chapter 3 network admittance and impedance … · the reduced admittance diagram obtained by...
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CHAPTER 3
NETWORK ADMITTANCE AND IMPEDANCE MATRICES
As we have seen in Chapter 1 that a power system network can be converted into an equivalent impedance diagram. This diagram forms the basis of power flow (or load flow) studies and short circuit analysis. In this chapter we shall discuss the formation of bus admittance matrix (also known as Ybus matrix) and bus impedance matrix (also known as Zbus matrix). These two matrices are related by
1−= busbus YZ (3.1) We shall discuss the formation of the Ybus matrix first. This will be followed by the discussion of the formation of the Zbus matrix.
3.1 FORMATION OF BUS ADMITTANCE MATRIX
Consider the voltage source VS with a source (series) impedance of ZS as shown in Fig. 3.1 (a). Using Norton’s theorem this circuit can be replaced by a current source IS with a parallel admittance of YS as shown in Fig. 3.1 (b). The relations between the original system and the Norton equivalent are
SS
S
SS Z
YZV
I 1 and == (3.2)
We shall use this Norton’s theorem for the formulation of the Ybus matrix.
Fig. 3.1 (a) Voltage source with a source impedance and (b) its Norton equivalent.
For the time being we shall assume the short line approximation for the formulation of the bus admittance matrix. We shall thereafter relax this assumption and use the π-representation of the network for power flow studies. Consider the 4-bus power system shown in Fig. 3.2. This contains two generators G1 and G2 that are connected through transformers T1 and T2 to buses 1 and 2. Let us denote the synchronous reactances of G1 and G2 by XG1 and XG2 respectively and the leakage reactances of T1 and T2 by XT1 and XT2 respectively. Let Zij, i = 1, …, 4 and j = 1, …, 4 denote the line impedance between buses i and j.
Then the system impedance diagram is as shown in Fig. 3.3 where Z11 = j(XG1 + XT1) and Z22 = j(XG2 + XT2). In this figure the nodes with the node voltages of V1 to V4 indicate the
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buses 1 to 4 respectively. Bus 0 indicates the reference node that is usually the neutral of the Y-connected system. The impedance diagram is converted into an equivalent admittance diagram shown in Fig. 3.4. In this diagram Yij = 1/Zij, i = 1, …, 4 and j = 1, …, 4. The voltage sources EG1 and EG2 are converted into the equivalent current sources I1 and I2 respectively using the Norton’s theorem discussed before.
Fig. 3.2 Single-line diagram of a simple power network.
Fig. 3.3 Impedance diagram of the power network of Fig. 3.2.
Fig. 3.4 Equivalent admittance diagram of the impedance of Fig. 3.3.
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We would like to determine the voltage-current relationships of the network shown in Fig. 3.4. It is to be noted that this relation can be written in terms of the node (bus) voltages V1 to V4 and injected currents I1 and I2 as follows
=
4
3
2
1
2
1
00
VVVV
YII
bus (3.3)
or
=
002
1
4
3
2
1
II
Z
VVVV
bus (3.4)
It can be easily seen that we get (3.1) from (3.3) and (3.4).
