THE PHYSICS OF... POWER

[wwhitney]

the kVAr's are indeed just lost power.

There are two problems with saying that:

1) While the VArs do represent power shuttling back and forth through the transmission system, the only loss associated with them is the transmission losses from the additional current, primarily the additional I^2*R heating losses. That loss will be a small fraction numerically of the VArs.

2) The value of VArs does not represent an actual 'rate of transfer of energy' at any point in the system over any natural period of time. For example, the amount of energy transferred over a quarter cycle (and soon to be returned over the next quarter cycle) by 1 VAr is 2/pi times the amount of energy transferred over a quarter cycle by 1 Watt of real power. (see math below)

Cheers, Wayne

The integral from 0 to pi/2 of sin(t)cos(t) is 1/2. The integral from 0 to pi/2 of sin(t)sin(t) is pi/4. The ratio is 2/pi.

 

[FionaZuppa]

There are two problems with saying that:

1) While the VArs do represent power shuttling back and forth through the transmission system, the only loss associated with them is the transmission losses from the additional current, primarily the additional I^2*R heating losses. That loss will be a small fraction numerically of the VArs.

2) The value of VArs does not represent an actual 'rate of transfer of energy' at any point in the system over any natural period of time. For example, the amount of energy transferred over a quarter cycle (and soon to be returned over the next quarter cycle) by 1 VAr is 2/pi times the amount of energy transferred over a quarter cycle by 1 Watt of real power. (see math below)

Cheers, Wayne

The integral from 0 to pi/2 of sin(t)cos(t) is 1/2. The integral from 0 to pi/2 of sin(t)sin(t) is pi/4. The ratio is 2/pi.

transmission? there is a line between src and load, thats the meter, everything before the meter is poco, which includes input to gen, the gen itself, and the transmission wires. everything before the meter accounts for the kVAr lost power. sum of all poco losses = P_Q+ other losses. the poco is limited to what is can manipulate to make things more economical, but they can indeed muck around with impedance matching, which is the easiest thing to do (add a cap) to manipulate the P_Q term, etc.

here, lets ask this way. how much work does it take to push 1kA_avg in real world? use whatever math you want, sum of instantaneous values, avg, whatever you like.

and just to be pita, the rate of energy at any given time is VA(t) where V is fixed and A is whatever your amp clamp says at t. that is the real energy at that time, J/s. you cannot and should not mixed terms energy and power w/o making proper distinction in the context of the statement.

 

[Ingenieur]

No, they are not.
I'm going to walk my dog.
That's a black and white issue.

tell that to the authors of the textbook
:lol:

 

[Ingenieur]

So what is your definition of power? Please be precise.

Cheers, Wayne

Please be precise in you question

I have stated it at least a dozen times
and posted excerpts from textbooks
proven the units are the same for P, Q and S
you can lead a horse to water....

 

[wwhitney]

Please be precise in you question

I have been. Please answer the question: what is your definition of power? None of the PDFs you posted define power.

Cheers, Wayne

 

[Ingenieur]

I have been. Please answer the question: what is your definition of power?

Cheers, Wayne

no you have not, please elaborate and be specific
I have posted the definitions of P, Q and S

 

[wwhitney]

no you have not, please elaborate and be specific
I have posted the definitions of P, Q and S

OK, let's try again. What is you definition of the term 'power'? Not P, Q, or S, but the term 'power'. After all, we can't determine the true or falsehood of the statements 'P is power', 'Q is power', or 'S is power' without defining 'power'.

Cheers, Wayne

 

[Smart $]

I am an EE, but my educational experience was a long time ago and my experience since then has been in other areas, so I freely admit my ignorance of detailed power factor analysis. I have been slogging through this thread and reading some of the contributions in more detail than others, as well as trying to ignore the little putdowns and snide remarks.

Anyway, enough disclaimers. Here's what I see: When I consider a pair of AC waveforms, one for voltage and one for current which are not in phase, as would be the case for a load with a reactive component, if I multiply the RMS values of voltage and current I get one number for "power". If I multiply all the instantaneous values of V and I, I get something different, and the waveform I would see if I plot the result over time has two differences from what I would get if the V and I waveforms were in phase i.e., when it and the RMS product would be the same. One is that there are four zero crossings instead of two and the other is that twice per cycle the VI product is negative. If I were to integrate that waveform over a full cycle I would get something less than I would if the load were purely resistive, where I understand the result would be equal to the product of RMS V and I. The ratio of the integrated V*I waveform to the RMS product is the power factor, correct?

It appears others have ignored your post.

Without getting all nitpicky, the answer to your question is yes.

 

[wwhitney]

It appears others have ignored your post.

You missed post 491. I don't blame you, though, as the signal to noise in this thread is very low.

