Question about calculating power factor...

Yes.

Yes again.

The instantaneous product of V*I has a DC (averaged) component, the true power, and a 120 Hz AC component, the 2F thing in your third graph; see the trig identity for the product of two sines. I think there's enough info in them to derive the power factor, at least for pure sine waves. I don't care to do the math this particular afternoon, but it looks fairly simple.

Interesting; I hadn't thought of this before.

John

Hi John,

Thanks for confirming this.

I want to make the 3 calculations on any set of samples, sinusoidal or not. Do you think it can be done on just the V*I samples in this case? I just see that clean graph showing clearly where power is being taken and power is being returned and if it is possible calculate the 3 values (true, apparent, and reactive) based on it.

Thanks,

Alan

Dunno. Too hard to think about for free.

But if you have the E and I samples, why not use them?

John

Hi John,

I probably will, I just like to figure out stuff so I know why something works or doesn't work.

My theory is that when you multiply v*i to end up with power that you lose something in that process which makes it impossible to calculate the other two values (apparent and reactive). Like 4x6=24 and 2x12=24, but once you have the 24, you don't know whether it was 4x6 or 2x12. It just seemed like the power graph had enough info to calculate it, but I've tried many ways without success.

Thanks for the help!

Alan

"Default User"

phi = F x 360

Apparent power is then = Watts / cos( phi)

Assuming all sine waves.

....... Phil

It's not bad for sines. Call the average value of the product waveform, the true power, P.

Call the 120 Hz AC componenent A, and measure it somehow; just subtract P to center it about zero, then compute the RMS or the mean-abs-value of what's left. P varies with the power factor, but A doesn't.

So PF = K * P / A, with constant K depending on how you scale the AC measurement.

John

If the voltage and current are 180 degrees apart, the product is always negative, so F = 1, so F*360 = 360. That doesn't look right.

If they're 90 apart, F=0.5, and F*360 = 180.

Is the correct expression phi = F * 180?

John

"John Larkin" "Phil Allison"

Since the " V*I " curve has half the period of the originals.

...... Phil

Want to have some fun? Run the same analysis for a current waveform has a harmonic component. In other words: sum a fundamental freq. sine wave (which you already have) with another of N times the fundamental frequency (plus some phase shift to make things fun).

The same principle holds true where the average VA over a fundamental waveform cycle (the DC offset) is the real power. But the whole idea of power factor angle becomes confused.

PF is poorly defined for non-sine and non-steady-state loads, or for unbalanced polyphase loads. Great range for argument.

John

Particularly when one person at the local utility reports 'poor power factor' and the knee jerk reaction is to hang power factor correction capacitors.

Hi,

Ok, so from the V samples, A samples, and VA samples, I am calculating

True Power Apparent Power Reactive Power --> SQR(Apparent^2-True^2) Power Factor --> True/Apparent Phase Angle --> ACOS(Power Factor)

Given that I have the samples for V, A, and of course VA, is there a way to find or technique out if the current is leading or lagging that does not rely on the waveform being sinusoidal?

Thanks,

Alan

"Default User"

In any case, PF is not related to phase angle UNLESS the waves are sine.

The basic formula is

PF = watts/VA

See, no phase angle involved.

...... Phil

A common issue nowadays is where voltage waveform is sinusoidal and current waveform is not.

Power factor in that situation requires determining RMS amps (and RMS supply voltage - easier) and load watts. RMS amps is usually not 1.111 (pi/sqr[8]) times average amps when the current waveform is not sinusoidal. When the current waveform is spikier than a sinusoid, RMS amps will generally be more than 1.111 times average amps.

Power factor is ratio of watts to product of RMS volts and RMS amps.

Leading and lagging are not so much the issue here as current waveform having harmonics.

==============================

If you have time for mathematical exercising, this can also be done in "frequency domain".

(Assuming sinusoidal voltage)

Determine amps in each frequency component - fundamental and all harmonics.

Total harmonic current is square root of the sum of the squares of RMS current for every frequency. If you calculate separately sine and cosine components, total RMS amps is still the square root of the sum of the squares of RMS value of each.

VA is that result times RMS voltage.

Watts is RMS voltage times RMS current for the fundamental frequency component part that is in phase with the voltage (sine subpart if you have voltage being sine, cosine subpart if you have voltage being cosine, or volts times amps times cosine of phase difference from voltage if you determine fundamental frequency component of current in total and its phase relationship to the voltage.).

Power factor is ratio of watts to VA.

- Don Klipstein ( snipped-for-privacy@misty.com)

Hi Phil,

Good point!!

Thanks for the help everyone,

Alan

Now, for some real fun, assume that the non sinusoidal load currents result in non sinusoidal supply voltages (due to supply impedance at each harmonic).

Now, multiply the Vi and Ii (for each harmonic i). In three phase systems, harmonics with multiples of three sum to non zero values. These represent real power and, in the case of motors connected to the circuit, real torque. The symptoms of this include mechanical vibration at the harmonic frequencies.