Consider node (bus) 1 that is connected to the nodes 2 and 3. Then applying KCL at this node we get
( ) ( )( ) 3132121131211
311321121111
VYVYVYYYVVYVVYVYI
−−++=−+−+=
(3.5)
In a similar way application of KCL at nodes 2, 3 and 4 results in the following equations
( ) ( ) ( )( ) 424323224231222112
4224322312122222
VYVYVYYYYVYVVYVVYVVYVYI
−−++++−=−+−+−+=
(3.6)
( ) ( ) ( )
( ) 4343342313223113
4334232313130VYVYYYVYVY
VVYVVYVVY−+++−−=−+−+−=
(3.7)
( ) ( )
( ) 43424334224
343424240VYYVYVY
VVYVVY++−−=−+−=
(3.8)
Combining (3.5) to (3.8) we get
+−−−++−−−−+++−
−−++
=
4
3
2
1
34243424
343423132313
24232423122212
1312131211
2
1
0
0
00
VVVV
YYYYYYYYYYYYYYYYY
YYYYYII
(3.9)
Comparing (3.9) with (3.3) we can write
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+−−−++−−−−+++−
−−++
=
34243424
343423132313
24232423122212
1312131211
0
0
YYYYYYYYYYYYYYYYY
YYYYY
Ybus (3.10)
In general the format of the Ybus matrix for an n-bus power system is as follows
−−−
−−−−−−
=
nnnnn
n
n
bus
YYYY
YYYYYYYY
Y
L
MOMMM
L
L
321
223212
113121
(3.11)
where
∑=
=n
jkjk YY
1
(3.12)
It is to be noted that Ybus is a symmetric matrix in which the sum of all the elements of the kth column is Ykk.
Example 3.1: Consider the impedance diagram of Fig. 3.2 in which the system parameters are given in per unit by
Z11 = Z22 = j0.25, Z12 = j0.2, Z13 = j0.25, Z23 = Z34 = j0.4 and Z24 = j0.5 The system admittance can then be written in per unit as
Y11 = Y22 = − j4, Y12 = − j5, Y13 = − j4, Y23 = Y34 = − j2.5 and Y24 = − j2 The Ybus is then given from (3.10) as
−−
−−
=
5.45.2205.295.24
25.25.13504513
jYbus per unit
Consequently the bus impedance matrix is given by
=
3926019740136701133.01974025650123601264013670123601531009690
0.1133126400969015310
...
....
.......
jZbus per unit
It can be seen that like the Ybus matrix the Zbus matrix is also symmetric.
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Let us now assume that the voltages EG1 and EG2 are given by
pu 01 andpu 301 21 °∠=°∠= GG EE The current sources I1 and I2 are then given by
pu 0949025.0
01
pu 6049025.0
301
22
22
11
11
°−∠=°∠
°∠==
°−∠=°∠
°∠==
ZEI
ZEI
G
G
We then get the node voltages from (3.4) as
°−∠°−∠
=
00
904604
3926019740136701133.01974025650123601264013670j123601531009690
j0.1133126400969015310
4
3
2
1
.j.j.jj
.j.j.j.j..j.j.
.j.j.j
VVVV
Solving the above equation we get the node voltages as
°∠°∠°∠°∠
=
56.139662.018.159659.0
11.550.967745.180.9677
4
3
2
1
VVVV
per unit
∆∆∆ 3.1.1 Node Elimination by Matrix Partitioning
Sometimes it is desirable to reduce the network by eliminating the nodes in which the current do not enter or leave. Let (3.3) be written as
=
x
AT
x
A
VV
MLLK
II
(3.13)
In the above equation IA is a vector containing the currents that are injected, Ix is a null vector and the Ybus matrix is portioned with the matrices K, L and M. Note that the Ybus matrix contains both L and LT due to its symmetric nature.
We get the following two sets of equations from (3.13)
xAA LVKVI += (3.14)
AT
xxAT
x VLMVMVVLI 10 −−=⇒+== (3.15) Substituting (3.15) in (3.14) we get
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( ) AT
A VLLMKI 1−−= (3.16) Therefore we obtain the following reduced bus admittance matrix
Treducedbus LLMKY 1−−= (3.17)
Example 3.2: Let us consider the system of Example 3.1. Since there is no current
injection in either bus 3 or bus 4, from the Ybus computed we can write
−
−=
=
−
−=
5.45.25.29
and25.2
04'
5.135513
jjjj
Mjj
jL
jjjj
K
We then have
−
−=−= −
8978.108978.68978.68978.101
jjjj
LLMKY Treducedbus per unit
Substituting I1 = 4∠−60° per unit and I2 = 4∠−90° per unit we shall get the same values of V1 and V2 as given in Example 3.1.
