I am not sure ?About power factor??

Question

By improving the power factor,can it really reduce the current flowing through the meter which decides our electrical bill, or doing the job of saving energy?

Answer 1

No. Unless you are a commercial user who is getting charged a premium for poor power factor.

A poor power factor DOES cause you to draw more current, but you get charged by the watt, which by definition is the power consumed. When your power factor is poor, you are constantly borrowing extra current but instantly returning it rather than burning it in wattage. But that extra current does have to flow from the source through distribution to your meter, and will incur additional and unnecessary losses which the utility company gets stuck with. Normally it's a very small amount, not worth messing with, but in some factories or other customers that run a lot of power at low power factor, such as somewhere where there are a lot of motors running, the utility company might draw the limit and charge extra to cover the loss they incur in supplying you out-of-phase current.

But no, just because you are reducing the current thru the meter does not mean your bill will be lower, because you only get charged for the power you burn, which is only the amount of in-phase current times the voltage.

[REPOST] The common electro-mechanical watt-hour meter found in most residences is a disc motor supplied by 2 signals, one a voltage proportional to the current, the other a voltage proportional to the voltage. There is also a braking magnet. At any time, the speed of the motor is proportioanl to the instantaneous product of the two. If you are drawing 100A at 100V, PF =1, it will spin at the product of the two, or 10kW. But if the PF is lower, the 2 signals will not be in phase, thus the product of the two will be lower, even if there's still actually 100A flowing. In fact, if the PF is 0, indicating a purely reactive load, the meter wil not spin at all. The current will be offset from the voltage by 90 degrees, and the instantaneous product of the two averaged over one cycle will be zero.

And this is rightly so. You are not consumming power at all, even though current is flowing. And it wouldn't be right for the utility to charge for power you don't consume. But it is fair that, if a heavy customer is drawing a large reactive load that causes losses during distribution, that that customer pay for the additional losses, which would be based on some fraction of the RMS current flow into the facility. But they are called kilowatt-hour meters for a reason - they measure kilowatt-hours, not kVA or kVAR,

Though I did build such meters (kWhr, kVAhr,, kVAR, etc) at a place I used to work, now I work at a place that takes it even further. We build both large industrial battery chargers, as for electric forklifts, and we also build electronic loads. The loads take a DC input, such as from the output of a battery charger being tested, and convert it to an AC current that is pumped back into the 3-phase utility line. This current is 180 degrees from the voltage, or what you might call a PF of -1. The result is that, when charger and load are operating at 90% effiiency, a typical value for both, the charger will be drawing 50kW, outputting 45kw into the load, and the load will pump 40 of those kilowatts back onto the line, indicating a total draw of only 10kW. When I first started working there, I tested this, and it does work as advertised, including at the meter. I'm a "show me" sort of person, and 20 years ago, when arguments like this came up about the way some electronic device or circuit worked, I would say, "ask the machine", because it never lies. If you ask the machine, you will see that it truly is a kilowatt-hour meter, not a kVA-hr meter.

Answer 2

Improving power factor really does reduce the incoming current, but the watt-hour meter measures only real power, so its reading will be unaffected. Power companies use a penalty rate increase if the power factor (measured by other means) drops below a certain value, usually 0.9. If you achieve a leading power factor, the watt-hour meter will actually run backwards. This can result in severe penalties when discovered by the power company.

Any improvement in power factor will result in an overall energy savings by reducing line losses. One must take care, though, not to establish resonant circuits (unity power factor) anywhere in the distribution system. The resulting current flow between inductive and capacitive loads can be many times what the devices and wiring are rated to handle.

Answer 3

Heres how a pf correcting capacitor works...Lets say you only have one motor in the whole plant and its a 10hp...thats 7460watts..in the morning you come in and turn on the motor the in-rush onto the motor might hit say 12000watts now i didnt give a voltage on purpose because if you throw the primary of any system into a demand (the 12kw) the voltage will drop so what we have here is a function of wattage as voltage follows the heat...lower the voltage and the amperage goes up....now the power company comes to see you after 30 days they have a bill for the 30 days they say you used 12kwh for 8 hours...well you argue that you only used 7.4kwh and they show you in a graph where your plant spiked at 12kwh in the morning . What they have done is charged you for the demand....not what you used....the pf correcting capacitors will be applied across the lines and will hold the voltage constant as the motor is started thereby reducing demand and reducing your biiiiilllllllllll.... Thats good news...they have a payback of about 3 - 8 month s in savings and after that they are paying you ....thats good news...they are easy to install and are eco friendly...thats good news too...now do they stop flow through the meter ??? only for a while in this example I only have one motor but if you have 40 and they stop and start at irregular times and your total load is say 40 kw but the demand is 100kw ( because for 1 minute during the day all 40 started at once)..well i'm sure you can see were i'm going ...you meter spins like crazy and you pay for it..Get the pf corecting capacitors from any good electrical supplier like WESCO Dist.com. Have a great day with this and be careful with any electrical installation ..Do good Work...From the E...

Answer 4

Here's how to compute and prove that raising or correcting your power factor will decrease the current of your connected load thereby decreasing your consumption of power.
Example:
KW load=100
P.F. or cosΘ=0.8
V, voltage 3Φ=460
Formula: KW=√3*(460)*I*(0.8)
I,current= KW/1.732*(460)*(0.8)
I= 100,000/637.37= 156.89 amps
improve your power factor to 0.95
I= 100,000/1.732*(460)*(0.95)
= 100,000/756.88= 132.12 amps
Difference is:
156.89-132.12= 24.77 amps(your saving)
Note:
The kilowatt meter actually records the current flowing and the voltages supplied, so it's not really a kilowatt-hour meter,
but they won't call it KVA-hour meter either.

Answer 5

If you lie on a stack of bibles, nothing happens. superstition makes his argument compelling. reason makes it laughable.