Electrical Power Systems

Characteristics and Performance of Transmission Lines 135
\ V S = (1 – 0.0123 + j0.00264) × 38.104 0° +
( 53. 809 77. 9°´ 218. 7 - 36. 87°
)
1000
\ VS = (0.9877 + j0.00264) × 38.104 0° + 11.76 4103 . °
\ V S = 37.63 + j0.1 + 8.87 + j7.72
\ VS = (46.5 + j7.82) = 47. 15 9. 54°
kV
\ VS(L – L) = 3 × 47. 15 9. 54°
kV = 8166 . 9. 54°
kV
| VS|
-|
VR|
Voltage regulation VR = || A
=
| V |
R
HG
8166 .
-
0. 9877 66
66
I KJ
= 25.26%
Power loss per phase = |I| 2 R = (218.7) 2 × 11.25 × 10 –6 MW = 0.538 MW
Per phase receiving end power
P R = 20
3 MW
Per phase sending end power
Transmission efficiency
P S = 20
h =
3
20 3
7. 204
+ 0. 538 = 7.204 MW
= 92.54%.
Example 6.4: Determine the voltage, current and power factor at the sending end of a 3 phase,
50 Hz, overhead transmission line 160 km long delivering a load of 100 MVA at 0.8 pf lagging
and 132 kV to a balanced load. Resistance per km is 0.16 W, inductance per km is 1.2 mH and
capacitance per km per conductor is 0.0082 m. Use nominal p method.
Solution:
R = 0.16 × 160 = 25.6 W
X = 1.2 × 10 –3 × 2p × 50 × 160 = 60.3 W.
Y = j2p × 50 × 0.0082 × 10 –6 × 160 = j4.12 × 10 –4 mho
rom eqn. (6.15), V S = 1
Phase voltage at the receiving end,
Z = R + jX = 25.6 + j60.3 = 65.51 67° W.
ZY
+ HG I VR + ZI
2 KJ
R
V R = 132
3
0° kV = 76.21 0° kV