Less than unity power factor?

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mbrooke said:
Question, is it possible for a load without an inductor or capacitor to have a PF less than 1.00? Here is why I ask:


http://www.lamptech.co.uk/Spec Sheets/D MB Jewel 750.htm

http://www.lamptech.co.uk/Spec Sheets/D MB DuroTest SB450R.htm


Classical thinking would tell me that both of these would be listed at 1.00PF...
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Yes, if it has distortion (harmonics). Inductance and capacitance contribute vars along the y-axis and distortion contributes vars along the z-axis.

The load obviously has one or more of the three.
 
mivey said:
Yes, if it has distortion (harmonics). Inductance and capacitance contribute vars along the y-axis and distortion contributes vars along the z-axis.

The load obviously has one or more of the three.
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And now you have peaked my interest. I can understand harmonics, but how do they produce vars? Ive always understood that you need something that "stores" energy like plates or a magnetic core to produce phase displacement of current in relation to voltage. But I admit my understanding is thin.
 
mbrooke said:
And now you have peaked my interest. I can understand harmonics, but how do they produce vars? Ive always understood that you need something that "stores" energy like plates or a magnetic core to produce phase displacement of current in relation to voltage. But I admit my understanding is thin.
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The association of VARs with short term energy storage (within each cycle of the voltage waveform) is ONLY applicable when the less than unity power factor is associated with a reactive component to the load. It applies only to the case of a (phase) displacement power factor.

The more general case of a power factor less than unity comes from its definition of Watts/(IRMSxVRMS).
You can have a low distortion power factor any time that the voltage waveform (shape) does not match the current waveform, regardless of the phase relationship. If you allow both the voltage and the current to be non-sinusoidal you can have a very simple example of a zero power factor if you consider a current waveform which is zero for the first half cycle while the voltage is a constant high level, then during the second half cycle the current is constant while the voltage is zero. At all times the product of voltage and current is zero, but the individual RMS values are not zero. (You might be hard pressed to come up with a physical load circuit that behaves that way, but this is just a thought experiment.
A more common situation is that the voltage is sinusoidal, but the current waveform is phase chopped, as in a solid state dimmer control. The fact that there are times when there is voltage but no current produces a low power factor.
In neither of these examples is there any energy storage.
 
181021-0625 EDT

mbrooke:

What is the definition of power factor?

PF = Real load power dissipation / (V load RMS * I load RMS)

As soon as the source is not DC, or V and/or I are not in phase sine waves, then PF is less than 1.

.
 
gar said:
181021-0625 EDT

mbrooke:

What is the definition of power factor?

PF = Real load power dissipation / (V load RMS * I load RMS)

As soon as the source is not DC, or V and/or I are not in phase sine waves, then PF is less than 1.

.
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That is a slightly overbroad statement. The applied voltage waveform need not be sinusoidal, as long as the current waveform has the same shape.
A resistive load will produce unity power factor regardless of how complex the applied voltage is.
 
GoldDigger said:
The association of VARs with short term energy storage (within each cycle of the voltage waveform) is ONLY applicable when the less than unity power factor is associated with a reactive component to the load. It applies only to the case of a (phase) displacement power factor.

The more general case of a power factor less than unity comes from its definition of Watts/(IRMSxVRMS).
You can have a low distortion power factor any time that the voltage waveform (shape) does not match the current waveform, regardless of the phase relationship. If you allow both the voltage and the current to be non-sinusoidal you can have a very simple example of a zero power factor if you consider a current waveform which is zero for the first half cycle while the voltage is a constant high level, then during the second half cycle the current is constant while the voltage is zero. At all times the product of voltage and current is zero, but the individual RMS values are not zero. (You might be hard pressed to come up with a physical load circuit that behaves that way, but this is just a thought experiment.
A more common situation is that the voltage is sinusoidal, but the current waveform is phase chopped, as in a solid state dimmer control. The fact that there are times when there is voltage but no current produces a low power factor.
In neither of these examples is there any energy storage.
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But wouldn't this be duty cycle rather than power factor? Sure your voltage is a sine way, but when multiplied by the average I being drawn at various points in the sine wave, you get watts. Watts=Irms average. There is no current slowing down and speeding up the generator. Only pulses which get converted to watts.
 
mbrooke said:
But wouldn't this be duty cycle rather than power factor? Sure your voltage is a sine way, but when multiplied by the average I being drawn at various points in the sine wave, you get watts. Watts=Irms average. There is no current slowing down and speeding up the generator. Only pulses which get converted to watts.
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There is a connection between duty cycle and power factor.

Power at any given instant is always voltage times current. Power factor comes into play only when you have some sort of time averaged measurements of voltage and current.

If your average power (calculated by taking the time average of (voltage * current) ) is different from your 'apparent power' (calculated by taking the ( rms time average of voltage) * (rms time average of current) ) then you have 'power factor'.

