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2007-03-12 22:06:57 · 4 answers · asked by bhagwant pareva 1 in Science & Mathematics Engineering

4 answers

Its the relationship between current and voltage in an electrical circuit. In normal resistive loads the power factor is 1 which means voltage is in phase with current, in motor loads and similiar loads the power factor can drop to say .65 and that is not good. you end up using more power and getting less good work from it. They put power factor correcting devices on the system to boots the power factor to > .9 or so.

2007-03-12 22:12:46 · answer #1 · answered by Anonymous · 0 0

The power factor of an AC electric power system is defined as the ratio of the real power to the apparent power, and is a number between 0 and 1. Real power is the capacity of the circuit for performing work in a particular time. Apparent power is the product of the current and voltage of the circuit. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power is equal to or greater than the real power. Low-power-factor loads increase losses in a power distribution system and result in increased energy costs.
Power factor=True Power (kW)/Apparent Power (kVA)


More the power factor better the system,because using phasor diagram it may be explained that as power P=V*I*Cos(phi),as power factor goes low high cos(phi) goes higher....so the real power obtained is more,and as Q=V*I*Sin(phi),that is the reactive power,goes lower.....so it is better to keep power factor as high as possible,otherwise more reactive power and less active power will cost higher tariff rate.
0.8 and more power factor is considered as good one.


For problems in India it is really too much to discuss here u may visit
http://www.indiacore.com/power4.html

2007-03-12 23:53:22 · answer #2 · answered by moon c 2 · 0 0

The answers provided are satisfactory. I just want to add the following:

For a resistive load the pf is 1. e.g. heaters.

For inductive loads the pf is lagging e.g. induction motors.

For capacitive loads the pf is leading e.g. capacitor bank.

The problem arises as in industry almost all drives are induction motors resulting in lower lagging pf. The solution is to connect capacitor banks either in individual motor terminal of at load centre. The capacitor can either be fixed type or switching type. In switching type the capacitor bank is either cut-in or cut-out depending on the pf of the system.

A lower pf means unnecessary loading the electrical power system without getting the real power for work. As a penalty the utility companys charge extra for reactive power consumption.

2007-03-14 21:16:19 · answer #3 · answered by Bishu 3 · 0 0

good technique and education. India under develop country and it have lake of good technique problem or lake of education level

2007-03-12 22:17:47 · answer #4 · answered by ruchi s 3 · 0 0

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