3 phase kwhr meter

Because the effective power depends on the phase relationship between the voltage and current (the power factor). The closer to one your power factor is, the more work you get out of each kilowatt. It the cosine of the angle that determines the power factor. At 0°, the cosine = 1, and maximum effective power is achieved.

At least, that's what I remember about it from school.

Old eddy current watt-hour meters did not register VARS -- only Watts. The power company supplied the complex power, so they were taking a loss and the user was getting power but only paying for the real part (watts) of it.

Companies who used synchronous motors and other machinery with capacitive reactance to balance out the power factor got better rates from power companies for helping them with imbalances on their transmission lines.

These days, electronic watt-hour meters can measure and record watts and VARS, and the customer pays for it. Power companies still offer incentives to companies who try to maintain a P. F. of 1.

The previous answers are correct.

By changing the power factor, you are changing the calibration of the meter

The Meter needs to be calibrated (or matched) with the load. In other words, the phase relationship of the load current to the voltage (power factor). This relationship will change depending on what is on the load. Standard electrical loads will simply delay the current, while other loads such as fluorescent lights will significantly change the phase relationship.

This is why when you change all your light bulbs to energy saving (fluorescent) bulbs, it is best to get your power meter re-matched. If you don't, you may never see the savings (but the power company will)

If the power factor is not correct for the load, the loser depends on where the actual phase of the load is. In one case, the meter will read high, in the other case it will read low.

"This is why when you change all your light bulbs to energy saving (fluorescent) bulbs, it is best to get your power meter re-matched. If you don't, you may never see the savings (but the power company will)"

Elucidate?

Off the top of my head

(don't quote the math, as the actual math has to do with cosines etc)

If the total load of your house causes a phase shift of the current of +45 Degrees your power factor may be +75

• If the power meter is set at +75 the meter will read accurately the amount of power being used. (this is ideal)
• If the power meter is calibrated to +95, it will record less power than you are using
• If the power meter is calibrated at +50 it will record more power than you are using. (or the other way around)

This is typical with all the changing loads in a house,

When you change the load to a highly capacitive, instead of inductive, the actual load may be something like -45. In this case, the recorded power is no where near the actual usage. The power meter could record much higher readings than is actually being used and you would be paying the power company for that power. (on the other hand it could also read much lower).

If it is too low,you can be sure the power company will investigate, however, if it is too high it will not be investigated (unless you are doing something stupid like a grow-op)

I guess I should have been more specific - I was wondering about the difference in PF between an incandescent and a fluorescent light bulb.

The PF for an incandescent bulb will be close to 1.0, as the only self-inductance it may have is that of the element and its connections to its terminals.

The PF for a fluorescent will depend on whether the on-board choke and capacitor are correctly selected to achieve 1.0, which depends upon design. Additional parallel capacitance is relatively easy to add in simple situations should the need arise, and for the domestic environment it could be that the one-off investment is not justified by the quarterly cost savings; only individual cases can determine this.

Interesting, I see what you're saying, but I always understood, perhaps wrongly, that kWh meters measure just that, and automatically take account of variations in power factor.

Codey

The newer meters in some cities have made an attempt to get more accurate readings by having two readings and related sensors. Some cities will get a bill showing KWhr and RKVA (the old fixed calibration of kwhr + a reactive reading.) Some Large Motor users actually have automated RKVAs installed that automatically switch the capacitance depending on the load to get more efficient use of the power.

Tecno.. Off the top of your head you sound like an idiot.

Watt hour meters read only in-phase power. ) other wise they are NOT a wattmeter.

A meter calibrated as you suggsted is NO LONGER A KWH meter. nor a RKVA meter.

You said "Watt hour meters read only in-phase power"

I explained what hapens when it is out of phase - It will not give an accurate reading

I may sound like an idiot to you. So why don't you take the time to calculate the cosines for my examples. The blog will get a better answer to his question.

Then you can explain the difference betwean a RKVA and a Watt meter. Constructive criticism is helpful. Calling names is un-productive, and un-welcome.

I apologize for the name calling, truly, it is unwelcome and unprofessional, however.

The rotating disk and all other PSC approved KWH meters only read the actual power and not the reactive KVA. Other wise it is not a KWH meter. General Electric company once introduced a meter like you were describing but PSC's would not approve it for domestic billing use. The residential customer cannot be penalized for power factor because he has no control over it. That is why the power company has the pole mounted capacitor banks on the circuits. It is their responsibility to correct the VARs at locations nearest to where they are generated.

Large industrial customers with large blocks or reactive power can be penalized and often have dual meters. In this case, it becomes prudent to correct it at the plant.

I have 35 years experience working in the standards calibration lab of an elecric meter department of a large Northern power company. I'm not going to go through your math, it may be correct but the conclusion is wrong.

I see that another contributer also does not understand power factors. He states that an incandescent bulb will be inefficient at low power factors.

He does not understand that the system power factor has no effect on the resistive load of the bulb. It is a resistive device and only can respond to the source voltage.

Again, forgive me for my rude comment, Lets keep this professional.

Note: I am also a member of IEEE and have authored several accepted IEEE papers on power factor control in electric systems. Not to mention, I am 79 years old and still active as a power consultant, (professionally)

another way of looking at it is: (if you want layman's terms), the power factor is going to be less efficient for an incandecent bulbs because it literally burns energy(watts) in the form of heat. fluorescent bulbs/lamps give a better power factor because they return most of the energy given to them, since they do not give off much heat.

The power factor can be defined two ways

1. The cosine of the phase of the current in relationship to the voltage

2. The relationship between the apparent Power (rms) and the actual power.

The power factor is an adjustment in the power meters used to calibrate the meters to give a more accurate reading.

The efficiency of the load has nothing to do with the power meters.

The capacitance or inductance of the load will affect the accuracy of the readings.

-The change in capacitance or inductance has the effect of delaying or advancing (changing) the relationship of the current to the voltage.

Watts is a measurement of power (volts X current)

If you change the load significantly (such as changing all your light bulbs to fluorescence), the fixed calibration meter is no longer accurate. (the newer meters that read RKVA will not need to be re-calibrated.)

The relationship between Kw, KVA and KVAr is represented by a right angle triangle shaped vector.

The True Power - kW is represented by the longer horizontal baseline, this is the resistive current component or pf 1 (Cos - 0 degrees)

The slope or gradient is represented the Apparent Power - KVA with the angle of the slope representing the PF - Cos degrees.

The smaller vertical side represents the Reactive Power - KVAr.

If the power factor is increased the angle between the KVA and KW is reduced and therefore the height of shorter side of the triangle is also reduced.

Therefore the reading on the KVAr meter will also be less.

Strictly speaking this component should not be called KWhr as wattage only relates to power where the current is in phase width the voltage.

Obliviously as the slope of the apparent power is reduced to near pf 1 then the actual kW reading will increase slightly. This one of reasons why power companies penalise customers for low power factor because they loose out on the artificially reduced KW reading

I have an interesting situation that I can't understand. I have a 600v-240v 50kva single phase transformer that should have a core loss of about 180w. When there is no load except for core losses I notice that my electrical meter doesn't read the core loss. In fact it's 0. My meter is a 3 ph 600v.