THE PHYSICS OF... POWER

[FionaZuppa]

Well, after the choice of a unit vector, a one dimensional vector space is "the same thing" as the scalar field. Not sure it is worth distinguishing in this case.


Nope, you're mistaken. Once you choose a direction to represent positive power transfer (e.g. from the "generator" to the "load"), then negative power just means that energy is being transferred in the reverse direction, i.e. from the "load" to the "generator". Which occurs every other quarter cycle with a purely reactive load, so that the net energy transfer is 0.

Cheers, Wayne

a generator never transfers energy back to itself :thumbsup:, the transfer direction of power is to the load, always. I(t) may change direction, transfer of power does not.

net energy transfer can never be zero, if it were you would have zero work.

 

[wwhitney]

a generator never transfers energy back to itself :thumbsup:, the transfer direction of power is to the load, always. I(t) may change direction, transfer of power does not.

Sorry, that's wrong.

net energy transfer can never be zero, if it were you would have zero work.

Indeed, ignoring transmission losses in the conductors, a purely reactive load does no work.

Cheers, Wayne

 

[wwhitney]

next is, show me negative AND positive power between generator and load where the load is consuming 100kW

Well, let's use this trigonometric identity:

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

So suppose I = sin(t) and V = sin(t + c) for some fixed phase angle c. From the identity above:

V = sin(t + c) = sin(t)cos(c) + cos(t)sin(c)

Remember that c is a constant, so this says that V is a mixture of a purely in-phase component [V(t) = sin(t)] and a purely out-of-phase component [V(t) = cos(t)]. We've already seen that the purely-out-of-phase component does no work and can have positive or negative energy transfer each quarter cycle. The purely in-phase component, on the other hand, has positive energy transfer each quarter cycle. [Integrate sin^2(t).]

So pick some value of c close to but not equal to pi/2. The load is almost entirely reactive, the power factor is near to zero. The net energy transfer is positive, but on the quarter cycles where the reactive component has negative energy transfer, that effect dominates. So the net energy transfer is negative for those quarter cycles; its magnitude is just less than that of the positive energy transfer during the alternate quarter cycles.

You asked for 100kW of real power delivered to the load, so now just scale the current or voltage as required. You can also scale time to get the conventional 60Hz waveform.

Cheers, Wayne

 

[FionaZuppa]

Sorry, that's wrong.


Indeed, ignoring transmission losses in the conductors, a purely reactive load does no work.

Cheers, Wayne

we dont ignore R in wires, not practical. but yes, purely reactive does no work. agree.

Mathematically representing power in an AC circuit is a challenge, because the power wave isn’t at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance.

and for clarity, power is energy consumed. purely reactive device consumes no power, and the generator needs to do no work, etc. there is no transfer of energy.


http://www.allaboutcircuits.com/tex...chpt-11/power-resistive-reactive-ac-circuits/

 

[wwhitney]

and for clarity, power is energy consumed.

Power is the rate of energy transfer. Energy is never consumed, just transformed from one variety to another: electrical energy to mechanical energy or heat, etc.

Cheers, Wayne

 

[FionaZuppa]

Power is the rate of energy transfer. Energy is never consumed, just transformed from one variety to another: electrical energy to mechanical energy or heat, etc.

Cheers, Wayne

Energy consumption refers to the amount of energy consumed by an individual or organization, or to the process or system of such consumption. Nearly every modern convenience increases the amount of energy consumed

if a generator is at poco, and in my house i have a big big capacitor with no charge, poco turns on the juice to charge my pure cap, did i get energy transfer and need to pay for that? but did i use any power or transform any energy?

"transfer" and "transformed" are not the same thing.

 

[GoldDigger]

if a generator is at poco, and in my house i have a big big capacitor with no charge, poco turns on the juice to charge my pure cap, did i get energy transfer and need to pay for that? but did i use any power or transform any energy?

"transfer" and "transformed" are not the same thing.

You got no net energy transfer over time except that there will be a very small amount of energy stored in the capacitor averaged over time. You get energy transferred into the capacitor, then transferred back out again, but the average will be non-zero.
Looking at metered consumption, the hypothetical 1kW for .08 seconds to start the process will amount to a negligible number of kWh.

