Electrical Power Systems

34 Electrical Power Systems
HG
HG
ln = 0.4605 Ia
1 1 1
log + I b log + I c log
D D D
1 1 1
\ ln = 0.4605 Ia log + Ib log + Ic
log
6 4 2
an
bn
cn
I
I KJ mWb–T/km
KJ mWb–T/km
\ l n = – 0.4605(0.778 I a + 0.602 I b + 0.301 I c ) mWb–T/km
Substituting the values of I a, I b and I c, we get
l n = –0.4605{–23.34 + j 18.672 – 12.04 + j15.652 + 15.05 – j 15.05} mWb–T/km
\ l n = –0.4605{–20.33 + j 19.274} mWb–T/km = 0.4605 (20.33 – j 19.274)mWb–T/km
(b) The voltage induced in the neutral wire is
V n = jwl n × 30 = j 2p × 50 × 30 × 0.4605 × 10 –3 (20.33 – j 19.274) volts
= 121.58 – 435 . º volts.
(c) rom eqn. (2.60), the flux linkages of the conductor a are
la = 0.4605 I I
r D I
1 1 1
a log b log c log
¢ 2D
+ + UVW mWb–T/km
Substituting I c = –(I a + I b )
RST
RST
1
log
r¢
1
log
D
I log( 2D) I log( 2D)
+ + +
2D
a log
r¢
I b log 2 + UVW mWb–T/km
\ la = 0.4605 I a I b a b
\ l a = 0.4605 I
Similarly,
RST
RST
D I
1 1 1
log log log
r¢
D
+ UVW mWb–T/km
lb = 0.4605 I a + b I c
D I
1 1
log log I log D I log D
r¢
+ +
\ lb = 0.4605 I a + b a b
\ l b = 0.4605. I b log D
r¢ mWb–T/km
RST
RST
lc = 0.4605 I
D I
D I
1 1 1
a log + b log + c log
2
r¢
\ lc = 0.4605 I D I D I
D I
1 1UVW
b log 2 + c log 2 + b log + c log
r¢
mWb–T/km
RST
UVW
UVW
UVW