THE PHYSICS OF... POWER

[FionaZuppa]

Volt-Amperes does not directly equal Joules per second.

what? it totally does. the definition of watt is not the same as "power dissipated" (aka "consumed"). if i take 10²⁰⁰eV and move it from inductor to cap via wire with zero R, i move it via watt units yet no power is consumed.

eV is a better unit to use when we get closer to what is really going on from the physics view.

This is an incandescent bulb filmed in slow motion, with a 50 Hz AC power source. You can see that the power is not steady, but instead comes in cycles. So the Wattage rating on the bulb, is really the average Watts over a cycle.
https://www.youtube.com/watch?v=eUprJS9sXYU

what you see is not what you get? if you place a sensor on generator shaft to monitor magnetic impulse jitter, does it match the frequency of the bulb filament as seen in that video? you tell me.

 

[Carultch]

what? it totally does. the definition of watt is not the same as "power dissipated" (aka "consumed"). if i take 10²⁰⁰eV and move it from inductor to cap via wire with zero R, i move it via watt units yet no power is consumed.

eV is a better unit to use when we get closer to what is really going on from the physics view..

Here's a similar idea.

Suppose you have a rectangle. We all know that the area is length*width.

Now suppose you have an irregular shape that is sort of rectangular. But it is irregular enough to matter. You can establish a bounding box to enclose this shape, to get a representative length and width. However, if you multiply the representative length and the representative width together, should you expect that you can calculate the area this way? You might get a good guess, but you will not accurately calculate the area by only accounting for representative length and width.

Volt-amperes work the same way. The inherent problem is that we are multiplying time "average" values of voltage and current, and expecting them to calculate power. In reality, it is only valid to calculate power by multiplying Volts and Amps, when you multiply the instantaneous voltages and currents that occur simultaneously. Multiplying time "average" values overlooks the phase shift and other irregularities.

 

[Besoeker]

In reality, it is only valid to calculate power by multiplying Volts and Amps, when you multiply the instantaneous voltages and currents that occur simultaneously.

And you get instantaneous power for that instant in time which isn't terribly useful for most purposes.

 

[Carultch]

And you get instantaneous power for that instant in time which isn't terribly useful for most purposes.

True. But your next step is intuitively obvious. Take the average value of that instantaneous power, which is the power we use in most purposes.

 

[wwhitney]

Volt-Amperes does not directly equal Joules per second.

what? it totally does.

You need to reread the rest of Carultch's post.

Instantaneously Volts * Amperes = Joules /second. So if you integrate Volts * Amperes over time, you will get real power integrated over time, or energy transferred. That is, you need to multiply first, then "average".

However, as a communications convention, when we choose to write VA instead of Watts, that means we are multiplying two "averages". Since we "averaged" first, before multiplying, we don't get the real power.

Cheers, Wayne

 

[FionaZuppa]

Here's a similar idea.

Suppose you have a rectangle. We all know that the area is length*width.

Now suppose you have an irregular shape that is sort of rectangular. But it is irregular enough to matter. You can establish an average length and an average width. However, if you multiply the average length and the average width together, should you expect that you can calculate the area this way? You might get a good guess, but you will not accurately calculate the area by only accounting for average length and average width.

Volt-amperes work the same way. The inherent problem is that we are multiplying time "average" values of voltage and current, and expecting them to calculate power. In reality, it is only valid to calculate power by multiplying Volts and Amps, when you multiply the instantaneous voltages and currents that occur simultaneously. Multiplying time "average" values overlooks the phase shift and other irregularities.

well, you say "Suppose you have a rectangle. We all know that the area is length*width."
no we do not, at least not by physics! you could only find scalar for a instantaneous observation, after that observation the area has changed
the physics says no such thing, you cannot create a true/fixed LxW rectangle in the physical world (it would indeed be a jagged rectangle when you look close, and would actually be changing shape with any delta t). if you wish to stay in math only then you are no longer using physics.

