Checking this Equation for Power Factor

[Guineafowl]

TL;DR

Is my expression for real power in an AC circuit, including distortion and displacement, valid? I’m almost certain it is, but just checking I’m not missing anything.

$$P=IV \left [ \frac {cos\phi} {\sqrt{1+THD^2}} \right ]$$

The reason for asking is, I can’t find the equation written out in full like this.

A friend wants to know why his generator is rated in kVA. Having done basic school physics, he’s familiar with P=IV in the DC steady state, so is expecting kW.

I was thinking the equation above would be a useful starting point, a sort of roadmap to point at, allowing me to pick out the term in square brackets, and deal with displacement PF and distortion PF using the top and bottom terms inside, respectively.

 

[Babadag]

PF = displacement power factor
θ = Difference between the phase of the voltage and the phase of the current (phase displacement) in degrees.
The distortion power factor is calculated as follows:
PFTHD=1/sqrt(1+TDH)
Then total power factor =PF*PFTHD=cos(θ)/sqrt(1+THD^2)
The power now will be:
I*V* cos(θ)/sqrt(1+THD^2)

 

[Babadag]

The synchronous generator- as a transformer also-may present any value of power factor from -1 to +1 then P=I*V*cos(θ) [in W ,kW or MW] may be any value from +V*I up -V*I so, only V*I [VA, kVA or MVA] does make sense.

 

[Guineafowl]

Thanks, so the equation as presented is correct, it seems. Oddly, I’ve never seen it written out in full.

With this chap, I thought I’d tailor the answer to his level of understanding, perhaps as follows:

With DC, P=IV, in watts. With AC, you have [post #1 equation], with the extra term because certain loads behave like bigger loads than they actually are. The extent to which they do this is measured by their power factor:
$$PF = \left [ \frac {cos \phi} {\sqrt {1+THD^2}} \right ]$$

Then go for $cos\phi$ leading/lagging, power triangle, brief bit on non-linear loads to cover THD.

With that in mind, the generator (or transformer) maker doesn’t know what load you’re driving, and therefore how much useful power you’re going to get out of it. So as you can see, particularly from the power triangle, they state their machine can make “this much” kVA, and leave you to work out the available, useful kW your load will get out of that, using the PF.

Something like that, anyway. The internet, wikipedia, etc are great resources, but sometimes an answer tailored to you is nice. It’s also nice to explain something well to someone, and watch their face as the penny drops.

 

[DaveE]

Much of what you see about power factor assumes sinusoidal wave forms. This is a common application and OK, but it's not the real definition. The real definition is conceptually simple, it's the ratio of power actually delivered to the load (the instantaneous product of voltage and current) to the product of the rms voltage and the rms current. The latter is the power that would have been delivered to a resistive load with that output.

$$ PF = \frac{1}{T} \int_0^T v(t)i(t) \, dt ~~/~~(V_{rms} I_{rms}) $$

The RMS (root mean square) operator is $\sqrt{\frac{1}{T} \int_0^T f(t)^2 \, dt}$ it is a description of the power that would be dissipated in a resistor exposed to that voltage or current.

$\frac{1}{T} \int_0^T v(t)i(t) \, dt$ is expressed as Watts

$V_{rms} I_{rms}$ is expressed as VA

VA is nearly always a better expression of the stress or capability of equipment like generators. In most applications the voltage is fairly constant, so it's really similar to a rms current limit. Suppose you connect just a big capacitor to the output of your friends generator, it won't consume any power but the generator still has to work hard, deliver voltage and current, get hot, burn gas, etc. No fair calling the manufacturer and complaining that the generator won't deliver any power to your capacitor, nothing could. They'll say tough, we gave you lots of VA.

PS: It is common for generators to specify a minimum power factor limit for your load. Most don't like reactive loads.

 

[Guineafowl]

PS: It is common for generators to specify a minimum power factor limit for your load. Most don't like purely reactive loads.

Yes, that was one thing I found confusing many years ago. A 30 kVA generator with a plated PF of 0.8, my understanding being that PF was a feature of the load, not the source. It was explained as a minimum load PF where full output can be guaranteed.

 

[DaveE]

It was explained as a minimum load PF where full output can be guaranteed.

Or not break things.

 

[Guineafowl]

Or not break things.

That might explain why it was totally burnt out when my friend got it home from a large machinery auction. It had come over from Spain to Scotland, and someone had thoughtfully disconnected all the wires from the alternator terminal box.

It’s a reasonably good Pramac 30 kVA ‘silent’ type, with Mecc-Alte alternator unit, and was otherwise in good condition and bristling with circuit breakers and RCDs. I wondered how they managed to cook the windings in those circumstances. Running for too long in a hot country with too low a PF would do it, I suppose.

One of those “could you just have a quick look at my…” jobs that turned out to be very interesting, swapping in a new (and VERY heavy) alternator into an enclosed box.

 

[Dullard]

From another angle: The 'extra' power dissipation associated with reactive currents is part of the reason that large industrial customers of Electric Utilities are typically (U.S.) charged for both real energy consumption and Power Factor. That extra heating will happen in a transformer secondary or generator output windings.