Electrical Power Systems

=(2Dd 1 · 2Dd 1) 1/4 = (2 Dd 1) 1/2
13 /
1/2 1/2 1/2
Hence Deq = ( Dd2) × ( Dd2) ( 2Dd1)
Resistance and Inductance of Transmission Lines 31
o t
1 1
\ Deq = 2 6 D2× d × d
...(2.72)
1
3
2
1
6
1
It may be noted that Deq will remain same for the section-II and section-III of the transposition
cycle as the conductors of each parallel circuit rotate cyclically.
Equivalent self GMD Ds can be given as
Ds = (Dsa Dsb Dsc ) 1/3 ...(2.73)
where Dsa = Self GMD in section-I of phase a (i.e., conductors a and a¢) is
=(r¢ d 3 r¢ d 3 ) 1/4 = (r¢d 3 ) 1/2
Dsb = (r¢d1 r¢ d1 ) 1/4 = (r¢d1 ) 1/2
Dsc = (r¢d3 r¢d3) 1/4 = (r¢d3) 1/2
o t
/
/ / /
3 12
1 12
13
12
3
\ Ds = ( rd ¢ ) × ( rd ¢ ) × ( rd ¢ )
=(r¢) 1/2 · (d 1 ) 1/6 · d 3 1/3 ...(2.74)
D s also remains same in each transposition section because of the cyclic rotation of the
conductors of each parallel circuit over the transposition cycle.
The inductance per phase is
HG
HG
R
S|
T|
R
S|
T| HG
HG
L = 0.4605 log D
D
= 0.4605 log
eq
= 0.4605 log 2 16 /
= 0.4605 log
= 1
2
L
NM
s
I
KJ mH/km
/
216
12 / 13 /
D d2 16 /
d1
12 /
( r¢ )
16 /
d1 13 /
d3
×
2 13 /
12 / I × HG ¢ KJ D
r HG
× D
r¢
I
KJ ×HG I
d
d
I
KJ mH/km
13 /
2I
U
3KJ
V|
W|
23 12 /
/ I U
KJ V|
W|
HG
d
d
2
3
13 /
2 D
0. 4605 log 0. 4605 log
r¢
KJ +
mH/km
d
d
mH/km
2
3
I
KJ
23 /
O
Q
P mH/km
= 1
2 [Ls – M] ...(2.75)