THE PHYSICS OF... POWER

[Besoeker]

tell that to the authors of the textbook
:lol:

Power is power.
It's a black and white issue.

The police love him.

 

[FionaZuppa]

We need to have only one definition.


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

are we talking physics or in poco nomenclature ??? the latter confuses many.
a transfer of energy is not power nor work

the only definition of power is work/time
P_Q thus cannot be power by this definition, because as you say P_Q does no work.

its counter intuitive to call something its not, but wait, thats why √-1 is handy, but only when your view is load only. if the view is the system then P_Q is very real in such system.

 

[winnie]

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....

Both of the references in your quoted post describe reactive power as 'the instantaneous power absorbed by the reactive part of the load,... is a double frequency sinusoid with zero average value'...

I agree: 'reactive power' is associated with instantaneous power.

The 'zero average value' is the point that we are all circling around from opposite sides. The 'zero average value' means (other than losses) that no net power is consumed at the generator, and no net power is delivered to the load.

Zero average value means that 'reactive power' does not on average deliver energy to a load.

-Jon

 

[winnie]

its counter intuitive to call something its not, but wait, thats why √-1 is handy, but only when your view is load only. if the view is the system then P_Q is very real in such system.

Question for you:
Assume a 100 kVA generator with 5% losses at full power. When I connect this generator to a 100 kW(electrical) resistive load, 105 kW(mechanical) must be supplied to the generator shaft.
If I instead connect this generator to a 100kVA 5% power factor load (a large inductor with low resistance), how much mechanical power would need to be supplied to the shaft?

-Jon

 

[wwhitney]

the only definition of power is work/time

That definition seems very limiting. When my radiant electric portable heater is drawing 1kW of power, where is the work (locally)?

Rather, I'd say 1kW of power is being converted from electrical energy to heat energy. If you want to make this breadth explicit in the definition, we could define power as 'the net time rate of transfer or conversion of energy'.

Cheers, Wayne

 

[FionaZuppa]

That definition seems very limiting. When my radiant electric portable heater is drawing 1kW of power, where is the work (locally)?

Rather, I'd say 1kW of power is being converted from electrical energy to heat energy. If you want to make this breadth explicit in the definition, we could define power as 'the net time rate of transfer or conversion of energy'.

Cheers, Wayne

must step away from the generic nomenclatures. "its all about the amps, the amps i say" :thumbsup:
the only way to get amps is to put work in to raise eV of electrons, you put the work in to move charge away from the opposite side, thus you now have potential energy, this creates the EMF. when EMF flows there is a counter force that must = input work in perfect harmony with amps value. how and where you choose to extract the eV(energy) is work. a heater of pure R does pure work :thumbsup:, not very useful work (just like kVAr), but perhaps R is useful load side, heat my house so i dont turn to ice in winter

you guys already pleaded the case that if the load is pure R then the input work on lossless gen side = the kW(work) on load side, doesnt really matter what the load is, its pure R (aka "real power"), etc.

from poco view, kVAr is just power they cant charge for because in real world the load (cust) is not consuming that power.

 

[GoldDigger]

the only way to get amps is to put work in to raise eV of electrons, you put the work in to move that charge away from the opposite side, thus you now have potential energy, this creates the EMF. how and where you choose to extract the eV(energy) is work. a heater of pure R does pure work :thumbsup:, not very useful work, perhaps useful, heat my house

you guys already pleaded the case that if the load is pure R then the input work on lossless gen side = the kW(work) on load side, doesnt really matter what the load is, its pure R (aka "real power"), etc.

Sooo many problems with that statement.

0. The eV of an electron, to a physicist, is the rest energy in electron volts which corresponds by E=mc² to the mass of the electron.
Beyond that, an energy of one eV is the energy given to a charge equal to that of one electron when it moves through a potential difference of one Volt.
1. When you start the current moving, the kinetic energy of the electrons involved is minuscule compared to the energy delivered to the load and the energy changed to heat in the material's resistance.
2. That is NOT what "creates" the EMF (short for ElectromMagnetic Field, but loosely used as an equivalent to voltage when what is really meant is that the space integral over any path of the EMF equals the voltage between the two points involved.)
3. And the problem with the classical mechanics definition of power is that creating heat in a resistance does not constitute work. Hence no power, by that definition, is involved. We need to use a different definition of power, and saying it is "anything with units of volts x amps" just does not work!

