Resistance and Inductance of Transmission Lines 47
Dbb¢ = Daa¢ = 3 m; Dab¢ = \
b
Dm = Daa¢ Dab¢ D ba¢ D bb ¢
\
L = 0.921 log
32 + 12 = 3.162 m = Dba¢
. g = b3 ´ 3162 g 1 4
1 2
= 3.08 m
FG 3.08 IJ mH/km = 1.375 mH/km H 0.099 K
Example 2.16: A three phase transmission line has an equilateral spacing of 6 m. It is desired to rebuild the line with same Deq and horizontal configuration so that the central conductor is midway between the outers. Find the spacing between the outer and central conductor. Solution: Using eqn. (2.59), we get L = 0.4605 log
FG D IJ mH/km H r¢ K
When conductors are placed in horizontal configuration (Fig. 2.26 (b)).
FG D IJ H r¢ K D = e2 d j .
Fig. 2.26 (=)
eq
L = 0.4605 log
eq
1 3 3
FG IJ d.2 J \ L = 0.4605 log G GH r ¢ JK mH/km. For both the cases, inductance is equal, therefore, FG IJ d.2 J FDI 0.4605 log G GH r ¢ JK = 0.4605 log GH r¢ JK 1 3
Fig. 2.26 (>)
1 3
\
d=
D 6 m = m = 4.762 m. 3 2 3 2
Example 2.17: Determine the inductance per km/phase of a double circuit three phase line as shown in Fig. 2.27. The radius of each conductor is 1.5 cm. Solution:
Fig. 2.27