Electrical Power Systems

Unbalanced ault Analysis 255
The boundary conditions at the fault point are
I a = 0 ...(10.29)
\ I a1 + I a2 + I a0 = 0 ...(10.30)
Vb = Vc = (Ib + Ic) Zf = 3Zf Ia0 The symmetrical components of voltages are given by
éVa1 ù
ê ú
ê
Va2
ú
êëVú a0 û
rom eqns. (10.32), we get,
= 1
3
é 2
1 b b ù é Va
ù
ê ú
2 ê ú
ê1b b ú
ê
Vb
ú
ê ú
ê1 1 1 ê
ë ú V c = Vú
b
û
ë û
...(10.31)
...(10.32)
V a1 = V a2 = 1
3 [V a + (b + b2 ) V b ] ...(10.33)
Va0 = 1
3 (Va + 2Vb) ...(10.34)
Using eqns. (10.33) and (10.34), we get,
V a0 – V a1 = 1
3 (2 – b – b2 ) V b = 3 Z f I a0
\ V a0 = V a1 + 3 Z f I a0 ...(10.35)
rom eqns. (10.33), (10.35) and eqn. (10.30) we can draw the connection of sequence nework
as shown in ig. 10.6.
ig. 10.6: Sequence network connection for double-line-to-ground fault.
rom ig. 10.6, we can write
Also,
I a1 =
I a2 =
Z
1
+ ( + 3 )
Ea
Z2 Z0 Zf
( Z + Z + 3 Z )
–(– Ea Z1Ia1) Z
2
2 0 f
...(10.36)
...(10.37)