Power Factor for AC Pumps

[wmcinnes]

Originally Posted by mrhasan

Yap residential meters record KWh, not KVA.

Thus your meter only counts the real power.


I'm with Steve and his research.

[mrhasan]

Originally Posted by wmcinnes

I'm with Steve and his research.

Then my question is, how would each of the real power of individual components in the household be calculated by the utility meter?

[wmcinnes]

Originally Posted by mrhasan

Then my question is, how would each of the real power of individual components in the household be calculated by the utility meter?

They simply measure kW and not kVA.

They consist of a motor with the stator current being the current into the building and the rotor curreent being proportional to the voltage at the building input. The motor turns a disc (the disc you see rotating in the window) that passes through a magnetic field generating eddy currents in the disc and this regulates the speed at which the disc turns so that the meter can be adjusted to read correctly.

[mrhasan]

Originally Posted by wmcinnes

They simply measure kW and not kVA.

They consist of a motor with the stator current being the current into the building and the rotor curreent being proportional to the voltage at the building input. The motor turns a disc (the disc you see rotating in the window) that passes through a magnetic field generating eddy currents in the disc and this regulates the speed at which the disc turns so that the meter can be adjusted to read correctly.

1. KW is not a direct unit so the meter actually does some conversions within itself

2. That is exactly how it is done. The rotor moves proportionally to the product of voltage and current to shows kWh consumption. No PF involved that's it. So when a outlet draws 1.85A @ 115V, that means the rotor is turning at a proportional rate directly to the product of those two.

The motor needs 82W to run and it is taking in 82W but due to its inefficient manner (age, water, etc.), the pump has to take in more power and hence 1.85*115V.

For this motor, you are bringing 1.85A "in" the house.

[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 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.

[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.

[sphelps]

Originally Posted by mrhasan

Then my question is, how would each of the real power of individual components in the household be calculated by the utility meter?

Because the utility meter measures the exact same thing as the energy monitor, it measures actual power in kWh. It doesn't measure apparent power or vars.