Inspecting the reduced Ybus matrix we can state that the admittance between buses 1 and 2 is −j6.8978. Therefore the self admittance (the admittance that is connected in shunt) of the buses 1 and 2 is −j4 per unit (= −j10.8978 + j6.8978). The reduced admittance diagram obtained by eliminating nodes 3 and 4 is shown in Fig. 3.5. It is to be noted that the impedance between buses 1 and 2 is the Thevenin impedance between these two buses. The value of this impedance is 1/(−j6.8978) = j0.145 per unit.
∆∆∆
Fig. 3.5 Reduced admittance diagram after the elimination of buses 3 and 4. 3.1.2 Node Elimination by Kron Reduction
Consider an equation of the form
bAx = (3.18) where A is an (n×n) real or complex valued matrix, x and b are vectors in either Rn or Cn. assume that the b vector has a zero element in the nth row such that (3.18) is given as
1.53
=
−−−−−
01
2
1
1
2
1
21
,12,11,1
22221
11211
n
n
n
nnnn
nnnn
n
n
b
bb
xx
xx
aaaaaa
aaaaaa
MM
L
L
MOMM
L
L
(3.19)
We can then eliminate the kth row and kth column to obtain a reduced (n − 1) number of equations of the form
=
−−−−−−
−
−
1
2
1
1
2
1
1,12,11,1
1,22221
1,11211
nn
new
nnnn
n
n
b
bb
x
xx
aaa
aaaaaa
MM
L
MOMM
L
L
(3.20)
The elimination is performed using the following elementary operations
nn
njknkj
newkj a
aaaa −= (3.21)
Example 3.3: Let us consider the same system of Example 3.1. We would like to
eliminate the last two rows and columns. Let us first eliminate the last row and last column. Some of the values are given below
6111.125.4225.13,5
44
42242222
44
41242121 jjj
YYYYYj
YYYYY newnew −=
×+−=−==−=
0,6111.35.4
5.225.244
4444444
44
43242323 =−==
×+=−=
YYYYYjjj
YYYYY newnew
In a similar way we can calculate the other elements. Finally eliminating the last row and last column, as all these elements are zero, we get the new Ybus matrix as
−−
−=
6111.76111.346111.36111.1254513
jjjjjj
jjjY new
bus
Further reducing the last row and the last column of the above matrix using (3.21), we obtain the reduced Ybus matrix given in Example 3.2.
∆∆∆ 3.1.3 Inclusion of Line Charging Capacitors
So far we have assumed that the transmission lines are modeled with lumped series impedances without the shunt capacitances. However in practice, the Ybus matrix contains the shunt admittances for load flow analysis in which the transmission lines are represented by its
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π-equivalent. Note that whether the line is assumed to be of medium length or long length is irrelevant as we have seen in Chapter 2 how both of them can be represented in a π-equivalent.
Consider now the power system of Fig. 3.2. Let us assume that all the lines are represented in an equivalent-π with the shunt admittance between the line i and j being denoted by Ychij. Then the equivalent admittance at the two end of this line will be Ychij/2. For example the shunt capacitance at the two ends of the line joining buses 1 and 3 will be Ych13/2. We can then modify the admittance diagram Fig. 3.4 as shown in Fig. 3.6. The Ybus matrix of (3.10) is then modified as
++−−−+++−−−−++++−
−+++
=
434243424
3433423132313
242322423122212
13121131211
0
0
ch
ch
ch
ch
bus
YYYYYYYYYYYYYYYYYYYY
YYYYYY
Y
(3.22) where
2
2
2
2
34244
3423133
2423122
13121
chchch
chchchch
chchchch
chchch
YYY
YYYY
YYYY
YYY
+=
++=
++=
+=
(3.23)
Fig. 3.6 Admittance diagram of the power system Fig. 3.2 with line charging capacitors.
