THE PHYSICS OF... POWER

[Besoeker]

you cannot generalize "power". in physics the term "power".

There is no need or requirement to generalise.
Power is power. That's it.

 

[Smart $]

https://xkcd.com/410/

Actual...

:thumbsup:

 

[FionaZuppa]

Once more, that's still nonsense.

for Φ=0 its not, for Φ !=0 then yes. in real world Φ !=0, thus VA in terms of watts is not so clear since now Cos(Φ) is involved, but yet you can still write the math in terms of watts or VA, its just that the terms must be kept separated to understand the perspective (load kW vs poco VA).

 

[wwhitney]

assume a current of 100/25 deg (90.6 + j42.3) flows to a load (pf is approximately 0.90)
assume line R is 1 ohm

loss S = I I R = 100/25 100/25 1/0 = 10000/50 = 6428 + j7666
loss P =6428 W
loss Q = 7666 var (not really lost as work)

only the active part of the current contributes to IR losses in watts, so 6428 W in this case
only 6428 of 'heating', not 10,000

I'm pretty sure your computation is wrong, and that the line experiences 10,000 W of heating. Your vector math is wrong.

Look at it this way: while the circuit current is out of phase with the circuit voltage, the voltage drop that occurs along the pure resistance of the wire is caused by the current and hence is in phase with the current.

That is, my understanding is that for resistance losses along a conductor, P = I^2 * R is a scalar equation. I is just the RMS current, and R is the resistance, not the complex impedance.

Cheers, Wayne

 

[FionaZuppa]

There is no need or requirement to generalise.
Power is power. That's it.

no its not. jX is not real, imaginary, not real.

using kW and VA and PF is in fact a big generalization of the system

there is no kW associated with jX when using the well-known physics definition of power.

post #316.

 

[Ingenieur]

I'm pretty sure your computation is wrong, and that the line experiences 10,000 W of heating. Your vector math is wrong.

Look at it this way: while the circuit current is out of phase with the circuit voltage, the voltage drop that occurs along the pure resistance of the wire is caused by the current and hence is in phase with the current.

That is, my understanding is that for resistance losses along a conductor, P = I^2 * R is a scalar equation. I is just the RMS current, and R is the resistance, not the complex impedance.

Cheers, Wayne

you may be right
I need to give it some thought
it may only involve magnitudes

 

[wwhitney]

P = I * V [instantaneously]
E_(a,b) = integral_(a,b) I * V [total energy transferred over the time interval from a to b]
P_(a,b) = E_(a,b) / (b - a) [average power over the time interval from a to b]

Say I = sin (t), and V = sin (t + pi/2) = cos (t) [pi/2 = 90 degrees in radians]

E_(a,b) = integral_(a,b) sin(t) cos(t)
= integral_(a,b) 1/2 sin(2t)
= - 1/2 cos(2t) |_(a,b)
= - 1/2 (cos(2b) - cos(2a))

That's a mistake right there, the integral of sin(2t) is -1/2 (cos 2t). I dropped the factor of 1/2. So the correct answer is

E_(a,b) = -1/4 (cos(2b) - cos(2a))

Consider a few different intervals:

"1st" quarter cycle
E_(0, pi/2) = -1/2 (cos(pi) - cos(0)) = -1/2 (-1 - 1) = 1
P_(0, pi/2) = 2 / pi

So that should be

E_(0, pi/2) = 1/2
P_(0, pi/2) = 1/pi

And likewise for the next quarter cycle. Regardless, the energy and average power over a half cycle are still 0.

Cheers, Wayne

 

[FionaZuppa]

That is, my understanding is that for resistance losses along a conductor, P = I^2 * R is a scalar equation. I is just the RMS current, and R is the resistance, not the complex impedance.

Cheers, Wayne

there is no reason to not use Z in all the math. as i explained earlier, if the physical system has only R then Z collapsed to a more simple term. in real world there is nothing that is just R, just X_C, or just X_L, in our real world everything is LRC, you can sometimes study each component and attempt to manipulate things (like adding X_C near a motor that has high X_L), but in order to characterize the "system" its usually described as a whole, a generalization, power guys talk in kW, VA, PF

 

[Ingenieur]

I'm pretty sure your computation is wrong, and that the line experiences 10,000 W of heating. Your vector math is wrong.

Look at it this way: while the circuit current is out of phase with the circuit voltage, the voltage drop that occurs along the pure resistance of the wire is caused by the current and hence is in phase with the current.

