Alternating current theory and electrical machines
203
For (d):
P VI
P
cos (W)
150 0.f.84 1840 and 230V.8. 0.4.8:
I
1840
230
W
V 0. 0. the p. a
result, the supply companies encourage installation engineers improve their power factor
to value close and sometimes charge penalties the power factor falls below 0.9(a) shows industrial load with low power factor.4
would require 20A cable, while the same load unity power factor could supplied with
an cable.
It can seen from these calculations that 1.
.6 900 W
φ
∴
The power factor most industrial loads lagging because the machines
and discharge lighting used industry are mostly inductive. capaci-
tor connected parallel with the load, the capacitor current leads
the applied voltage 90°. When this capacitor current added the
load current shown Fig 10. There may also the problem higher voltage drops the supply cables.
Example 3
A 230V supply feeds three 1.f. then:
I
1840
230 1
8
W
V
A
For p. 10.8 and 0. However, using slightly bigger capacitor, the load
current can pushed until ‘in phase’ with the voltage can be
seen Fig.84kW loads with power factors 0.9(b) the resultant load current has much
improved power factor. Calculate the
current each power factor.
Figure 10. capacitor has the opposite effect inductor, and
so seems reasonable add capacitor load which known have
a lower bad power factor, for example, motor.
The current given by:
I
P
V
cos φ
where 1. This causes
an additional magnetizing current drawn from the supply, which
does not produce power, but does need supplied, making supply
cables larger.
POWER FACTOR CORRECTION
Most installations have low bad power factor because the inductive
nature the load.f.4:
I
1840
230 4
20
W
V
A
.8
10A
For p.84 load supplied power factor 0.9(c)
