impedance matrix
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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
iXiXiXiRiRiRjiXiXiXiRiRiR
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
)sin()sin(sin 232323
322121
22
1211
22
22
2 αθαθα −+−+=Z
VVZ
VVZVP (3-16)
)sin()sin(sin 323232
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.
Take Z13 =0.02+j0.04p.u and Z23 = 0.015+j0.030 p.u.
(b)If θ12 is maintained as30°, plot the loss with corresponding values ofθ13 and θ23.
Take Z13 =0.02+j0.04p.u and Z23 = 0.04+j0.08 p.u.
Problem3.8.
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.
Take Z13 =0.02+j0.04p.u and Z23 = 0.015+j0.030 p.u.
(a)If θ12 is maintained as 30°, but K23/K13 is changed to 1.5 find the difference value of
dPL12/dP1.
Problem3.9.
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.
Take Z13 =0.02+j0.04p.u and Z23 = 0.015+j0.030 p.u.
(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
With intermediate load ,)tanK)(tanK()tanK)(tanK(
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
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