Basic Electrical Installation Work

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Figure 10.9(a) shows industrial load with low power factor.84kW loads with power factors 0.9(b) the resultant load current has much improved power factor.84 1840 and 230V.f.4: I 1840 230 4 20 W V A .4 would require 20A cable, while the same load unity power factor could supplied with an cable.Alternating current theory and electrical machines 203 For (d): P VI P cos (W) 150 0. POWER FACTOR CORRECTION Most installations have low bad power factor because the inductive nature the load. capaci- tor connected parallel with the load, the capacitor current leads the applied voltage 90°. then: I 1840 230 1 8 W V A For p. 10.6 900 W φ ∴ The power factor most industrial loads lagging because the machines and discharge lighting used industry are mostly inductive. Calculate the current each power factor.8.4. However, using slightly bigger capacitor, the load current can pushed until ‘in phase’ with the voltage can be seen Fig. It can seen from these calculations that 1.8 10A For p. Example 3 A 230V supply feeds three 1. 0. This causes an additional magnetizing current drawn from the supply, which does not produce power, but does need supplied, making supply cables larger.84 load supplied power factor 0. capacitor has the opposite effect inductor, and so seems reasonable add capacitor load which known have a lower bad power factor, for example, motor. 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(c). The current given by: I P V cos φ where 1. 0. When this capacitor current added the load current shown Fig 10.8 and 0. the p. There may also the problem higher voltage drops the supply cables.f.8: I 1840 230 W V 0.f