Electrical Power Systems

58 Electrical Power Systems
rom ig. 3.6, we can write
V ab = V an – V bn
Also
V ac = V an – V cn
...(3.21)
...(3.22)
V bn = V an -120º ...(3.23)
V cn = V an -240º ...(3.24)
Adding eqns. (3.21) and (3.22) and substituting
Vab + Vac = 3Van rom eqns. (3.20) and (3.25), we have,
\
3qa
2pÎ
o
qa
2p Î
o
HG
HG
V bn = V an -120º and V cn = V an -240º , we have
I
KJ
I
KJ
Deq
ln = 3Van r
...(3.25)
Deq
ln = Van ...(3.26)
r
The capacitance per phase to neutral of the transposed transmission line is then given by
or
Can = qa
HG
2p Î
=
Van D
ln
r
HG
Can = 00242 .
eq
log D
r
I
KJ
o
eq
I
KJ
/m ...(3.27)
m/km ...(3.28)
or equilateral spacing, D 12 = D 23 = D 31 = D, and D eq = D. Therefore,
I
HG KJ
Can = 00242 .
log D
r
The line charging current for a three phase transmission line
m/km ...(3.29)
I (line charging) = jwC an V LN A/phase/km. ...(3.30)
3.5 BUNDLED CONDUCTORS
As mentioned in chapter-2 (section-2.13), the bundle usually comprises two, three or four
conductors. Geometric mean radius of the bundle conductor calculated earlier for the inductance