1.55
3.2 ELEMENTS OF THE BUS IMPEDANCE AND ADMITTANCE MATRICES
Equation (3.1) indicates that the bus impedance and admittance matrices are inverses of each other. Also since Ybus is a symmetric matrix, Zbus is also a symmetric matrix. Consider a 4-bus system for which the voltage-current relations are given in terms of the Ybus matrix as
=
4
3
2
1
44434241
34333231
24232221
14131211
4
3
2
1
VVVV
YYYYYYYYYYYYYYYY
IIII
(3.24)
We can then write
01
111
432 ===
=VVVV
IY (3.25)
This implies that Y11 is the admittance measured at bus-1 when buses 2, 3 and 4 are short circuited. The admittance Y11 is defined as the self admittance at bus-1. In a similar way the self admittances of buses 2, 3 and 4 can also be defined that are the diagonal elements of the Ybus matrix. The off diagonal elements are denoted as the mutual admittances. For example the mutual admittance between buses 1 and 2 is defined as
02
112
431 ===
=VVVV
IY (3.26)
The mutual admittance Y12 is obtained as the ratio of the current injected in bus-1 to the voltage of bus-2 when buses 1, 3 and 4 are short circuited. This is obtained by applying a voltage at bus-2 while shorting the other three buses.
The voltage-current relation can be written in terms of the Zbus matrix as
=
4
3
2
1
44434241
34333231
24232221
14131211
4
3
2
1
IIII
ZZZZZZZZZZZZZZZZ
VVVV
(3.27)
The driving point impedance at bus-1 is then defined as
01
111
432 ===
=IIII
VZ (3.28)
i.e., the driving point impedance is obtained by injecting a current at bus-1 while keeping buses 2, 3 and 4 open-circuited. Comparing (3.26) and (3.28) we can conclude that Z11 is not the reciprocal of Y11. The transfer impedance between buses 1 and 2 can be obtained by injecting a current at bus-2 while open-circuiting buses 1, 3 and 4 as
1.56
02
112
431 ===
=IIII
VZ (3.29)
It can also be seen that Z12 is not the reciprocal of Y12.
3.3 MODIFICATION OF BUS IMPEDANCE MATRIX
Equation (3.1) gives the relation between the bus impedance and admittance matrices. However it may be possible that the topology of the power system changes by the inclusion of a new bus or line. In that case it is not necessary to recompute the Ybus matrix again for the formation of Zbus matrix. We shall discuss four possible cases by which an existing bus impedance matrix can be modified.
Let us assume that an n-bus power system exists in which the voltage-current relations are given in terms of the bus impedance matrix as
=
n
orig
n I
IZ
V
VMM11
(3.30)
The aim is to modify the matrix Zorig when a new bus or line is connected to the power system. 3.3.1 Adding a New Bus to the Reference Bus
It is assumed that a new bus p (p > n) is added to the reference bus through an impedance Zp. The schematic diagram for this case is shown in Fig. 3.7. Since this bus is only connected to the reference bus, the voltage-current relations the new system are
=
=
p
nnew
p
n
p
orig
p
n
II
I
Z
II
I
Z
Z
VV
VMM
L
MM111
000
0
(3.31)
Fig. 3.7 A new bus is added to the reference bus.
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3.3.2 Adding a New Bus to an Existing Bus through an Impedance
This is the case when a bus, which has not been a part of the original network, is added to an existing bus through a transmission line with an impedance of Zb. Let us assume that p (p > n) is the new bus that is connected to bus k (k < n) through Zb. Then the schematic diagram of the circuit is as shown in Fig. 3.8. Note from this figure that the current Ip flowing from bus p will alter the voltage of the bus k. We shall then have
( ) nknpkkkkkk IZIIZIZIZV ++++++= LL2211 (3.32) In a similar way the current Ip will also alter the voltages of all the other buses as
( ) kiIZIIZIZIZV ninpkikiii ≠++++++= ,2211 LL (3.33) Furthermore the voltage of the bus p is given by
( ) pbkknknkkkkk
pbkp
IZZIZIZIZIZ
IZVV
+++++++=
+=
LL2211
(3.34)
Therefore the new voltage current relations are
=
+
=
p
nnew
p
n
bkkknk
kn
orig
k
p
n
II
I
Z
II
I
ZZZZZ
ZZ
VV
VMM
L
MM11
1
11
(3.35)
It can be noticed that the new Zbus matrix is also symmetric.
