus one, starting with plus one in upper

left-hand corner, as shown in the following scheme.

(d) Divide each resulting element by the determinant of the whole original matrix.

3.Starring or Pivotal Condensation. Assume that the original set equations is given

by AP =C

From the composite matrix we obtain

kk

ikkjijij A

AAAA −=*

4. iterative procedure. The iterative procedure involves a method of successive

approximations which rapidly converge to the correct solution. From equation

4-1 the exact coordination equations are given by

λλ =∂∂

+nn

n

PP

dPdF 1 (4.19)

Fmn Pn + fn + λ∑ mmn PB2 =λ (4.20)

Collecting all coefficient of Pn we obtain

Pn ( Fmn +λ2Bnn)= - λ (∑ mmn PB2 )-fn+ λ (4.21)

Solving for Pn we obtain

75

nnnn

nmmn

n

n

BF

PmBf

P2

21

+

−−=

∑≠

λ

λ

The number of required iterations in general is quite small, since the diagonal

terms are generally much larger than the off-diagonal terms.

This iterative procedure is illustrated for a simple for a simple two-plant system.

Example 4.1

Assume that the loss-formula coefficients in 1/ Mw units are given by

M n Bmn

1 1 +0.001

1 2 -0.005

2 2 +0.0024

Also assume that

lincrementadPdF

1

1 = production cost of plant 1

= F11 R1 +f1

lincrementadPdF

2

2 = production cost of plant 2

= F22 R2 +f2

where F11= +0.01 f1= 2.0

F22= +0.01 f2= 1.5

Find a point I the generation schedule.

SOLUTION

Substituting the above numbers into equation 6-22

0.0020.01/λ0.001P(2.0/λ21.0P 2

1 ++−

=

0.00480.01/λ0.001P(1.5/λ11.0P 1

2 ++−

=

To find a point in the generation schedule we choose a λ, say 2.5, and iterate

unit the Pn have converged to sufficient accuracy. The calculating form is then

76

006.0001.02.0

002.004.0001.080.00.1 22

1PPP +

=++−

=

0.0060.001P0.2

0.0020.040.001P0.801.0P 11

2+

=++−

=

The calculations are started by first assuming all Pn= 0; and as new values are

calculated they are used immediately as follows:

Iteration No. Calculation

1 3.33006.0

02.0P1 =+

=

2.490088.0

0333.04.0P2 =+

=

2 5.41006.0

0492.02.0P1 =+

=

2.500088.0

0415.04.0P2 =+

=

3 3.4

006.00502.02.0P1 =

+=

2.500088.0

0417.04.0P2 =+

=

4 7.41006.005022.0P1 =

+=

2.500088.0

0417.04.0P2 =+

=

The iterative solution is completed after four iterations, since the results of the

fourth iteration agree with the preceding results to a sufficient degree of accuracy.

The number of required iterations is small. Even for a much larger system , as the

terms which are independent of the other plants are the dominant ones.

If a knowledge of system transmission losses is desired, they may be evaluated

by a total loss formula

P1= PmBmnPn = B11 P1P1 + 2 B12P1 P2+B22P2

= 0.0010(41.7)2-(2)(0.005) (41.7)(50.2)+0.0024(50.2)2= 5.7 Mw

77

The total generation is obtained by

∑ =+== 91.9Mw50.241.7PP nT

The received power is calculated

PR = PT - Pl= 91.9-5.7 = 86.2 Mw

Thus , one entry in a table used for scheduling is

Total Received

P1 P2 generation Loss Load λ

41.7 50.2 91.9 5.7 86.2 2.5

This type of problem is well suited to the use of digital computer; and generation

schedules may be economically calculated by this iterative method either with

small –or large scale, general purpose digital computers. One computer toward

the small end of the spectrum which has been used for a substantial amount of

scheduling is the card-programmed calculator (CPC) . The control unit of the

CPC senses holes punched in cards, which are the basis of operation sequencing.

Thus a deck of these cards is a program of operations, as the name of the device

implies.

The larger scale computer operates in a similar manner , but the numbers

which the control unit senses may be stored in its more extensive memory,

permitting more rapid access to these instructions and thus a higher effective

operating speed. Many such machines, with variations in the techniques of

storage in memory and data-handling equipment, are in existence. The use of

either size of equipment may be justified for scheduling, depending on the

circumstances.

It is desirable that the computer programmer see the logical sequenc of

operations in a problem represented graphically. This is usually called a flow

diagram, and one form is illustrated in Figure 4.5 . The sequence of basic operations

is established by following the arrow through operation boxes and taking the correct

branch when an operation box has two output paths. The operations pertaining to data

handing were committed, since they are not he same for all computers for the

scheduling of generation.

78

This digital computer method also provides the automatic selection of the

approximate incremental cost characteristics. As previously discussed, the

representation of the incremental cost characteristics is achieved by breaking the

incremental –cost curves in to several sections

Much may be accurately represented by straight line segments. The computer

can now select side of the break point on which the previously iterated generation of

that plant occurred and find the correct slope and intercept to use in the current

iteration. Also, the plants have physical maximum capacities and some practical

minim um load points. These limitations must compose and a digital computer can do

this very well by testing the generation calculated. If resulting value is outside the

limits, the computer substitutes the appropriate limiting value.