Since residential meters only register real power, the existence of the capacitor, and the high constant (AC) load current associated with it will not change your bill in any way.

With a commercial service where there is often a penalty for high peak kVA it could increase your bill slightly.

To sum up using your words, you get energy transfer and you have a net power consumption for a fraction of the first cycle to account for that.
You transferred electrical energy transmitted over the wire into electrical energy stored in the capacitor, but it is a semantic question whether or not that constitutes transformation. It is still electrical energy, just stored in the form of an electric field inside the capacitor.

 

[jumper]

You all are too smart for this dumb bunny, so let me ask a simple question:

I like useful stuff like the ability to get something done, so is kVars or kVA gonna help get the field plowed like kW can? Nope since I am an AC feller. PF is in play.

Yes the word 'Power" gets into semantics, but what really brings the crops in?

Power is the ability to "Git it done". Watts IMO. The rest is just rhetorical/academic bluster.

I am probably gonna get crucified for this post. Oh well.

 

[FionaZuppa]

You got no net energy transfer over time except that there will be a very small amount of energy stored in the capacitor averaged over time. You get energy transferred into the capacitor, then transferred back out again, but the average will be non-zero.
Looking at metered consumption, the hypothetical 1kW for .08 seconds to start the process will amount to a negligible number of kWh.

Since residential meters only register real power, the existence of the capacitor, and the high constant (AC) load current associated with it will not change your bill in any way.

With a commercial service where there is often a penalty for high peak kVA it could increase your bill slightly.

To sum up using your words, you get energy transfer and you have a net power consumption for a fraction of the first cycle to account for that.
You transferred electrical energy transmitted over the wire into electrical energy stored in the capacitor, but it is a semantic question whether or not that constitutes transformation. It is still electrical energy, just stored in the form of an electric field inside the capacitor.

V*C = what ?? energy, right?
my cap used a lossless full bridge rectifier, once the big cap filled i quickly disconnected it. what am i left with, did i have to pay for that energy that has not yet been consumed, did i use any power ??

 

[GoldDigger]

V*C = what ?? energy, right?
my cap used a lossless full bridge rectifier, once the big cap filled i quickly disconnected it. what am i left with, did i have to pay for that energy that has not yet been consumed, did i use any power ??

With an ideal, zero resistance, zero lead inductance capacitor you had better throw your switch at the moment of a voltage zero crossing. If you then open the switch at the voltage maximum no diodes or other hardware is needed.
And the energy stored in the capacitor is 1/2 CV² (Looks a lot like 1/2 mv² doesn't it?)

You are left with that much energy stored in the capacitor, and you need to find something useful to do with it, or just let it sit there slowly leaking away (in the case of a real rather than a perfect capacitor.)

You have to pay for this energy, but it is so small that it will not be noticeable as even a one cent increase in your electric bill for that month. It will have no effect on later months bills either, even though you are keeping that energy around in storage. You pay to get it, not to keep it. Note that it has been transferred to you and is in your possession even though you chose not to transform it (use it.)

One other thing that you might do is go back to your switch and make the connection at the voltage maximum point in a cycle and open it at the first voltage zero crossing. If you do that you have just sold that energy back to POCO and are left with an empty capacitor.

PS: This is fun. :happyyes:

 

[K8MHZ]

What if I charge up a capacitor and then use it like a battery to supply power to a light bulb?

Is there no power transfer from the generation source to the cap, but then there is some from the cap to the light bulb?

 

[FionaZuppa]

What if I charge up a capacitor and then use it like a battery to supply power to a light bulb?

Is there no power transfer from the generation source to the cap, but then there is some from the cap to the light bulb?

you initially had transfer (no power), then you had transformation (power). :thumbsup:

here's another riddle for the team, during the transfer did the generator do any work?

 

[GoldDigger]

you initially had transfer (no power), then you had transformation (power). :thumbsup:

here's another riddle for the team, during the transfer did the generator do any work?

Actually, using the definition that POCO relies on, you did get power from them while you were charging the cap.

