Electrical Power Systems

\ V 0 = 110.15 KV(rms)
V n = 220
3
KV = 127 KV
\ Pc = 244
2 001 .
(50 + 25) (127 – 110.15)
0983 .
\ P c = 472.73 KW/phase
Total corona loss = 3 ´ 472.73 KW = 1418.19 KW.
1
I 2
´ 10 HG 5 KJ –5 ´ 200 KW/phase
Corona 369
Example 14.4: Calculate the disruptive critical voltage for a three phase line with conductors
of radius 1 cm and spaced symmelrically 4m apart.
Solution:
Using eqn. (14.31(b)), disruptive critical voltage
\
\ V 0 =
HG
6
3´ 10 Deq
V0 = r ln
2 r
volt/phase KJ
r = 1 cm = 0.01 m, Deq = 4 m
6
3´ 10 4 I
´ 0.01 ln
2 HG 001 . KJ
\ V 0 = 127.1 KV (line-to-neutral)
\line-to-line disruptive critical voltage
= 3 ´ 127.1 = 220.14 KV.
I
Example 14.5: A 220 KV three phase transmission line with conductors radius 1.3 cm is built
so that corona takes place if the line voltage exceeds 260 KV(rms). ind the spacing between
conductors.
Solution:
Disruptive critical voltage V rms = 260
rom eqn. (14.35)
HG
3
KV = 150.11 KV
6
3´ 10
Deq
V0 = rdm ln
0
2
r
volts/phase KJ
Assuming d = 1, and m0 = 1 (smooth conductor),
I
r = 0.013 m, V 0 = 150.11KV = 150.11 ´ 10 3 volts
\ = 150.11 ´ 10 3 =
3´ 10
2
6
HG
D eq
´ 1 ´ 1 ´ 0.013 ´ ln
0. 013
I
KJ