December 2012
Beginner
528 pages
17h 31m
English
Content preview from Introduction to Electrical Power and Power Electronics
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270 Introduction to Electrical Power and Power Electronics
Solution:
Using Equation 10.14, we have
kVAR
3ph
new
200
2400
3300
50
60
0
4408
2
=
⋅
= . .
Therefore,kVAR
3ph
new
=×=0 4408 200 88
16..
.
The kVAR capacity is reduced to 0.4408 pu or 44.08% primarily due to significant
reduction in the voltage.
The capacitor improves the pf as explained in Figure 10.5. Without the capacitor
in Figure 10.5a, the load kilowatt and kVAR both are supplied by the source. With
the capacitor in Figure 10.5b, the capacitor kVAR leads the voltage by 90°, whereas
inductive load kVARs lag the voltage by 90°. Thus, they have opposite polarity:
when one is going away from ...
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