Electrical Power Systems

252 Electrical Power Systems
or
I
2 Vb = b E Z Z I
Z I
a a
a - . I I 1 + b - .
2 + HG 3KJ HG 3KJ a
- 0.
3
...(10.8)
Using eqns (10.8) and (10.7), we get
2 2 2
f 2 0
I
HG KJ
Similarly
Vb = Ea 3 b Z + Z ( b – b) + Z ( b – 1)
( Z1 + Z2 + Z0) + 3Zf
...(10.9)
Vc = bVa1 + b2 \
Va2 + Va0 Vc = b E Z I
Z I
Z I
(
aIa a
a - . 2
I I 1 + b - .
2 + - 0.
HG 3KJ HG 3KJ HG 3KJ
...(10.10)
Using eqns (10.7) and (10.10), we get
\ Vc = Ea 2
3 bZf + Z2 ( b– b ) + Z0(
b–1)
Z + Z + Z + 3Z )
...(10.11)
10.3 LINE-TO-LINE AULT
1 2 0 f
igure 10.3 shows a three phase synchronous generator with a fault through an impedance Z f
between phase b and c. It is assumed that that generator is initially on no load condition.
ig. 10.3: Line-to-line fault between phase b and c.
The boundary conditions at the fault point are:
Vb – Vc = Zf .Ib ...(10.12)
Ib + Ic = 0 ...(10.13)
Ia = 0 ...(10.14)