Page 793 - Industrial Power Engineering and Applications Handbook

Power capacitors: behaviour, switching and improvement of power factor 231749

352
Actual, p.f., cos $2 = -
525
132133 kV
= 0.67
.. $2 = 47.9"
R = 1.95Q
Let us improve the apparent load p.f. to 0.98
i.e. cos q3 = 0.98
and $3 = 11.48'
From Example 24.3 and Figure 24.25(b) inductive reactance
of line = 6 Q per phase.
- suppress 5th harmonic Reactance of each transformer = 3.63 (2.
quantities
I L I I I 117
4 Total inductive reactance of the network = 13.26 (1 per phase
:. 13.26
27c f
line inductance L = ~
= 42.23 mH/phase
Since the p.f. is to be improved from 0.88 to 0.98
:. Shunt reactive compensation required
Ixc = 400 [sin 9, - sin $3]
= 400 [sin 28.36 - sin 11.481
1111111 h,
TTTTTTTl = 400 [0.475 - 0.1991
= 110.4 A

and kVAr required = & x 33 x 11 0.4
= 6310 kVAr
Say, 6300 kVAr
The shunt capacitors can be provided on the LT or the HT
Arrangement of capacitors in double star, 11 20 kVAr in each arm. side, whichever is more convenient. In the above case, since
it is large, the HT side of the load-end transformer will be
more convenient. The receiving-end transformer, however,
Figure 23.18(b) Network of Figure 24.25 shown with shunt will now be operating under more stringent conditions that
compensation must be taken in account or the capacitors may be provided
on the LT side to relieve this transformer also from excessive
currents. In Figure 23.18(b) we have considered them on the
LT side.
0.98
Improved actual p.f., cos = ~ x 0.67 = 0.75
0.88
4; = 41.4"
and improved actual current
-- 0.67 x 525
-
0.75
= 469 A (Figure 23.18(a))
and improved apparent line current = 400 x 0.88 = 359 A
0.98
kV' x 1000
and X, =
kVAr
- 33' x 1000
6300
= 172.86 R
:. c= 106 KF
Figure 23.18(c) Reducing the actual loading of line by tuning 27c x 50 x 172.86
for the fifth harmonic = 18.42 pF