Fig. 3.8 A new bus is added to an existing bus through an impedance. 3.3.3 Adding an Impedance to the Reference Bus from an Existing Bus
To accomplish this we first assume that an impedance Zb is added from a new bus p to an existing bus k. This can be accomplished using the method discussed in Section 3.3.2. Then to add this bus k to the reference bus through Zb, we set the voltage Vp of the new bus to zero. However now we have an (n + 1) × (n + 1) Zbus matrix instead of an n × n matrix. We can then remove the last row and last column of the new Zbus matrix using the Kron’s reduction given in (3.21).
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3.3.4 Adding an Impedance between two Existing Buses
Let us assume that we add an impedance Zb between two existing buses k and j as shown in Fig. 3.9. Therefore the current injected into the network from the bus k side will be Ik − Ib instead of Ik. Similarly the current injected into the network from the bus j side will be Ij + Ib instead of Ij. Consequently the voltage of the ith bus will be
( ) ( )( ) bikijninkikjijii
ninbkikbjijiii
IZZIZIZIZIZIZ
IZIIZIIZIZIZV
−+++++++=
++−+++++=
LL
LL
2211
2211 (3.36)
Similarly we have
( ) bjkjjnjnkjkjjjjjj IZZIZIZIZIZIZV −+++++++= LL2211 (3.37) and
( ) bkkkjnknkkkjkjkkk IZZIZIZIZIZIZV −+++++++= LL2211 (3.38)
Fig. 3.9 An impedance is added between two existing buses.
We shall now have to eliminate Ib from the above equations. To do that we note from Fig. 3.9 that
jkbbbbjk VVIZIZVV +−=⇒=− 0 (3.39) Substituting (3.37) and (3.38) in (3.39) we get
( ) ( ) ( )( ) ( ) bkkjkjjnknjn
kkkjkjkjjjkjbb
IZZZIZZ
IZZIZZIZZIZ
+−+−+
+−+−++−+=
20 111 LL
(3.40)
We can then write the voltage current relations as
=
−−−
−
=
p
nnew
p
n
bbknjnkj
knjn
orig
kj
n
II
I
Z
II
I
ZZZZZZZ
ZZZ
V
VMM
L
MM
11
11
111
0
(3.41)
where
1.59
kkjkjjbbb ZZZZZ +−+= 2 (3.42) We can now eliminate the last row and last column using the Kron’s reduction given in (3.21). 3.3.5 Direct Determination of Zbus Matrix
We shall now use the methods given in Sections 3.3.1 to 3.3.4 for the direct determination of the Zbus matrix without forming the Ybus matrix first. To accomplish this we shall consider the system of Fig. 3.2 and shall use the system data given in Example 3.1. Note that for the construction of the Zbus matrix we first eliminate all the voltage sources from the system.
Step-1: Start with bus-1. Assume that no other buses or lines exist in the system. We add this bus to the reference bus with the impedance of j0.25 per unit. Then the Zbus matrix is
25.01, jZbus = (3.43)
Step-2: We now add bus-2 to the reference bus using (3.31). The system impedance diagram is shown in Fig. 3.10. We then can modify (3.43) as
=
25.00025.0
2, jj
Zbus (3.44)
Fig. 3.10 Network of step-2.
