Electrical Power Systems

Solution:
rom eqn. (14.38), air density factor,
d =
0. 392p
273 + t
Given p = 740 mm, t = 30ºC
b g
0. 392 ´ 740
\ d = = 0.957
273 + 30
rom eqn. (14.35), disruptive critical voltage,
6
3´ 10
V0 = rdm0 ln
2
D eqI
volts HG r KJ
Given that m0 = 0.92, d = 0.957, r = 1 cm = 0.01 mt, Deq = 6 m
\ V 0 =
3´ 10
2
6
´ 0.01 ´ 0.957 ´ 0.92 ´ ln 6
001 .
\ V0 = 119.475 KV (rms value)
rom eqn. (14.37), visual critical voltage
V v =
\ V v =
3´ 10
2
3´ 10
2
6
6
HG
mvrd 1 00301 .
+
rd
I
KJ
HG
ln D
r
I
HG KJ volts
eq
HG
´ 0.83 ´ 0.01 ´ 0.957 ´ 1 +
I
KJ volts
00301 .
001 . ´ 0957 .
I KJ
6
´ ln
001 .
Corona 367
I volts HG KJ
\ Vv = 140.95 KV (rms value).
Example 14.2: A three phase 220 KV, 50 Hz, 200 Km long transmjission line consists of three
stranded aluminium conductors spaced triangularly at 4.8 m. Radius of each conductor is 1.5
cm. The air temperature is 27ºC and pressure is 740 mm of Hg. If the breakdown strength of air
is 21.1 KV (rms) per cm and the surface factor is 0.85, compute disruptive critical voltage. Also,
determine the visual critical voltages for local and general corona if the irregularity factors are
0.72 and 0.80 for visual corona (local) and visual corona (general) respectively.
Solution:
Given parameters are:
r = 1.5 cm = 0.015 m, p = 740 mm, t = 27ºC
\ d =
0. 392p
273 + t
= 0 392 740 . ´
= 0.967.
273 + 27
D eq = 4.8 m, m 0 = 0.85, G 0 = 21.1 KV (rms)/cm = 2.11 ´ 10 6 v/m
We know, disruptive critical voltage (rms)
HG
V0 = G0m0rd ln Deq
r
I
KJ volts