THE PHYSICS OF... POWER

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dionysius said:
Anyone brave enough to summarize the content of this thread???????:angel:
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Sure. It's a protracted and occasionally vitriolic argument about how many angels can dance on the head of a pin where the participants cannot agree on a numbering system or the definition of "angel" or "pin". :D
 
Zeno Paradox

Zeno Paradox

Here is Zeno's paradox in Energy version: Energy is an integral of power. So energy at any given instant is zero. Time is made up of instants. Hence energy is always zero even when power is nonzero.
 
winnie said:
A 1W load consumes 1 J per second.
A 60Hz 1VA pure reactive load involves something on the order of 0.003J shuttling back and forth.

-Jon
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i not following what you mean here, can you explain with more details.
 
winnie said:

A 1W load consumes 1 J per second.
A 60Hz 1VA pure reactive load involves something on the order of 0.003J shuttling back and forth.
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I wanted to highlight this, as it is so important. The calculation is (2/pi) * (1/4) * (1/60) = 0.00265 joules shuttling back and forth from 1VA at 60Hz

FionaZuppa said:
i not following what you mean here, can you explain with more details.
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At 60 Hz, there are 60 cycles a second, or 240 quarter cycles a second.

A 1W load receives 0.00417 joules (1/240 J) every quarter cycle. Add them up over a second, and you get 1 joule to the load.

With a 1VA (edit: purely reactive) load, in the first quarter cycle, 0.00265 joules flows towards the load. In the second quarter cycle, 0.00265 joules flows away from the load. Etc. Add up them up over a second, and you get 0 joules to the load. At any give point in time, the net energy surplus/deficit to the load is at most 0.00265 joules. The amount of energy shuttling back and forth is 0.00265 joules.

Cheers, Wayne
 
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Smart $ said:
Mathematical fabrications to determine what is not power. :lol:
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Actually this is a good conclusion. A different plane in multi-dimensional space has been selected to express that different quantity. It has been called the complex plane.
 
wwhitney said:
At 60 Hz, there are 60 cycles a second, or 240 quarter cycles a second.

A 1W load receives 0.00417 joules (1/240 J) every quarter cycle. Add them up over a second, and you get 1 joule to the load.

With a 1VA (edit: purely reactive) load, in the first quarter cycle, 0.00265 joules flows towards the load. In the second quarter cycle, 0.00265 joules flows away from the load. Etc. Add up them up over a second, and you get 0 joules to the load. At any give point in time, the net energy surplus/deficit to the load is at most 0.00265 joules. The amount of energy shuttling back and forth is 0.00265 joules.

Cheers, Wayne
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thanks wayne for the details.
the view however is a load in a box and you are describing the box, which is perfectly valid, but is not the system view.

with any given voltage how much work (amps) does it take to push that much energy into a reactive XL component?
 
wwhitney said:
Energy is conserved, so if 0.00265 joules is shuttling back and forth, then it takes 0.00265 joules to get the process started.

Cheers, Wayne
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what real system is this model of yours? this J/s moves around free of penalty?
 
Here's a semi-realistic example, which I'm going to work out because I'm curious about the answer:

Say we have a portable 120V 60 Hz AC gasoline generator connected to a 1 kW load through conductors with a total resistance of 0.1 ohms. If the load has power factor 1, the load current is 1000/120=8 1/3 amps, and the resistive loss in the conductors is I^2*R = (25/3)^2 * 0.1 = 6.94 W. The total generator electrical output is 1007 W.

If instead the load power factor is 2/3 (0.667), then the load current is 1000/(120*2/3) = 12.5 amps, and resistive loss in the conductors is I^2*R = (25/2)^2*0.1 = 15.6 W. The total generator electrical output is 1016 W, the increase is 5.7 W. The generator fuel consumption goes up by 5.7/1007 = 0.57% compared to power factor 1.

So in this particular case, supporting the 1.12 kVAr when the power factor is 2/3 requires 5.7 W of additional power due to the resistive losses.

Cheers, Wayne
 
FionaZuppa said:
what real system is this model of yours? this J/s moves around free of penalty?
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Yes, 0 cost is a good approximation. But I can be a bit more accurate:

1VA load at 120V 60 Hz, with power factor 0 or 1. Say the transmission conductors have a resistance of 0.1 ohm. The load current in either case is 1/120 amps, so the resistive losses are (1/120)^2 * 0.1 = 6.94 microWatts. Each quarter cycle (1/240 second), that's 28.9 nanoJoules of heating loss.

