Electrical Power Systems

Resistance and Inductance of Transmission Lines 23
These currents set up magnetic field lines that links between the conductors as shown in
ig. 2.3.
Inductance of conductor 1 due to internal flux
is given by eqn. (2.21). As a simplifying assumption
we can assume that all the external flux set up by
current in conductor 1 links all the current upto
the centre of conductor 2 and that the flux beyond
the centre of conductor 2 does not link any current.
This assumption gives quite accurate results
especially when D is much greater than r1 and r2. Thus, to obtain the inductance of conductor 1 due ig. 2.3: Single phase two wire lines.
to the external flux linkage, substituting D1 = r1 and D2 = D in eqn. (2.26).
HG
L1(ext) = 2 × 10 –7 ln D
r1 The total inductance of conductor 1 is then
\ L 1 = 1
2
where r 1¢ = re
1
L 1 = L int + L 1(ext)
I
KJ
-7 -7 D
ln
´ 10 + 2 ´ 10
HG
HG
HG
HG
-7
1
= 2 ´ 10 + ln
4
D
r
H/m ...(2.27)
1
I
KJ
= 2 × 10 –7 D
ln e4
+ ln
r
= 2 × 10 –7 ln
1
re
1
D
-14
I
I
KJ
1
r
1
I
KJ
H/m ...(2.28)
= 0.4605 log D
mH/km ...(2.29)
r1 ¢ KJ
- 1/ 4
= 0.7788 r1
The radius r 1¢ is the radius of a fictious conductor which has no internal inductance but has
the same total inductance as the actual conductor.
Similarly, the inductance of conductor 2 is
HG
L 2 = 0.4605 log D
The total inductance of the circuit is
L = L 1 + L 2
I
mH/km ...(2.30)
r2 ¢ KJ