Step-3: We now add an impedance of j0.2 per unit between buses 1 and 2 as shown in Fig. 3.11. The interim Zbus matrix is then obtained by applying (3.41) on (3.44) as
−−=
7.025.025.025.025.00
25.0025.0
3,
jjjjj
jjZ in
bus
Eliminating the last row and last column using the Kron’s reduction of (3.31) we get
=
1607.00893.00893.01607.0
3, jjjj
Zbus (3.45)
Step-4: We now add bus-3 to bus-1 through an impedance of j0.25 per unit as shown
in Fig. 3.12. The application of (3.35) on (3.45) will then result in the following matrix
1.60
Fig. 3.11 Network of step-3.
=
4107.00893.01607.00893.01607.00893.01607.00893.01607.0
4,
jjjjjjjjj
Zbus (3.46)
Fig. 3.12 Network of step-4.
Step-5: Connect buses 2 and 3 through an impedance of j0.4 per unit as shown in Fig. 3.13. The interim Zbus matrix is then formed from (3.41) and (3.46) as
−−−
−
=
7928.03214.00714.00714.03214.04107.00893.01607.0
0714.00893.01607.00893.00714.01607.00893.01607.0
5,
jjjjjjjj
jjjjjjjj
Z inbus
Fig. 3.13 Network of step-5. Using the Kron’s reduction we get the following matrix
=
0.28040.11820.13180.11821543.00.09570.13180.09571543.0
5,
jjjjjjjjj
Zbus (3.47)
1.61
Step-6: We now add a new bus-4 to bus-2 through an impedance of j0.5 as shown in Fig. 3.14. Then the application of (3.35) on (3.47) results in the following matrix
=
0.65430.11821543.00.09570.11820.28040.11820.1318
1543.00.11821543.00.09570.09570.13180.09571543.0
6,
jjjjjjjjjjjjjjjj
Zbus (3.48)
Fig. 3.14 Network of step-6.
Step-7: Finally we add buses 3 and 4 through an impedance of j0.4 to obtain the network of Fig. 3.3 minus the voltage sources. The application of (3.41) on (3.48) results in the interim Zbus matrix of
−
−=
09821536001622.00360003600536000.65430.11821543.00.09571622.00.11820.28040.11820.1318036001543.00.11821543.00.0957
036000.09570.13180.09571543.0
7,
.j.jj.j.j.jjjjj
jjjjj.jjjjj
.jjjjj
Z inbus
Eliminating the 5th row and column through Kron’s reduction we get the final Zbus as
=
3926019740136701133.01974025650123601264013670j123601531009690
j0.1133126400969015310
7,
.j.j.jj
.j.j.j.j..j.j.
.j.j.j
Zbus (3.49)
The Zbus matrix given in (3.49) is the as that given in Example 3.1 which is obtained by inverting the Ybus matrix.
3.4 THEVENIN IMPEDANCE AND Zbus MATRIX
To establish relationships between the elements of the Zbus matrix and Thevenin equivalent, let us consider the following example.
Example 3.4: Consider the two bus power system shown in Fig. 3.15. It can be seen that the open-circuit voltages of buses a and b are Va and Vb respectively. From (3.11) we can write the Ybus matrix of the system as
1.62
Fig. 3.15 Two-bus power system of Example 3.4.
+−
−+
=
+−
−+=
bbab
bbab
ab
ababaa
abaa
bbabab
ababaabus
ZZZZ
Z
ZZZZZ
ZZZ
ZZZY 1
1
111
111
The determinant of the above matrix is
bbabaa
bbabaabus ZZZ
ZZZY ++=
Therefore the Zbus matrix is
+
+
== −
abaa
abaa
ab
abbbab
bbab
busbusbus
ZZZZ
Z
ZZZZZ
YYZ 1
111
Solving the last two equations we get
( )
( )
+++
++
+++++
=
=
bbabaa
abaabb
bbabaa
bbaa
bbabaa
bbaa
bbabaa
bbabaa
bus
ZZZZZZ
ZZZZZ
ZZZZZ
ZZZZZZ
ZZZZ
Z2212
1211 (3.50)
Now consider the system of Fig. 3.15. The Thevenin impedance of looking into the
system at bus-a is the parallel combination of Zaa and Zab + Zbb, i.e.,
( )11, Z
ZZZZZZZ
bbabaa
bbabaaath =
+++
= (3.51)
Similarly the Thevenin impedance obtained by looking into the system at bus-b is the parallel combination of Zbb and Zaa + Zab, i.e.,
( )22, Z
ZZZZZZZ
bbabaa
abaabbbth =
+++
= (3.52)
Hence the driving point impedances of the two buses are their Thevenin impedances.