 

[Carultch]

well, you say "Suppose you have a rectangle. We all know that the area is length*width."
no we do not, at least not by physics! you could only find scalar for a instantaneous observation, after that observation the area has changed
the physics says no such thing, you cannot create a true/fixed LxW rectangle in the physical world (it would indeed be a jagged rectangle when you look close, and would actually be changing shape with any delta t). if you wish to stay in math only then you are no longer using physics.

You are making up red herrings to intentionally miss my point.

 

[Besoeker]

True. But your next step is intuitively obvious. Take the average value of that instantaneous power, which is the power we use in most purposes.

And taking that a step further, we don't normally use instantaneous values of volts and amps either.

Sqrt(3)*V_L*I_L*cos(phi) with RMS values for V and I is how most of us would do it for three phase systems.

Somehow, this thread has served to complicate that simple formula out of all proportion.

 

[Carultch]

And taking that a step further, we don't normally use instantaneous values of volts and amps either.

Sqrt(3)*V_L*I_L*cos(phi) with RMS values for V and I is how most of us would do it for three phase systems.

Somehow, this thread has served to complicate that simple formula out of all proportion.

How does it work when each phase has a different power factor and value of phi?

 

[Besoeker]

How does it work when each phase has a different power factor and value of phi?

Then treat each phase individually.
But my point was about how infrequently we need instantaneous values.

 

[Phil Corso]

Then treat each phase individually.

Treating each phase individually to find PF is incorrect!

Phil

 

[FionaZuppa]

Treating each phase individually to find PF is incorrect!

Phil

just add some kVAR caps and call it a day :lol:

 

[Besoeker]

Treating each phase individually to find PF is incorrect!

Phil

Really?
How would you do it?
And how would instantaneous values feature?

 

[Phil Corso]

just add some kVAr caps and call it a day :lol:

"Anche Tu, Brute"

 

[Phil Corso]

Really?
How would you do it? And how would instantaneous values feature?

Here's a hint... it ain't the average of the 3 individual phase PF's!

Phil

 

[ggunn]

At some point what was physics becomes philosophy, but then doesn't everything?

 

[GoldDigger]

Here's a hint... it ain't the average of the 3 individual phase PF's!

Phil

If you had exactly matched leading and lagging PFs on two phase lines and a PF of 1 on the third line the "overall" PF (kW/kVA) might end up as 1 but POCO would still treat it as an non unity PF when calculating for possible penalties. Just how to calculate the kVA term could be a little difficult when combining leading and lagging kVARs.

The "average" of a leading PF of .8 and a lagging PF of .8, on the other hand, would be .8, and for some purposes that would be a good measure of the reactive nature of the load(s).

I do not see a simple, one size fits all, measure that fully represents such a single or three phase circuit.
When dealing with an unbalanced PF load, I wonder just how the POCO meter captures kVA for the purposes of demand penalty?
Simple sum of kVA over all phases with no consideration of leading or lagging?

 

[FionaZuppa]

but i ask again, a different way

i have a motor that says 1kW output, PF=0.8

i now wish to build my generator/engine so that my motor can do the work i need it to. how much kW (approx) does the generator/engine need to produce assuming the motor leads are directly attached to the generator leads (no wire between)?

1) 1kW
2) 1.25kW
3) 1.5kW
4) 0.8kW

 

[wwhitney]

So is there a definition of power factor for an unbalanced polyphase load? There may not be a neutral present.

Cheers, Wayne

 

[GoldDigger]

but i ask again, a different way

i have a motor that says 1kW output, PF=0.8

i now wish to build my generator/engine so that my motor can do the work i need it to. how much kW (approx) does the generator/engine need to produce assuming the motor leads are directly attached to the generator leads (no wire between)?

1) 1kW
2) 1.25kW
3) 1.5kW
4) 0.8kW

Leaving starting out of consideration and assuming a 100% efficient motor, so that the mechanical output power equals the electrical input power:

The engine (prime mover of the generator) needs to be able to produce 1kW continuously if you need the motor to work at full rated load. The one answer you ask for is (1).
The generator itself needs to source 1kW but also needs to be rated for 1.25kVA output (continuous).