That is, my understanding is that for resistance losses along a conductor, P = I^2 * R is a scalar equation. I is just the RMS current, and R is the resistance, not the complex impedance.

Cheers, Wayne

you are correct
the whole ckt and load must be considered as a system

 

[FionaZuppa]

could you possibly characterize the system as Z_source and Z_load and then look at Delta(Z_source,Z_load)

hmmm, an impedance mismatch, sounds familiar.

cheers, i exit this thread here, but i leave you with something to ponder over, is all the Power_reactive wasted by the Z of the gen and transmission?

carry on.

 

[FionaZuppa]

i promise, my last post. 33 pages summarized right here.
in real world, poco VA is their input work, kW (customer work) is a portion of VA that is paid for, $kW must be greater than $VA otherwise poco is bankrupt :thumbsup:

too bad we cant "ID" power on the gen side and then charge customer for their ID'd power use. hmmmm, tagging and routing energy, sounds very Star Trek'ish.

 

[Besoeker]

no its not. jX is not real, imaginary, not real.

using kW and VA and PF is in fact a big generalization of the system

there is no kW associated with jX when using the well-known physics definition of power.

post #316.

What units are a measure of power?

 

[Carultch]

could you possibly characterize the system as Z_source and Z_load and then look at Delta(Z_source,Z_load)

hmmm, an impedance mismatch, sounds familiar.

cheers, i exit this thread here, but i leave you with something to ponder over, is all the Power_reactive wasted by the Z of the gen and transmission?

carry on.

Reactive power is not wasted.


Reactive power causes a need for more real power generated, than real power delivered to the load. What you really want to look at, is (real power delivered to load)/(real power supplied at source).


And if you want to determine power lost because of the reactive nature of the load, what you want to compare is (real power generated) - (real power delivered) for the load as is, compared to what that difference would be, if you had an resistive load of equivalent power in its place.

 

[GoldDigger]

What units are a measure of power?

A reminder, slightly off the focus of this thread, but relevant:

The fact that two measurable physical quantities have the same dimensional units is not an argument that the two quantities are the same or even related.
A prime example is the fact that work is expressable in foot-pounds (or Newton-metres) and so is torque.

 

[FionaZuppa]

What units are a measure of power?

units are J/s
but let me make it very clear, J/s by itself does not mean heat or work. its already been stated a few times, lossless X_C coupled to a lossless X_L
you can pre-charge the cap or run a magnet over the inductor to get things moving, after that if you observe any part of the system you will observe J/s of energy moving back and forth. adding charge to cap or moving magnet over the inductor requires some input work, but after that the observation is J/s at any given time, yet there is no power at any given time. add 10^{-100000000000000000000000000} of R and then and only then will your input work slowly dissipate into heat.

the use of the term "power" should be dropped, talk about everything in eV, then you'll see the clarity.

Reactive power is not wasted.

all of the reactive power is wasted because there exists R in real world. the only way reactive "power" is not wasted is if the system is lossless gen on one side, and a lossless X connected to gen, you would need work to push charge into the X, that work is then returned, net zero over time. we see this with super conductors where the load side is pure X. it requires work to push charge across R, no R then no work. real world is LRC. it requires zero avg work to push 10^{100000000000000} Amps_continuous through a wire that has zero R.

 

[Besoeker]

A reminder, slightly off the focus of this thread, but relevant:

The fact that two measurable physical quantities have the same dimensional units is not an argument that the two quantities are the same or even related.
A prime example is the fact that work is expressable in foot-pounds (or Newton-metres) and so is torque.

You can have torque and no work performed.

 

[Sahib]

Once more, that's still nonsense.

To avoid confusion between active, reactive and apparent powers, they are measured in watt, var and va units. Each such unit is equivalent to joules/second. Hence the units measure the same thing: POWER.

 

[Phil Corso]

Being imaginary power makes it not real... just like anything else imagined.

But, Besoeker is the only contributer(?) to this topic thus far, that understood my thesis! More later!


Smart A--, who do you think is giving you your :thumbsup:!

Phil

 

[GoldDigger]

To avoid confusion between active, reactive and apparent powers, they are measured in watt, var and va units. Each such unit is equivalent to joules/second. Hence the units measure the same thing: POWER.

See Post #334

 

[Besoeker]

To avoid confusion between active, reactive and apparent powers, they are measured in watt, var and va units. Each such unit is equivalent to joules/second. Hence the units measure the same thing: POWER.

Incorrect.