1.63
Let us now consider the Thevenin impedance while looking at the system between the buses a and b. From Fig. 3.15 it is evident that this Thevenin impedance is the parallel combination of Zab and Zaa + Zbb, i.e.,
( )bbabaa
bbaaababth ZZZ
ZZZZ+++
=,
With the values given in (3.50) we can write
( ) ( )
[ ]bbababaabbabaa
bbabaa
bbaa
bbabaa
abaabb
bbabaa
bbabaa
ZZZZZZZ
ZZZZZ
ZZZZZZ
ZZZZZZZZZ
+++
=
++−
+++
++++
=−+
1
22 122211
Comparing the last two equations we can write
122211, 2ZZZZ abth −+= (3.53) ∆∆∆
As we have seen in the above example in the relation V = ZbusI, the node or bus
voltages Vi, i = 1, …, n are the open circuit voltages. Let us assume that the currents injected in buses 1, …, k − 1 and k + 1, …, n are zero when a short circuit occurs at bus k. Then Thevenin impedance at bus k is
kkk
kkth Z
IVZ ==, (3.54)
From (3.51), (3.52) and (3.54) we can surmise that the driving point impedance at each bus is the Thevenin impedance.
Let us now find the Thevenin impedance between two buses j and k of a power system. Let the open circuit voltages be defined by the voltage vector V° and corresponding currents be defined by I° such that
°=° IZV bus (3.55) Now suppose the currents are changed by ∆I such that the voltages are changed by ∆V. Then
( )IIZVVV bus ∆+°=∆+°= (3.56) Comparing (3.55) and (3.56) we can write
IZV bus∆=∆ (3.57)
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Let us now assume that additional currents ∆Ik and ∆Ij are injected at the buses k and j respectively while the currents injected at the other buses remain the same. Then from (3.57) we can write
∆+∆
∆+∆∆+∆
∆+∆
=
∆∆
=∆
knkjnj
kkkjkj
kjkjjj
kkjj
k
jbus
IZIZ
IZIZIZIZ
IZIZ
II
ZV
M
M
M
M
11
0
0
(3.58)
We can therefore write the following two equations form (3.58)
kjkjjjojj
ojj IZIZVVVV ∆+∆+=∆+=
kkkjkj
okk
okk IZIZVVVV ∆+∆+=∆+=
The above two equations can be rewritten as
( ) ( )kjjkjjkjjojj IIZIZZVV ∆+∆+∆−+= (3.59)
( ) ( ) kkjkkkjkj
okk IZZIIZVV ∆−+∆+∆+= (3.60)
Since Zjk = Zkj, the network can be drawn as shown in Fig. 3.16. By inspection we can
see that the open circuit voltage between the buses k and j is
0, j
okkjoc VVV −= (3.61)
and the short circuit current through these two buses is
kjkjsc III ∆−=∆=, (3.62) Also during the short circuit Vk − Vj = 0. Therefore combining (3.59) and (3.60) we get
( ) ( ) 02 , =−−+−=− kjsckkjjkjoj
okjk IZZZVVVV (3.63)
Combining (3.61) to (3.63) we find the Thevenin impedance between the buses k and j as
kjkkjjkjsc
kjockjth ZZZ
IV
Z 2,
,, −+== (3.64)
The above equation agrees with our earlier derivation of the two bus network given in (3.53).
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Fig. 3.16 Thevenin equivalent between buses k and j.