 

[FionaZuppa]

i meant EMF as Electro Motive Force

ok, if i do work to build charge up to say 10kV, and the electrons from that charge field now escape to rush back home to 0v, what electric field did it traverse? 10kV is the answer. during the trip home you'll strip away all the eV that went in each electron that made it home! so how much energy would that be, depends on how many amps there were and for how long those amps were there. the charged carrier (electron) has energy associated with it.

a pure R on load side, be it a stick of iron, a non-existent zero jX induction motor, widget XYZ, the pure R is pure real power aka "work", useful or not depends on if its useful or not. and induction motor that has no inductance makes for a kind of useless motor, yet it still produces nameplate HP.

its all about the amps, and no free ride for anyone, gotta pay to get on the bus, and the quality of the ride depends on who's driving :thumbsup:


Electric power

Electric power

V is the voltage. Work is defined by:

Therefore

 

[FionaZuppa]

i give up here. if anyone thinks kVAr is power that the poco gets back, then so be it.
post #1 is explained by every classical example of impedance mismatching.
if you wish to understand it in detail under the governance of physics, you need study the problem using terms of eV with LCR components. pushing amps is never free, ever.

cheers, happy physics to all :blink:

 

[wwhitney]

a heater of pure R does pure work

Since we haven't defined work, things are a little unclear. Most definitions of work would hold that heating is not work, but of course it is a transfer of energy. So if you want include 'heating' in your definition of power (as both of us do), you need a definition of power that is broader than just "work/time".

Cheers, Wayne

 

[wwhitney]

pushing amps is never free, ever.

Isn't it free in a superconductor?

Regardless, no one is saying that "pushing amps" is free. What I am saying is that the cost of pushing amps is not measured by vars.

Cheers, Wayne

 

[Carultch]

Since we haven't defined work, things are a little unclear. Most definitions of work would hold that heating is not work, but of course it is a transfer of energy. So if you want include 'heating' in your definition of power (as both of us do), you need a definition of power that is broader than just "work/time".

Cheers, Wayne

Power in general, is any time rate of change, transfer, or conversion, in a quantity of energy. It is not restricted to energy that can classify as work.

 

[jumper]

67 more posts to break into the 600 count list.

 

[Carultch]

Isn't it free in a superconductor?

Regardless, no one is saying that "pushing amps" is free. What I am saying is that the cost of pushing amps is not measured by vars.

Cheers, Wayne

In a perfect superconductor, if you initiate the flow of electrons in a short circuit and then cease the EMF that initially caused their flow, then the electrons will flow continuously with constant current. Just like a body in motion in a theoretical frictionless environment, will continue at a constant velocity. The electrons will not initiate the flow on their own, as there is inertia to get them moving due to both their mass and due to the magnetic fields they create. This is what we measure with the quantity called inductance. The inductance is a circuit's reluctance to change its current, and the resistance is a circuit's reluctance to maintain a current.


In a standard conductor, the Ohm's law voltage is what is required to maintain a steady current, once the initial acceleration of the charges is complete. The cost (power) of pushing amps is not directly equal to the VARS, but it is caused by the VARS. See my example with the 1 kW motor at 120V that has an 80% power factor, supplied by a 1 ohm distribution circuit. The efficiency of the circuit goes from 96.1% (pf = 1) to 92.5% (pf=0.8), due to the lower power factor. The corresponding power loss in the line resistance goes from 68 Watts to 108 Watts due to the power factor, even though it is an example of a load that has 750 VArs.

 

[wwhitney]

Power in general, is any time rate of change, transfer, or conversion, in a quantity of energy. It is not restricted to energy that can classify as work.

To be clear, I agree with you. FionaZuppa proposed it is just "work/time".

Cheers, Wayne

 

[GoldDigger]

In a perfect superconductor, if you initiate the flow of electrons in a short circuit and then cease the EMF that initially caused their flow, then the electrons will flow continuously with constant current. Just like a body in motion in a theoretical frictionless environment, will continue at a constant velocity.

And, entertainingly, one way to get a DC current in a superconducting loop (that you can carry around with you as long as you keep it refrigerated) is to break the superconductivity (not the superconductor) by heating or applying a large localized magnetic field, insert a magnet, restore the superconductivity, and remove the magnet.
No leads or terminals needed.

 

[FionaZuppa]

Isn't it free in a superconductor?

Regardless, no one is saying that "pushing amps" is free. What I am saying is that the cost of pushing amps is not measured by vars.

Cheers, Wayne

there is no pushing in free wheel super conducting, there is no R to convert any of the amps. if you wish to remove energy by way of magnetic coupling, like what is done in a induction motor, then so be it, you'll extract energy until amps = 0. no free ride here either, you have to put some work in to get the trons to start racing. its like a battery at that point, make them race at 200kA, put it in your pocket, power your laptop on the moon.

over, and out.

 

[dionysius]

Anyone able to summarize the content of this thread???????

Anyone able to summarize the content of this thread???????

Anyone brave enough to summarize the content of this thread???????:angel:

 

[jtinge]

Anyone brave enough to summarize the content of this thread???????:angel:

My head hurts!

 

[GoldDigger]

Anyone brave enough to summarize the content of this thread???????:angel:

This is a highly reactive thread whose net information content averages close to zero, but dissipates a lot of heat in the process.

More seriously, this thread has two firmly opposed camps, each of which is content with their own justification and rejects the arguments of the other camp, so I would not want to have to summarize either for fear of being seen by both camps as misstating them.