For the case of power factor 1, each quarter cycle the load receives 4.166667 milliJoules of energy, the source proves 4.166696 milliJoules of energy, and 29 nanoJoules of heating occurs.

For the case of power factor 0, each quarter cycle 2.652582 milliJoules of energy is being shuffled one way or the other. The source is provideing 29 nanoJoules of energy for the resistive losses.

As you can see, in this case the losses are a rounding error. 0 is a good answer.

Cheers, Wayne
 
wwhitney said:
Yes, 0 cost is a good approximation. But I can be a bit more accurate:

1VA load at 120V 60 Hz, with power factor 0 or 1. Say the transmission conductors have a resistance of 0.1 ohm. The load current in either case is 1/120 amps, so the resistive losses are (1/120)^2 * 0.1 = 6.94 microWatts. Each quarter cycle (1/240 second), that's 28.9 nanoJoules of heating loss.

For the case of power factor 1, each quarter cycle the load receives 4.166667 milliJoules of energy, the source proves 4.166696 milliJoules of energy, and 29 nanoJoules of heating occurs.

For the case of power factor 0, each quarter cycle 2.652582 milliJoules of energy is being shuffled one way or the other. The source is provideing 29 nanoJoules of energy for the resistive losses.

As you can see, in this case the losses are a rounding error. 0 is a good answer.

Cheers, Wayne
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If a load has a power factor of zero, then by definition, it has zero resistance, and thus has no zero real power associated with it. In practice, this kind of load only exists in superconductors. So most highly reactive loads only have near-zero power factor, and not actually zero.
 
wwhitney said:
Here's a semi-realistic example, which I'm going to work out because I'm curious about the answer:

Say we have a portable 120V 60 Hz AC gasoline generator connected to a 1 kW load through conductors with a total resistance of 0.1 ohms. If the load has power factor 1, the load current is 1000/120=8 1/3 amps, and the resistive loss in the conductors is I^2*R = (25/3)^2 * 0.1 = 6.94 W. The total generator electrical output is 1007 W.

If instead the load power factor is 2/3 (0.667), then the load current is 1000/(120*2/3) = 12.5 amps, and resistive loss in the conductors is I^2*R = (25/2)^2*0.1 = 15.6 W. The total generator electrical output is 1016 W, the increase is 5.7 W. The generator fuel consumption goes up by 5.7/1007 = 0.57% compared to power factor 1.

So in this particular case, supporting the 1.12 kVAr when the power factor is 2/3 requires 5.7 W of additional power due to the resistive losses.

Cheers, Wayne
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PF =1, gen power is 1007W
PF =0.667, gen power is 1016W

the diff is 9W. you said only +5.7W was needed??
 
FionaZuppa said:
PF =1, gen power is 1007W
PF =0.667, gen power is 1016W

the diff is 9W. you said only +5.7W was needed??
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Good catch, I made a subtraction error. The correct answer is 8.7W of resistive loss to support the 1.12 kVAr in my example, (edit: and 0.87% increased fuel consumption).

Cheers, Wayne
 
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Carultch said:
So most highly reactive loads only have near-zero power factor, and not actually zero.
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Sure, I agree, any non superconductive load will have some resistance. But including that in the example would just complicate things without clarifying anything. So let's just assume the load resistance << the conductor resistance (0.1 ohms in my example), so we can treat it as zero.

Cheers, Wayne
 
wwhitney said:
Good catch, I made a subtraction error. The correct answer is 8.7W of resistive loss to support the 1.12 kVAr in my example, (edit: and 0.87% increased fuel consumption).

Cheers, Wayne
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right, so the diff bewteen PF's means you needed to do more work on the gen side so you can deliver same Pwatts of real power to the load. so in reality, the energy associated with XL is wasted power. bad PF simply means more work to take a "kW package" from src to load.

its like UPS delivering to me daily a box. the box contains Ywork of energy everytime. UPS says "we will charge you $1 to deliver this box to you". everyone is happy until suddenly my office is 10 stories up and there is only stairs to get there. now all of a sudden UPS has to do more work to deliver the same box.

its takes work for UPS guy to climb the stairs against force of gravity, he reached 10th floor, now has a bunch of potential energy. that energy is also wasted energy because on his way down every step down is just a tad of heat, when he gets back to ground level all of the potential is gone, he is not standing there with the energy he put in to climb the stairs, its all wasted. he put in 1kJ to climb the stairs, he loses 1kJ on his way down.

climbing stairs and back down again is same as the reactance power Q. wasted, gone.
 
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