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If power factor comes in leading & showing -0.95, is there any effect on electrical system? If yes, then please tell me what would be the effects and how long we keep power factor in leading??

Keeping power factor in leading shows more KW & less KVA? Is it correct or not.

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Keeping power factor in leading shows more KW & less KVA? Is it correct or not.

This is incorrect. If the real power load is 10kW, at unity power factor the kVA is 10kVA. If PF = +/-0.95, on a real load power of 10kW, kVA is 10.52kVA.

Leading PF is caused by a net capacitive load and the effect of it is the same as a lagging PF (inductive load); the supply current is higher than for a truly resistive load.

You can keep PF leading (or lagging) as long as you want but it isn't an ideal situation because some electricity companies have a policy to charge you for use of reactive power in order to supplement the cost of putting bigger cables into their feeders to factories - higher current means bigger cables.

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Leading power factor means that the current leads the voltage, that is, the load is capacitive. If the load is inductive then the power factor is lagging and its sign is positive.

When calculating kW from kVA use absolute value of the power factor because it will be the same whether the PF is leading or lagging.

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The effect on power system of "leading" power factor is that there is more capacitive power. It depends whether you analyse a source (a generator) or a load. If the load has capacitive properties, it can be used to compensate inductive reactive power, however not with such big PF. Usually it is done by inserting capacitors (in parallel or series), and they have very low resistance relatively to their reactance and it is very close to 0.

Considering the generator, by changing its excitation current, it is also possible to make it work as a compensator. There are even special machines designed for compensation work.

Compensating (both by capacitors or compensators) allows to feed inductive loads (like motors) without transmission losses. We don't need to send inductive power to loads.

Answering the question how long: this is a steady state so it does not matter how long PF can be with such value. If load changes or topology of the network will be modified, this PF can change too.

While calculating active power you don't need to know what type of reactive power you take/get. The equation for active power is $$P = Re(S) = |S| \cos \phi $$ where $$\cos \phi = -0.95$$ is your power factor.

In this case, however, the fact that PF is negative means that active power is negative too. When you are considering a load, this means that the load outputs energy to the system, so it works as a generator. If you would consider a generator, the negative active power means it works as a load, so it takes energy from the system. This can happen in fact in one and only state, when the generator's turbine failed and rotation of the rotor is created by rotating magnetic field of the system (the generator works as a motor and it is rotating the turbine).

The apparent power will never be smaller than active neither reactive powers. Because these are complex numbers we can't say which one is larger. The minus sign on your meter means that the power factor, which is calculated as cosine of angle of current phasor related to voltage phasor is in some kind of opposition (so it flows in opposite direction), so the power is.

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  • \$\begingroup\$ Wouldn't phi be the angle between voltage and current, while the power factor is (cos phi)? \$\endgroup\$ – Pentium100 Jun 20 '13 at 11:31
  • \$\begingroup\$ Of course, you are right -- I corrected. Thank you, I was probably typing too fast. \$\endgroup\$ – Voitcus Jun 20 '13 at 11:40
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If by inadvertent mistake too many capacitors are placed in service on the distribution system during heavy load periods, not only will the distribution voltage rise to an intolerable level, but the total apparent power flow through the transformer[s] could exceed its [their] rating[s], as the excess reactive power supplied by the capacitors but not drawn by the load will flow out of the transformer to the supply side.

If by mistake the load is shed but the capacitors remain in service, the main problem to deal with will be high distribution system voltage, which can indeed rise to dangerous levels depending on the circumstances.

Transformers with on load tapchangers and automatic ditribution voltage regulating schemes may well go to bottom tap in an attempt to buck the voltage down into the normal range , and if while at bottom tap load is lost the voltage may rise unacceptably. Provided the facilities to do so exist, operators will routinely and pre-emptively remove capacitors from service in this situation.

Too much capacitance will raise the voltage. Back in the day when bulk correction was common, The voltage could be driven high enough to cause very early failure of the lighting in a plant on the weekends when the lighting was almost the only load.

Depending how the network operator charges for reactive power, the plant might be subject to unexpected power factor charges.

Highly capacitive loads are harder to interrupt than resistive loads. It might be possible switching device may not operate as expected. For example, a lineman might take a current reading at a 15 kV fused cutout before deciding whether to it is safe to use a loadbuster tool. Since the ammeter does not indicate power factor, he might not realize that the plant load was highly capacitive. One particular model of loadbuster can interrupt 900A of normal load, but only 120 amps to a capacitor bank.

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protected by Dave Tweed Aug 3 '14 at 15:18

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