Electrical Power Systems

om eqn. (6.6),
Characteristics and Performance of Transmission Lines 131
|V S| » |V R| + |I|(R cos d R + X sin d R)
Here |V R | = 10.2 kV = 10200 Volt
cosd R = 0.8, sind R = 0.6
\ |V S| = 10200 + 551.47 (0.39 × 0.8 + 4.52 × 0.6)
\ |V S | = 11.867 kV
( 11867 . - 10. 2)
\ Voltage regulation = ´ 100 = 16.34%
10. 2
(b) Voltage regulation desired = 0.60 × 16.34 = 9.804%
Therefore, under this condition we can write
| VS| - 10. 2
= 0.09804
10. 2
\ |V S| = 11.2 kV
igure 6.6 shows the equivalent circuit of the
line with a capacitor in parallel with the load.
Assuming combined power factor of the load and
capacitor = cos d R¢
By using eqn. (6.6), we can write,
(11.2 – 10.2) × 10 –3 = |I R | (R cos d R ¢ + X sin d R ¢) ...(i)
Since the capacitance does not draw any real power, we have,
|I R| =
rom eqns. (i) and (ii), we get
4.52 tan d R ¢ = 1.876
\ tan d R¢ = 0.415
\ d R ¢ = 22.5°
4500
\ cos d R ¢ = 0.9238
\ |I R| = 477.56 Amp.
Now I C = I R – I,
¢ ...(ii)
10. 2 cosdR IR = 477. 56 - 22. 5°
= 441.2 – j182.75
I = 551. 47 - 36. 87°
= 441.2 – j330.88
\ I C = 441.2 – j182.75 – 441.2 + j330.88
\ I C = j148.13 Amp.
1 VR
10. 2 ´ 1000
Now XC =
= =
2p´ 60 ´ C I 148. 13
C
ig. 6.6