In using the computer for scheduling one may arbitrarily select a set of values

for λ and proceed to calculate a schedule point for each value of λ . This mode of

operation is quite efficient. For example, for one given value of λ, the time on the

IBM 650 computer to calculate the economic allocation of generation total generation

, losses , and received load for a ten-plant system would be approximately 10 minutes.

In other situation it may be convenient to calculate the λ which

1. The net generation of a given plant to be a specified value.

2. The total generation to be a specified value.

3. The received power to be a specified value.

In any of these forms a new value of λ is calculated for each iteration in

addition to new generation amounts , and all quantities converge to

constant values.

It is desirable in defining a schedule with a minimum number of points to

obtain solutions corresponding to those outputs at which individual plants

encounter their maximum, break, and minimum generations. In this case

the defending equation is

mmnm

nnnn

PBfPF

∑−+

=21

λ

where Pn is the load on plant n that is specified rather than calculated.

In scheduling on a specified total generation basis the λ are calculated by

there superscript I indicates the iteration being started

i-l indicates the iteration just completed

79

i-2 indicates the preceding iteration

PT – total generartion (Pn)

=dTP desirable total generation

Scheduling by this scheme is initiated by selecting two values of λ . Iteration is

carried out convergence for both these λ value to establish the =−1diTP and

=−1diTP required in equation 4.23.

Scheduling on a specified received load may be achieved by substituting PR=

for Pt in equation 23. This implies a substantial increase in calculation, as total losses

must be calculated for iteration. Since most utilities schedule on the basis of total

generation rather than received, this form would not usually desired except in making

economic comparisons.

The iterative digital method described has been found to be an extremely

valuable tool in obtaining more economical operation of power systems. This method

of solution of co-ordinations offers several distinct advantages over previous methods:

1. The flexibility of this method allows a generation schedule point to be

obtained for a given value of incremental cost of received power, or a specific

value of local load, a specific of total generation , or a specific load on a

number of solutions.

2. Once the loss formula coefficients are developed, the time required to

transcribe the use in the computer is very small. Since the program of

calculation is general, a single routine is maintain in the library which will

permit scheduling of any size system in a most efficient manner.

3. For convenience in comparing schedules the incremental cost of received

power, total transmission losses, and received load may be calculated along

with the allocation and summation of generation, If desired, the fuel input to

each plant, as well as the total fuel input , may also be calculated

4. The desired method may be applied to all general- purpose digital

computers.

80

Figure 4.4 Digital computer flow diagram for solution of coordination equations

.

81

4.6 PROCEDURE FOR EVALUATING ANNUAL SAVINGS

The following general procedure for evaluating the annual saving incurred by

including transmission –loss considerations in the scheduling of generation is

suggested:

1. Determine the generation schedule for operation at equal incremental

production costs, neglect ion the effect of transmission losses.

2. Determine the generation schedule, including the effects of both

incremental production costs and incremental transmission losses.

3. Calculate cost of received power by the following method:

(a) Determine PL by summing the i2 R losses , line –by –line or from a

transmission losses formula

Thus PL = kk Ri∑ 2

Or PL= Pm Bmn Pn

(b) Then the received power PR is given by

∑ −= LR PPP n

(c) The fuel input to a given plant n (Fn) in dollar per hour may be

determined by reference to plant heat rate or input –output data.

An alternative procedure is to integrate the incremental fuel cost

data . Thus , if

11111

1 fPFdPdf

+=

F1= 11

2111

2pfPF

−

This expression for F1 does not include the fuel input incurred for zero

output . If the same units are in operation for both schedules, the no-

load , fuel input intercept is no required since the input for zero output

disappears when subtracting the fuel input for such schedules. The

total fuel input (F1) is, of course, given by

∑= nFF1

(d) the cost fo the received power in dollars per MW-hr id given by

82

F/PR

4. Plot Ft/PR vs PR of schedules 1 and 2 (Figure 4.5)

Figure 4.5 Plot of dollars per mw-hr of received load for two schedules.

5.Plot a curve of difference between the two curves of point 4 vs PR (Fig. 4.6).

6.From 5 plot dollars per hour vs PR,

where dollars per hour = dollars per Mw- hrx PR

7. With a load duration curve (as indicated in Figure 4.7a) plot dollars per hour vs :

hours integral of this curve is dollars saved (Figure 4.7b)

In general, transmission loss considerations in the scheduling of generation are not

important a metropolitan system. Transmission loss consideration in system

scheduling usually prove be significant in a widespread system in which there is a

significant difference in the incremental production costs between various parts of the

power system. Typical annual savings obtained for widespread systems are fifty

thousand dollars per 1000 Mw of installed capacity

Figure4.6 Difference in dollars per Mw-hr of received load for two schedules.

83

Figure4.7(a)Load duration curve. (b) Plot of dollar per hour vs time.

PROBLEMS

Problem4.1

Assume that the loss formula coefficients in Mw-1units are given by

M n Bmn

1 1 +0.001

1 2 -0.005

2 2 +0.0024

Also assume that 1

1

dPdF

= incremental production cost of plant 1 = F11P1 +f1

2

2

dPdF = incremental production cost of plant 1 = F22P2 +f2

where F11 = +0.01 f1 =2.0 F22 = +0.01 f2 = 1.5

Find a point I the generation schedule.

Problem4.2

Describe the required condition of the economic scheduling of system generation.

Problem4.3

Draw the digital computer flow diagram for solution of co-ordination equation.

84

Documents

Impedance Matrix