What's the difference between power factor of 1 VS .7 ?

[Ratchit]

KISS,

...The point I was trying to make is a source with a higher current capability is required.

Why is that? A higher line voltage will be required to do that, and the utility is not going to raise their voltage so they can pump more current into your service. Take the extreme case. Suppose you attach a humongous capacitor to your electrical service. For the most part, you will not be dissipating any energy that can be billed to you. And yet, the utility has to supply a lot of current back and forth and swallow the IR line losses. No matter what the voltage is, they will not be able to bill you for the energy transfer of the reactive load, which corresponds to a PF of 0. They don't like that, so they will bill you extra for having such a lousy PF. The bottom line is that they can deliver more billable power with the same current load when the PF is close to 1.

Ratch

 

[MrAl]

Hi,

I think what he means is that say you have a load that at 120v uses 2280 watts but it has a power factor of 0.7. Although that would only be 19 amps at a power factor of 1, it is much more current at the power factor of 0.7 . You would not be able to run that from a single 'regular' 20 amp outlet without blowing the 'normal size' house fuse although you could probably get away with it if it had a power factor of 1.

 

[Ratchit]

MrAl,

I think what he means is that say you have a load that at 120v uses 2280 watts but it has a power factor of 0.7. Although that would only be 19 amps at a power factor of 1, it is much more current at the power factor of 0.7 . You would not be able to run that from a single 'regular' 20 amp outlet without blowing the 'normal size' house fuse although you could probably get away with it if it had a power factor of 1.

And what I mean is that the line voltage is fixed at 120 volts. Assuming the load resistance remains the same, a decrease of the PF to 0.7 means that additional reactance has been added to the load. This will increase the impedance of the load, so with a constant voltage, the current will decrease, not increase.

Ratch

 

[ronsimpson]

The line is 120 volts AC, fixed, we all agree on that.
With a resistance load the PF=1. The current and watts are simple. 120 ohms makes 1A and 120 watts.

With a motor the load has a "impedance" that is complex. Resistance & inductive &/or capacitive.
So a 120 watt motor will pull about 1.4A.
You are receiving 120W of work. Paying for 120W. But the power company has to make 1.4A not 1.0A. The 1.4A is not occurring at the same time as the 120V. (phase shift)

 

[Ratchit]

ronsimpson,

With a motor the load has a "impedance" that is complex. Resistance & inductive &/or capacitive.
So a 120 watt motor will pull about 1.4A.
You are receiving 120W of work. Paying for 120W. But the power company has to make 1.4A not 1.0A. The 1.4A is not occurring at the same time as the 120V. (phase shift)

As I explained previously, the resistance of the motor does not change. So any increase in reactance will cause the current to decrease. The only way to increase the current is to increase the line voltage, which we agree does not happen.

Ratch

 

[ronsimpson]

As I explained previously, the resistance of the motor does not change. So any increase in reactance will cause the current to decrease. The only way to increase the current is to increase the line voltage, which we agree does not happen.

The resistance of the motor is made much lower so to get the 120 watts. The total impedance is such that it uses 120 watts.
The motor was never designed to run at DC. It is built make X horse power, or to draw 1A from 120V @ 60hz.
We should not say the motor has 120 ohms because that means almost nothing. The current is dependent on frequency and thus not pure resistance.
The motor has all the information you need to know on the plate.
120V 60hz 1A xHP. (that does not indicate 120 ohms)
DC resistance is never rated on a AC motor.

 

[Ratchit]

ronsimpson,

Let's not use a motor for an example because it changes its impedance dynamically, depending on the mechanical load.

The resistance of the motor is made much lower so to get the 120 watts.

OK.

The total impedance is such that it uses 120 watts.

And it won't use 120 watts unless it is connected to the proper mechanical load.

The motor was never designed to run at DC.

Who is running an AC motor at DC?

It is built make X horse power, or to draw 1A from 120V @ 60hz.

Provided it is connect to a proper mechanical load.

We should not say the motor has 120 ohms because that means almost nothing.

Correct, a heavy mechanical load will cause the motor to decrease its reactance, thereby increasing its current draw.

The current is dependent on frequency and thus not pure resistance.

The frequency does not change.

120V 60hz 1A xHP. (that does not indicate 120 ohms)

Correct, the load determines the impedance.

DC resistance is never rated on a AC motor.

Correct, DC and AC resistance are the same, and impedance depends on the load.

Ratch

 

[ronv]

PF

The whole thing doesn't work in my mind.

But I can understand that uncorrected the first trace needs 3X the current from the power source as the one with power factor correction.

 

[ronsimpson]

ok maybe the motor is a bad example.

3KVA X power factor of 1 = 25 amps
3KVA Watts X power factor .7 = 17.5 amps

There is a recurring theme that if you had a 3kw load and it had a PF of 0.7 it would not be a 3kw load anymore.
If I designed a 3kw load for DC it would use 3kw. (25A from the example in post #8)
If I designed a 3kw load with a pf of 0.7 and used 60hz it would use 3kw. (35.7A not 17.5A in the example)
I see a 3kw load (under rated conditions) uses 3kw.