Consider what 'duty cycle' does to RMS value. Compare a duty cycle of 100% with a current of 1A and a duty cycle of 10% with a current of 10A. The RMS current in the first case is 1A, in the second case it is 10A (10^2 * 0.1). Of course you have to do your rms averaging over the full 'on-off' time period.

-Jon
 
181021-0744 EDT

mbrooke:

Instantaneous power into any arbitrary load is p = v*i. Average power over some time period is the integral of v*i over that time period divided by the time period.

RMS values of voltage and current are independent of each other except in some special cases (a sine wave applied to a pure resistance). RMS is special kind of average of one variable.

In general you can not take Vrms * Irms over some time interval and expect this product to be equal to the average over that time interval of the integral of the instantaneous product of v and i with respect to time.

.
 
Take a close look at the shape of the electrode. Does it remind you of anything ?

A typical wire wound resistor also has inductance wound in. They can be specially wound to negate the inductive effect of the coil.
 

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181021-0922 EDT

GoldDigger:

Correct I tried to simplify my statement too much.

To continue --- I reread both my post and yours trying to see what I should have said. Possibly this --- As soon as the source is not DC, or V and/or I are not in phase identical waves, then PF is less than 1.
.
 
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SG-1 said:
Take a close look at the shape of the electrode. Does it remind you of anything ?

A typical wire wound resistor also has inductance wound in. They can be specially wound to negate the inductive effect of the coil.
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I doubt though that would make much of a contribution- but I am less on stance about this one.
 
winnie said:
There is a connection between duty cycle and power factor.

Power at any given instant is always voltage times current. Power factor comes into play only when you have some sort of time averaged measurements of voltage and current.
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Well, by definition. But we are taught its in relation to phase displacement.

If your average power (calculated by taking the time average of (voltage * current) ) is different from your 'apparent power' (calculated by taking the ( rms time average of voltage) * (rms time average of current) ) then you have 'power factor'.
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RMS I would have to be higher then RMS V such that IxV yielded a higher magnitude then watts actually dissipated.



Consider what 'duty cycle' does to RMS value. Compare a duty cycle of 100% with a current of 1A and a duty cycle of 10% with a current of 10A. The RMS current in the first case is 1A, in the second case it is 10A (10^2 * 0.1). Of course you have to do your rms averaging over the full 'on-off' time period.

-Jon
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Of course, however, I would imagine current as being in phase with the voltage when it is being drawn in the sign wave.


gar said:
181021-0744 EDT

mbrooke:

Instantaneous power into any arbitrary load is p = v*i. Average power over some time period is the integral of v*i over that time period divided by the time period.

RMS values of voltage and current are independent of each other except in some special cases (a sine wave applied to a pure resistance). RMS is special kind of average of one variable.

In general you can not take Vrms * Irms over some time interval and expect this product to be equal to the average over that time interval of the integral of the instantaneous product of v and i with respect to time.

.
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Clarify, I'm following you, but not quite.
 
mbrooke said:
Well, by definition. But we are taught its in relation to phase displacement.
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If you take a three phase six-pulse uncontrolled rectifier puting out level DC there is no phase displacement but the power factor is not unity. It is 0.958. If the DC is not level, and it isn't likely to be, the PF is lower.
 
winnie said:
Consider what 'duty cycle' does to RMS value. Compare a duty cycle of 100% with a current of 1A and a duty cycle of 10% with a current of 10A. The RMS current in the first case is 1A, in the second case it is 10A (10^2 * 0.1).
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10 would be the mean square, RMS is sqrt(10).

Cheers, Wayne
 
181021-116 EDT

I see more problems in my post numbered 5.

PF = Real load power dissipation / (V load RMS * I load RMS)

This should have read:

For a stationary process, and arbitrary load

PF = Average Real power input to the load / (V load RMS * I load RMS)

where the averaging time period is the same for power, voltage, and current.

These changes allow the load to contain energy storage, like a battery.

.
 
181021-1139 EDT

mbrooke:

What would you like me to clarify?

I assume that you are OK with p = v*i for any real world load.

Is
Average power over some time period is the integral of v*i over that time period divided by the time period
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OK?

.
 
mbrooke said:
Well, by definition. But we are taught its in relation to phase displacement.
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Absolutely, I agree. IMHO that is an error in teaching. 'Displacement power factor' is one thing, 'distortion power factor' is something else, both related by the fact that the RMS current carried by the wires (which causes heating) is greater than the minimum required to deliver the power consumed. Displacement power factor might be the first type of PF that someone encounters; they should not be taught that it is the only kind.

(responding to my duty cycle example)
mbrooke said:
Of course, however, I would imagine current as being in phase with the voltage when it is being drawn in the sign wave.
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Agreed, in most common resistor 'on-off' regulated loads. This is in some sense 'distortion power factor', but only over multiple AC cycles; during a single AC cycle the PF is 1 or no current is flowing. Maybe it deserves a different name.

But consider a dimmer which functions by turning the load off for part of the AC cycle. This can introduce both distortion and phase displacement, changing the power factor.

-Jon
 
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