how electricity consumption is reduced when power factor is improved ?

Question

this for domestic use of electricity

Answer 1

You will not reduce your electricity bill by improving your power factor.
Domestic consumers are billed on kWhrs NOT on kVAhrs......
However, there is a proviso......
I²R (watts) losses due to heating of wiring/equipment in your domestic situation will be billed. However, the extra reactive amps that cause heating losses in wiring & equipment will be so small as to be insignificant......
Source: Electrical Engineer (ret.) with 45+yrs. experience

Answer 2

Power factor can be mysterious to some. I'll try to simplify if possible. The things we plug in to our outlets can be very different electrically speaking. A fan, for example, has a motor with coils (or inductors) in it. This kind of load is "inductive". A computer with a switching power supply may be a "capacitive" type of load. An iron which has a resistive wire in it which gets hot is neither inductive or capacitive just purely resistive and this kind of a load is said to have a power factor of 1. The inductive loads and the capacitive loads have power factors less than 1, say .7 leading (capacitve load) or .7 lagging (inductive load). When the total power factor in your house is different from 1 you use more electricity because it's less efficient. Most of our loads are inductive so you could correct your inductive power factor by connecting capacitors acros the input cables to your house. Most homes don't lose too much power to power factor causes so this isn't usually practical (adding capacitors) but if you can get somebody to determine your power factor and add some capacitors it would be an interesting experiment. To the power companies power factor is important and if you look up near the top of certain power poles you will see a bank of square metal boxes hooked to the high voltage lines. These are capacitors and are being used to improve the power factor and hence improve power transmission. Hope this is clear enough.

Answer 3

I'm assuming you mean electrical energy or power when you say electricity. We can measure current, voltage, energy and power, but I'm not aware of any "electricity" meters. The power factor is the cosine of the phase angle shift between the current and voltage. The real power used in your house is P = VxIxpf. To deliver the same real power at the same voltage as power factor decreases, the current must increase. Even though this current increase doesn't perform real work, it contributes to I**2 x R loss in your wiring which becomes part of your real energy consumption.

Electric motors contribute to a lagging power factor because the magnetic field in the rotating armature must lag behind the rotating field in the stator in order for the stator to apply torque to the armature. I don't know power factors for typical houses, but I do know that such loads are variable since any large motors run intermittently, and that would make general pf correction impractical. Conceivably, a larger motor, on a heat pump for example, could be pf corrected with capacitors internal to the unit, but I don't know if that is done. If a motor was locally pf corrected with a capacitor, this portion of the current that is out of phase with the voltage would be passed back and forth between the motor windings and the capacitor and would not affect the remainder of the house hold wiring.

Power factor correction is more important in an industrial setting with many large motors and inductive loads. Failure correct the power factor would also reduce the capacity of a power transmission line because it is limited by current, not real power transmission.

Answer 4

The answer to your question is kind of a "Yes & No" scenario. Assuming your old PSU was working fine, a higher output power supply will have internal losses corresponding with it's size. However, the rest of the computer will draw exactly what it did with the previous power supply. When you first turn your PC on, all the HD's & DVD/CD drives spin up & the motors can suck upto 7x their normal current (power) requirements. Low rated PSU's often choke at this point, because they just can't maintain the inital inrush currents. A large power supply will ride this initial inrush out fine, because they are rated high enough Add up the components in your PC & I suspect you will be surprised if it even reaches 200W. For an approximate example see below: CPU 70w Fans (5W x 5) 25W {80mm fans are generally less than 1W, 120mm =7W} HD (2x25W) 50W {Seagate 200GB 7200RPM SATA state max 13.5W when seeking) Motherboard & RAM (2 x 10W) 20W Sound card 20W Miscellaneous 30W That would make a total of 215 Watts usuage. It'll suck that, no matter what PSU you have High End GPU's (such as video cards) can also suck a bit, possibly 100W extra, which is why people often need a better PSU in the first place.

Answer 5

With improved power factor the electrical appliances and machinery are at the peak of the performance efficiency therefore the production or consumption being optimum the electricity bill get reduces

Answer 6

If the power factor is not 1, then the voltage and current are not in phase. This leads to more current being drawn than is necessary for the power that is needed. Excess current means more losses in the power line leading to the appliance. Adjusting the power factor will cut the power line losses, without affecting the amount of power being used by the appliance.

Answer 7

thee r 2 components of an electric current..
Icosa and I sina.
where a= angle by which I lags voltage V.

Only icosa is the useful component that means only Icosa does useful work.
cosa is called the power factor of the circuit.
If cosa Increases less electricity will be required cuz Icosa will increase