my power meter operating a 240vac fan shows a power factor of 0.56 and current of 0.414 amps. I added a capacitor to fix the power factor using an online calculator. The online calculator asks you to enter the real power in kw, the power factor and the desired power factor. An internet enquiry states that the unit of real power is watts and apparent power is VA.
The meter showed 58 watts or .058 in kw. I entered 0.9 as the desired power factor. The answer in KVAR units was 0.06. KVAR=2pfCufV^2x10^-9. p is pye, f is frequency C is capacitance in uf. Substituting to get the capacitance I got 3.3uf. I used a 3.2uf.
Using this capacitor I got a new power factor reading of 0.94 and current reading of 0.24 amps.
What I don't understand is the meter was still showing 58 watts even though the current was much less with the capacitor. Can this be explained? Nothing seems to be gained by using a capacitor.

 

What I don't understand is the meter was still showing 58 watts even though the current was much less with the capacitor. Can this be explained?

Of course.
The real power of 58W does not change with power factor, only the reactive power changes.
The current consists of two components: real power (in phase with the voltage) and reactive power (90° out of phase with the voltage).
You cannot reduce the real power current unless you reduce the real power (which for the fan stays fixed).
But you reduce the reactive current with the capacitor as its leading reactive current cancels most of the inductive lagging reactive current of the motor.

Nothing seems to be gained by using a capacitor.

If you expected to reduce the real power consumed by the fan, than yes, nothing is gained.

So what you gain is a reduction in the power line current which reduces resistive losses in the power line resistance back to the power station.
But since power companies generally only charge residential users for real power used, improving the power factor helps them, but has negligible effect on your bill.

Commercial power users are somethings charged extra for a high power factor (the reactive current) so it can be to their advantage to provide PF compensation.

A nit: p is pi, not pye.

 

Of course.
The real power of 58W does not change with power factor, only the reactive power changes.
The current consists of two components: real power (in phase with the voltage) and reactive power (90° out of phase with the voltage).
You cannot reduce the real power current unless you reduce the real power (which for the fan stays fixed).
But you reduce the reactive current with the capacitor as its leading reactive current cancels most of the inductive lagging reactive current of the motor.
If you expected to reduce the real power consumed by the fan, than yes, nothing is gained.

So what you gain is a reduction in the power line current which reduces resistive losses in the power line resistance back to the power station.
But since power companies generally only charge residential users for real power used, improving the power factor helps them, but has negligible effect on your bill.

Commercial power users are somethings charged extra for a high power factor (the reactive current) so it can be to their advantage to provide PF compensation.

A nit: p is pi, not pye.

I think I get it. The current in phase with the voltage does not change, only the current at 90 out of phase changes with the power factor. That result seems a little strange. If you increase the power factor closer to one you would think the current in phase with the voltage should be increasing and thus increasing the real power.
Thanks for the 'pi' tip. No symbol for it on the keyboard.

 

Thanks for the 'pi' tip. No symbol for it on the keyboard.

Click the symbol that looks like a Greek temple halfway along the top of the reply box - you'll get all the useful symbols so you can write Ωπ√± etc.

A domestic electricity meter measures "real power" (in most countries), so you don't save money correcting your power factor, but you reduce the current in the cables, and reduce the power company's losses, saving them money.
The result of your altruism is a reduction in CO2 emissions!

Many LED lamps (generally the cheap ones) have leading power factors, so about 50 Watts of LED lighting might power factor correct your fan!

 

Click the symbol that looks like a Greek temple halfway along the top of the reply box - you'll get all the useful symbols so you can write Ωπ√± etc.

A domestic electricity meter measures "real power" (in most countries), so you don't save money correcting your power factor, but you reduce the current in the cables, and reduce the power company's losses, saving them money.
The result of your altruism is a reduction in CO2 emissions!

Many LED lamps (generally the cheap ones) have leading power factors, so about 50 Watts of LED lighting might power factor correct your fan!

Of course.
The real power of 58W does not change with power factor, only the reactive power changes.
The current consists of two components: real power (in phase with the voltage) and reactive power (90° out of phase with the voltage).
You cannot reduce the real power current unless you reduce the real power (which for the fan stays fixed).
But you reduce the reactive current with the capacitor as its leading reactive current cancels most of the inductive lagging reactive current of the motor.
If you expected to reduce the real power consumed by the fan, than yes, nothing is gained.

So what you gain is a reduction in the power line current which reduces resistive losses in the power line resistance back to the power station.
But since power companies generally only charge residential users for real power used, improving the power factor helps them, but has negligible effect on your bill.

Commercial power users are somethings charged extra for a high power factor (the reactive current) so it can be to their advantage to provide PF compensation.

A nit: p is pi, not pye.

Found this piece of wisdom on the internet which explains why the real power stays the same even when the power factor is changed by a capacitor for a mains AC powered electric motor. Quote " real power measured in watts defines power consumed by resistive part of a circuit ." The resistive part of the electric motor is a constant. This is why I get 58 watts reading on my power meter regardless of whether I use a capacitor or not in parallel with the motor.

 

Found this piece of wisdom on the internet which explains why the real power stays the same even when the power factor is changed by a capacitor for a mains AC powered electric motor. Quote " real power measured in watts defines power consumed by resistive part of a circuit ." The resistive part of the electric motor is a constant. This is why I get 58 watts reading on my power meter regardless of whether I use a capacitor or not in parallel with the motor.

Unfortunately, the internet is not so wise.
The real power represents the rate that it is doing work. If you imagine the a perfect one-horse-power motor lifting 33000 lb one foot in a minute, it will use 746 Watts of real power.
If it only has to lift 16500 lb, it will only use 373 Watts of real power.

A real motor will have resistance - real power will be used by the current flowing in its resistance, which will add to the 746 Watts.
The gearbox which transforms the motor shaft rotation speed into a pulley that moves the rope only 1ft in a minute will have friction. Real power will be used to overcome that friction, adding to the 746 Watts.
A different grade of oil in the gearbox will result in a different amount of friction and different power consumption.

If you take your fan, and reduce the length of the blades by 50%, then it can move less air (a quarter, I think) and it will do less work and use less real power, but its resistance will remain the same.

 

No symbol for it on the keyboard.

Just an FYI, Alt+227(Must use the numpad I believe) will print the pi symbol. It doesn't work if you enable the Unicode UTF-8 worldwide language support checkbox under Locale settings in Windows 10 though.