Electrical Power Systems

368 Electrical Power Systems
\ V 0 = 2.11 × 10 6 × 0.85 × 0.015 × 0.967 × ln
\ V 0 = 150.06 KV.
or local corona, mv = 0.72,
We know, visual critical voltage (rms)
HG
00301 .
Vv = G0mvrd 1 +
rd
I
KJ
HG
eq
ln D
r
\ V0 = 2.11 ´ 10 6 ´ 0.72 ´ 0.015 ´ 0.967 ´ 1 +
\
V v = 158.87 KV
or general corona, m v = 0.82
\ Vv = 158.87 ´ 082 .
I
KJ
KV = 180.93 KV
072 .
HG
48 . I volts HG 0. 015KJ
00301 .
0. 015 ´ 0. 967
I KJ
48 . I ´ lnHG
KJ
0. 015
Actual operating voltage to neutral = 220
3 = 127 KV, which is less than Vv and there is no
corona.
Example 14.3: Determine the corona loss of a three phase, 220 KV, 50 Hz and 200 Km long
transmission line of three conductors each of radius 1 cm and spaced 5 m apart in an equilateral
triangle formation. The air temperature is 30ºC and the atmospheric pressure is 760 mm of Hg.
The irregularity factor is 0.85.
Solution:
rom eqn. (14.39),
Pc = 244
d (f + 25) (Vn –V0) 2
HG
r
D
I
KJ ´ 10 –5 KW/Km/phase
b g
0. 392p
0. 392 ´ 760
f = 50 Hz, d = =
t + 273 273 + 30 = 0.983
r = 1 cm = 0.01 m, D = 5 m.
V 0 =
\ V 0 =
3´ 10
2
3´ 10
2
6
6
´ r ´ d ´ m 0 ´
I
HG KJ
D
ln
r
´ 0.01 ´ 0.983 ´ 0.85 ´ ln 5
001 .
I
HG KJ volts