Cheers, Wayne

 

[Ingenieur]

OK, let's try again. What is you definition of the term 'power'? Not P, Q, or S, but the term 'power'. After all, we can't determine the true or falsehood of the statements 'P is power', 'Q is power', or 'S is power' without defining 'power'.

Cheers, Wayne

again, please be precise
mechanical as in Work/time = (F x D)/time
electric heating I^2R
chemical reaction?
influence over people and events?
raising a number to an exponent?
???

you need to ask a question that can be answered
what is YOUR definition of 'power'

 

[Ingenieur]

You missed post 491. I don't blame you, though, as the signal to noise in this thread is very low.

Cheers, Wayne

and I suppose you consider yourself 'signal' :lol:

 

[wwhitney]

how much work does it take to push 1kA_avg in real world? use whatever math you want, sum of instantaneous values, avg, whatever you like.

Your question is very poorly defined. So I'm going to give you a contrived answer to make my point: If I have a single phase AC generator, connected through conductors with a total resistance of 0.01 ohms, to a purely reactive load, and the system has a load current of 1000 amps RMS, then the I^2 R losses are 10,000 watts (average). So after reaching the steady state, every second the generator has to do 10,000 joules of work to overcome these transmission losses. That's it, no other work is being done.

and just to be pita, the rate of energy at any given time is VA(t) where V is fixed and A is whatever your amp clamp says at t.

V is not fixed, V varies in time, just like A does. Your multimeter is doing an RMS average, so you can't use it to measure V or A instantaneously.

you cannot and should not mixed terms energy and power w/o making proper distinction in the context of the statement.

Thank you, I was slightly sloppy. I should have said "While the VArs do represent _energy_ shuttling back and forth through the transmission system".

Cheers, Wayne

 

[wwhitney]

again, please be precise

I have been. Let's try this. You are writing a physics dictionary. You need an entry for the bare term 'power.' No modifiers, nothing. Just 'power'. What is your definition?

Cheers, Wayne

 

[GeorgeB]

From the perspective of instructing someone who is going to use or employ your information, for example loading on a UPS rated 100 kVA 80 kW where the VA rating is bolded on the equipment labels and marketing promotions but the kW rating is in fine print on the nameplate or specs ...

Would you say VA = watts ?

Is it true or is it false?

True or false?

For DC, true.
for AC, Power = E*I*cos(angle of lead or lag)

It's been a long time since engineering school, but aren't P, E, and I magnitudes, and p, e, and i instantaneous values? P=e*i(integrated over an integral number of cycles)

 

[Ingenieur]

I have been. Let's try this. You are writing a physics dictionary. You need an entry for the bare term 'power.' No modifiers, nothing. Just 'power'. What is your definition?

Cheers, Wayne

what is yours?
we are discussing electrical engineering, not Newtonian Mechanics

 

[Ingenieur]

For DC, true.
for AC, Power = E*I*cos(angle of lead or lag)

It's been a long time since engineering school, but aren't P, E, and I magnitudes, and p, e, and i instantaneous values? P=e*i(integrated over an integral number of cycles)

yes, P = VI cos ang = VI pf

caps are used to denote phasors at times
in this S = VI* = P + jQ

 

[wwhitney]

what is yours?

My definition of 'power' is 'the time rate of transfer of energy'. Do you agree to that? If not, please propose your own definition.

Cheers, Wayne

 

[Ingenieur]

My definition of 'power' is 'the time rate of transfer of energy'. Do you agree to that? If not, please propose your own definition.

Cheers, Wayne

that is one
work/time = (F x D)/time is another
I^2R is another
P = Torque x w (w = 2 Pi n rev/sec) is another
P = VI* = P + jQ is another
there are many

by your definition Q is power if the time interval is 1/4 cycle (or even a whole cycle even though it nets to 0, energy is transferred back AND forth over a time period)

 

[Smart $]

You missed post 491. ...

Yes I did. :slaphead:

 

[wwhitney]

that is one

We need to have only one definition.

work/time = (F x D)/time is another
I^2R is another
P = Torque x w (w = 2 Pi n rev/sec) is another
P = VI* = P + jQ is another

Those aren't definitions. Those are physics theorems or results.

by your definition Q is power if the time interval is 1/4 cycle

As I've pointed out, Q is not the power over a 1/4 cycle. For a purely reactive power Q with sinusoidal current and voltage, the average power transfer over a standard 1/4 cycle is 2Q/pi. Which is one good reason not to call Q power, since it is off by a factor of 2/pi (in the purely reactive case, for a standard 1/4 cycle).

(or even a whole cycle even though it nets to 0, energy is transferred back AND forth over a time period)

I take 'rate' to mean a signed sum. If you prefer, we can change the definition of power to 'the net time rate of transfer of energy' to make that more explicit.

Cheers, Wayne