Power Factor for AC Pumps

[wmcinnes]

If the current and voltage are in phase, the motor speed is the product of voltage and current (PF=1). If they are not in phase the motor in the meter will run fast for part of a cycle and slower for the other part with the average speed being the product of the in-phase voltage and current. Thus measuring real power.

[sphelps]

According the producers of the energy monitor:

In AC power measurements, there actually three separate power components that can be measured:

Apparent Power: Measured in VA (volt - amperes). This is measured by taking the RMS voltage and RMS current readings and multiplying them together. This is what the eMonitor reports for power. All electrical circuits must be sized to handle apparent power because instantaneously, this is the maximum amount of power that can be flowing in a circuit.

Real Power: Measured in watts. This is the actual power that is being consumed by the load. This is what the utility company measures on the meter, as is what customers see on their electrical bill.

Reactive Power: This is kind of like imaginary power. It is really the difference between the Apparent Power and the Real Power. It is power that flows back and forth while the voltage and current are out of phase and is caused by the inductive load. Your utility company does not charge for reactive power, but it must handle reactive power, and reactive power does cause real current to run in your wires, and can generate heat. Reactive power is measured in VAR (volt - amperes reactive).

[mrhasan]

Originally Posted by wmcinnes

If the current and voltage are in phase, the motor speed is the product of voltage and current (PF=1). If they are not in phase the motor in the meter will run fast for part of a cycle and slower for the other part with the average speed being the product of the in-phase voltage and current. Thus measuring real power.

Yes. So how would the meter know which device has what amount of PF? It must do an equivalent resistance scenario for all the loads in the house with one common PF for the whole house and thus multiplying the voltage * current going into house * common PF. The meters can't judge individual PF of the devices.

[mrhasan]

Originally Posted by sphelps

According the producers of the energy monitor:

Yap that's right. Like I said in the very beginning, the utility companies either consider pf=1 or maybe they just do some calculations to find a relative pf for the houses to multiply the RMS volt and current. The meter's simply can't find the individual PF of each appliances. Alongside, households doesn't generally have heavy PF lowering stuffs and hence its more or less above 0.9. Two or three pumps may have low PF in your house but in the end, it won't effect much to the pf of your whole house.

I have sent an email to enmax regarding this. The theory and the claims are just not going through and would love to clear it up too.

Another thing, if possible, if to measure the resistance of the motor (disconnect it and then measure) and then putting in the current (1.85) and resistance into the formula:
I^2/R.

[sphelps]

Well my understanding of AC power isn't great, not my specialty by a long shot. However it seems obvious to me that the term reactive power (the difference between apparent and real) when referred to "imaginary power" isn't something you're going to billed for. It's not being used despite the fact it goes through the circuit. While I might find it amusing to see a line on my power bill that said "Imaginary power usage = 958 kWh (this is the power you didn't use)" I just can't see it.

It's why everything online states a utility meter that measures in kW measures real power and not apparent power. It's clear power factors have virtually zero impact on your power bill which is why all those power factor compensation devices are a scam, also well documented.

I've also sent an inquiry to Enmax and GE (meter manufacturer) for conformation.

[mrhasan]

Alright Steve you don't have to break the motor. I came across an energystar article and they say that in NA grid system, if you have a low pf equipment in domestic facility, you draw in more current but you are only charged for the amount of power that you are consuming and the extra power due to the charge is actually put into the penalty of the industries who have to pay for pf corrections.

So I guess you are actually getting charged for 82W :P

Sorry for the confusions; sometimes theory and practical stuffs doesn't match I will still post the response I get from enmax.

[mrhasan]

Originally Posted by sphelps

Well my understanding of AC power isn't great, not my specialty by a long shot. However it seems obvious to me that the term reactive power (the difference between apparent and real) when referred to "imaginary power" isn't something you're going to billed for. It's not being used despite the fact it goes through the circuit. While I might find it amusing to see a line on my power bill that said "Imaginary power usage = 958 kWh (this is the power you didn't use)" I just can't see it.

It's why everything online states a utility meter that measures in kW measures real power and not apparent power. It's clear power factors have virtually zero impact on your power bill which is why all those power factor compensation devices are a scam, also well documented.

I've also sent an inquiry to Enmax and GE (meter manufacturer) for conformation.

Please do post what they say. It will help me get my theories straight too.

World should revert back to DC grids like Edison's time. AC stuffs are just too complicated

[sphelps]

Originally Posted by mrhasan

Yap that's right. Like I said in the very beginning, the utility companies either consider pf=1 or maybe they just do some calculations to find a relative pf for the houses to multiply the RMS volt and current. The meter's simply can't find the individual PF of each appliances. Alongside, households doesn't generally have heavy PF lowering stuffs and hence its more or less above 0.9. Two or three pumps may have low PF in your house but in the end, it won't effect much to the pf of your whole house.

I have sent an email to enmax regarding this. The theory and the claims are just not going through and would love to clear it up too.

Another thing, if possible, if to measure the resistance of the motor (disconnect it and then measure) and then putting in the current (1.85) and resistance into the formula:
I^2/R.

That's really not what my quote said by the way, but at this point we'll have to agree to disagree until we get conformation either way.

[sphelps]

Originally Posted by mrhasan

Alright Steve you don't have to break the motor. I came across an energystar article and they say that in NA grid system, if you have a low pf equipment in domestic facility, you draw in more current but you are only charged for the amount of power that you are consuming and the extra power due to the charge is actually put into the penalty of the industries who have to pay for pf corrections.

So I guess you are actually getting charged for 82W :P

Sorry for the confusions; sometimes theory and practical stuffs doesn't match I will still post the response I get from enmax.

Good news so far, I'll wait for conformation from Enmax before celebrating though

[mrhasan]

Originally Posted by sphelps

That's really not what my quote said by the way, but at this point we'll have to agree to disagree until we get conformation either way.

I am convinced that utilities charge households just for amount consumed so you are getting charged for 82W. Maybe that's why household businesses are becoming popular now a days?