Electrical Power Systems

72 Electrical Power Systems
Example 3.9: A three phase, 50 Hz overhead line has regularly transposed conductors are
horizontally spaced 4m apart. The capacitance of such line is 0.01 m/km. Recalculate the
capacitance per km to neutral when conductors are placed equilaterally spaced 4 m apart and
are regularly transposed.
Solution. When conductors are horizontally spaced, applying eqn. (3.28),
log D eqI
HG r KJ m/km.
b ´ ´
1
g3m = 5.04 m
Can = 00242 .
D eq = 4 4 8
C an = 0.01 m/km
504 . I \ log = HG r KJ 00242 .
= 242 .
001 .
\ r = 0.019 m
In the 2nd case conductors are placed equilaterally, therefore D eq = D.
\ C an =
I
HG KJ = I
HG KJ m/km
00242 .
00242 .
D
4
log log
r 0019 .
\ C an = 0.0104 m/km Ans.
Example 3.10: Calculate the capacitance to neutral of a
single phase line composed of four equal strands as
shown in ig. 3.18. The radius of each subconductor is
0.5 cm.
Solution.
Applying eqn. (3.10)
C1n = C2n = 00242 .
log D
I
HG DsKJ m/km.
1
1
4
b aa ab ac adg
e j
Ds = d d d d 4 = r. 2r. 2 2. r. 2r
\ D s = 1.834 r = 1.834 × 0.005 m
\ D s = 0.00917 m
\ C 1n = C 2n =
HG
00242 .
45 .
log
0. 00917
I KJ m/km.
ig. 3.18