Electrical Power Systems

Resistance and Inductance of Transmission Lines 35
2DUVW
\ lc = 0.4605 Ib log 2 + Ic
log
r¢
mWb–T/km
Therefore,
RST
-3
HG
DVa = 0.4605 × jw´ 10 I
I KJ ´
D
I
r¢
+
2
log log 230
\ DVa = j 4.34 I
G
4
log . . + I
7788 ´ 75
100
log 2
H
a b volts.
I
KJ
a b volts.
= j 4.34 (2.835 I a + 0.301 I b) volts.
\ DV a = j 4.34 {2.835 (–30 + j24) + 0.301(–20 + j26)}
\ DV a = j 4.34 {–85.05 –6.02 + j68.04 + j7.826}
= j 4.34 {–91.07 + j75.866}
= (–329.25 – j395.24)
= –(329.25 + j395.24) volts.
= 514.4 230. 2ºvolts.
DV b = jw × 30 × 0.4605 × 10 –3 (–20 + j26) × log
\ DV b = j11 (–20 + j26) volts.
= 11 (–26 – j20) = 360.8 217. 56º
volts.
DVc = jw × 30 × 0.4605 × 10 –3 2D
log 2.
Ib + Ic
log
r¢
R
S|
T|
4
\ = j4.34 0. 301( - 20 + j26) + ( 50 - j50)log
. 7788 ´ . 75
100
= j4.34 {–6.02 + j 7.826 + 141.77 – j 141.77}
= j4.34 {135.76 – j 133.944}
= (581.31 + j 589.2} = 827.7 45. 4ºvolts.
HG
2
. 7788 .
volts.
´ 75
100
Example 2.3: A single-phase 50 Hz power line is supported on a horizontal cross-arm. The
spacing between the conductors is 4 mt. A telephone line is supported symmetrically below the
power line as shown in ig. 2.14. ind the mutual inductance between the two circuits and the
voltage induced per km in the telephone line if the current in the power line is 120 amp.
I KJ
U
V|
W|