Electrical Power Systems

or
L
V
1
V
M
NM
2
V
0
n
O
QP
=
L
old
ZBUS
M
dZni - Znji
Zb + Zii
+
dZi1 -Zj1i dZin -Zjni
d
L
Z - 2Z
NM
dZ1i - Z1jiO
dZ2i - Z2ji
jj ij
iQP
L
NM
I
I
I
1
2
M
n
I K
Symmetrical ault 221
O
QP
...(8.34)
Eliminating I k in equation (8.34) and following the same procedure for Type-2 modification,
we get,
new old
ZBUS = ZBUS
1
-
dZb + Zii + Zjj - 2 Ziji
L
dZ1i dZ2i - Z1ji
- Z2ji
dZni M
- Znji
NM
[(Z i1 – Z j1) (Z i2 – Z j2) ... (Z in – Z jn)] ... (8.35)
With the use of above mentioned four modifications bus impedance matrix can be formulated
by a step-by-step technique considering one branch at a time.
Example 8.19 ig. 8.26 shows a three bus network. Obtain impedance matrix Z BUS.
ig. 8.26: Three bus network.
Solution:
Step-1: Add branch Z1r = 0.50 (from new bus 1 to reference bus r)
\ ZBUS = [0.50] ... (i)
Step-2: Type-2 modification. That is add branch Z21 = 0.20 (from new bus 2 to old bus 1)
\ Z BUS = 1
2
L
NM
. .
. .
050 050
050 070
Step-3: Add branch Z 13 = 0.20 from new bus 3 to old bus 1. This is type-2 modification.
\ Z BUS =
L
NM
. . .
050
050 .
050
070 .
050
050 .
050 050 070
. . .
O
QP
O
QP
O
QP
×
...(ii)