Electrical Power Systems

Sending end power factor angle = 8.18° – (–33.06°) = 41.24°
Sending end power factor = cos (41.24°) = 0.752
Characteristics and Performance of Transmission Lines 137
Example 6.5: A long transmission line delivers a load of 60 MVA at 124 kV, 50 Hz, at 0.8 power
factor lagging. Resistance of the line is 25.3 W, reactance is 66.5 W and admittance due to
charging capacitance is 0.442 × 10 –3 mho. ind (a) A, B, C, D constants (b) Sending end voltage,
current and power factor (c) regulation (d) efficiency of the line.
Solution:
R = 25.3 ohm, X = 66.5 ohm, Z = (25.3 + j66.5) ohm
Y = j0.442 × 10 –3 mho.
(a) g l = zy l = zl. yl = ZY
ZY = ( 25. 3 + j665 . )( j0.
442 ´ 10 )
= (. 0 0327 + j0.
174)
e j
\ A = D = cosh(g l) = cosh ZY
B = Z C sinh (g l) =
Z
Y
= cosh (0.03217 + j 0.174)
= 0.986 032 . °
HG
Z
Y
I sinh ZY
KJ e j
= (393 – j72.3)
\ B = 70.3 69. 2°
C = 1
Z C
sinh (g l) =
HG
Y
Z
I
–3
ZY
KJ sinhe j
= 4.44 × 10 –4 90° = j4.44 × 10 –4
(b) Load at 60 MVA at 124 kV (line-to-line)
\ Load current,
I R =
60 ´ 1000
Amp = 279.36 Amp
3 ´ 124
Power factor is 0.80 (lagging)
\ IR = 279.36 - 36. 87°
Amp
Now
V R = 124
3
V S = AV R + BI R
= 716 . kV (phase voltage)