Electrical Power Systems

Since from symmetry D b1= D b2, l 12(I b) = 0.
Resistance and Inductance of Transmission Lines 45
Total flux linkages between conductors T 1 and T 2 due to all currents
l12 = l12(Ia) + l12(Ib) = 0.4605 Ia log Da2
HG
D
a1
I
KJ mWb–T/km
6. 708I
\ M12 = 0.4605 log mH/km = 0.05877 mH/km
HG 5 KJ
\ Magnitude of the voltage drop in the telephone line
|V rms| = w · M 12 · I a = 2p × 50 × 0.05877 × 10 –3 × 200 Volt/km
|V rms| = 3.692 Volt/km.
Example 2.13: Determine the self inductance, mutual inductance and the inductance per phase
of a three phase double circuit transposed transmission line as shown in ig. 2.10. Radius of
each conductor is 1.266 cm and D = 5 m, d 1 = 4 m, d 2 = 6.403 m and d 3 = 10.77 m.
Solution: Using eqn. (2.72)
D eq = 2 1/6 · D 1/2 · d 2 1/3 · d1 1/6 = 2 1/6 · (5) 1/2 (6.403) 1/3 (4) 1/6 m
\ D eq = 1.1222 × 2.236 × 1.857 × 1.26 = 5.871 m
Using eqn (2.74)
Ds = (r¢) 1/2 · (d1) 1/6 · (d3) 1/3 = 0. 7788
\ D s = 0.0992 × 1.26 × 2.208 = 0.276 m
Using eqn. (2.76), self inductance is
. log
L s = 0 4605
\ L s = 0.4605 log
Using eqn. (2.77), mutual inductance is
HG
2 13 /
D
HG
M = 0.4605 log d
d
r¢
I
KJ
63 .
0. 00985
23
3
2KJ
/
\ M = +0.0696 mH/km
Using eqn. (2.75)
\ L = 1
2
( L M)
HG
I
HG
S - = 0.6112 mH/km
1266 .
´
100
mH/km = 0.4605 log
12 / I ´ KJ
HG
I KJ mH/km = 1.292 mH/km
mH/km = 0.4605 log
16 / 13 /
( 4) ( 1077 . )
13 /
2 ´ 5
0. 7788 ´ 1266 .
100
10. 77I
HG 6. 403KJ
23 /
I
KJ