Electrical Power Systems

24 Electrical Power Systems
= 0.921 log
If the two conductors are identical, i.e.,
HG
r 1 ¢ = r 2 ¢ = r¢, then
L = 0.921 log D
r¢
D
r¢ 1 r¢
2
I
KJ
mH/km
...(2.31)
I
HG KJ mH/km ...(2.32)
Eqn. (2.32) gives the inductance of a two wire line in which one conductor acts as a return
conductor for the other. This is known as loop inductance.
rom eqn. (2.29), the inductance of conductor 1 can be written as:
L 1 =
Similarly, the inductance of conductor 2,
1
D
0. 4605 log 0. 4605 log
r¢
1
+ I
HG KJ mH/km ...(2.33)
1
D
. log 0. 4605 log
r¢
1
+ I mH/km ...(2.34)
HG KJ
L 2 = 0 4605
As both the conductors are identical, hence we can write L 1 = L 2 = L. Therefore, inductance
per phase per km length of the line is given by
I mH/km ...(2.35)
HG KJ
1
D
L = 0. 4605 log 0. 4605 log
r¢
1
+
rom eqn. (2.35), it is clear that the first term is only a function of the fictious radius of the
conductor. The first term of eqn. (2.35) can be defined as the inductance due to both the internal
flux and that external to conductor 1 to a radius of 1 mt. The second term of eqn. (2.35) is
dependent only upon the conductor spacing and this is known as inductance spacing factor.
2.7 SEL AND MUTUAL INDUCTANCES
The inductance per phase for the single-phase two wire line (ig. 2.3) can also be expressed in
terms of self inductance of each conductor and their mutual inductances. Let us consider the
single phase circuit represented by two coils characterized by the self inductances L 11 and L 22
and the mutual inductance M 12.
igure 2.4 shows the single-phase line viewed as two
magnetically coupled coils and the magnetic polarity is
shown by dot symbols.
The flux linkages l 1 and l 2 can be written as:
l1 = L11I1 + M12I2 l2 = M21I1 + L22I2 Since I2 = –I1 , we get
l1 = (L11 – M12)I1 l2 = (–M21 + L22 )I2 ...(2.36)
...(2.37)
...(2.38)
...(2.39)
ig. 2.4: The single phase two wire
lines viewed as two magnetically
coupled coils.