 

[FionaZuppa]

Actually, using the definition that POCO relies on, you did get power from them while you were charging the cap.

my cap (load) used no power :happysad:

 

[jumper]

Okay, no response to my my post. Hmmmm.

I will just fade away since I guess I am way off track, not IMO.

True, real, active power is what makes world go round.

That's my story and I'm sticking to it.

 

[wwhitney]

my cap (load) used no power :happysad:

Power is just the rate of transfer of energy. The meter just measures the power transfer at that point in the circuit. Your cap was charged up, meaning power was transferred from the POCO, to the cap, through the meter. So the POCO bills you for that.

Cheers, Wayne

 

[GoldDigger]

my cap (load) used no power :happysad:

It all depends on the meaning of "used" you used*.

You did not transform the energy you got from POCO into any other form of energy or do "useful" work. However you did do work on the electrons in the capacitor, to store energy in potential form, just as you do work on a weight to raise it in a gravitational field.

At some point it does not make sense to argue about whether or not you used or transformed the energy, by applying a narrow definition of power.

For me the possession of that energy and the ability to exploit it in the future has been transferred from POCO to you. And POCO has paid for fuel for its generators to allow that transfer to take place. From their point of view they have delivered power in the process of transferring that energy. No other way to interpret it unless you want your "power" bill, which is actually an energy bill, to be held in suspension while POCO determines what use you have made of the energy they sent you.

A definition of power as the time rate of doing work is far too limiting for our modern world in which there are so many ways of doing work other than just mechanical work. I prefer the definition of power as the rate of transfer of energy (across some specified boundary or interface), regardless of the source or ultimate use of that energy.




*With apologies to Bill Clinton.

 

[Carultch]

It all depends on the meaning of "used" you used*.

You did not transform the energy you got from POCO into any other form of energy or do "useful" work. However you did do work on the electrons in the capacitor, to store energy in potential form, just as you do work on a weight to raise it in a gravitational field.

At some point it does not make sense to argue about whether or not you used or transformed the energy, by applying a narrow definition of power.

For me the possession of that energy and the ability to exploit it in the future has been transferred from POCO to you. And POCO has paid for fuel for its generators to allow that transfer to take place. From their point of view they have delivered power in the process of transferring that energy. No other way to interpret it unless you want your "power" bill, which is actually an energy bill, to be held in suspension while POCO determines what use you have made of the energy they sent you.

A definition of power as the time rate of doing work is far too limiting for our modern world in which there are so many ways of doing work other than just mechanical work. I prefer the definition of power as the rate of transfer of energy (across some specified boundary or interface), regardless of the source or ultimate use of that energy.




*With apologies to Bill Clinton.

On net over the course of a cycle, the ideal capacitors release just as much energy that is used for charging them. Same for the inductors via energy stored in the magnetic field.

Think about it like carrying a heavy backpack up and down a mountain. You put energy in to the backpack as you climb, and the backpack returns that energy back to you as you descend. On net, the work you actually do, is delivered to heat in some form or another. Human muscle stopping forces and friction.

 

[FionaZuppa]

It all depends on the meaning of "used" you used*.

You did not transform the energy you got from POCO into any other form of energy or do "useful" work. However you did do work on the electrons in the capacitor, to store energy in potential form, just as you do work on a weight to raise it in a gravitational field.

my device did no work at all, zero. the generator may have done some work though.

 

[GoldDigger]

On net over the course of a cycle, the ideal capacitors release just as much energy that is used for charging them. Same for the inductors via energy stored in the magnetic field.

Think about it like carrying a heavy backpack up and down a mountain. You put energy in to the backpack as you climb, and the backpack returns that energy back to you as you descend. On net, the work you actually do, is delivered to heat in some form or another. Human muscle stopping forces and friction.

A useful description, but not relevant to the specific case that FZ and I were discussing, namely connecting a capacitor just long enough to charge it and then disconnecting.
Perhaps carrying the backpack to the top of the hill and leaving it there. Since you just left the backpack lying around when you were done, did you do work? Of course. Was work done on the backpack? Of course.
But FZ would seem to be arguing that since that energy was not actually used these energy transfers did not involve any power. I disagree.