Power factor and VA vs Watts

[Besoeker]

I understand power is not energy (and vice-versa). See post #357
Yet you make this comment:

How much of that power is realized energy (i.e. converted to another form of energy

I think this is where we are having a major snafu...

One volt times one ampere occuring at the same instant is one volt-ampere. Until we include such conditions or parameters which indicate work is performed or energy converted at that same instant, it cannot be one watt.

It is one Watt.
Let me try yet again.....
Suppose you have one Volt DC and one Amp dc you would agree, would you not, that this is one Watt?
Does the duration of that 1V and 1A make it anything other than 1W?

 

[ggunn]

Let me try yet again.....
Suppose you have one Volt DC and one Amp dc you would agree, would you not, that this is one Watt?
Does the duration of that 1V and 1A make it anything other than 1W?

Here's your hammer and nails. Please affix this here lump of Jello to yonder wall. :grin:

 

[Smart $]

Exactly. Or do you not understand how it supports my previous statement, hand-in-glove?.

Oh, I understand... and practical vs. theoretical is another matter not yet discussed. I have not said any of my assertions are meant to be a matter of practicality. And I am aware of how samples outside of standard deviation are determined. Yes, measurement instruments have limitations on accuracy and consistency, but we don't use them in theoretical discussions.

Yet it seems I am not the only one that looks at instantaneous power as meaningful. Ham' has quoted or paraphrased (as has Gar) and attached images from several texts Additionally, though I do not give wikipedia any major creedance, but its page on Power (physics) puts this statement up top in the opening section...

The instantaneous power is then the limiting value of the average power as the time interval Δt approaches zero.

 

[Smart $]

I understand power is not energy (and vice-versa). See post #357
Yet you make this comment:

How much of that power is realized energy (i.e. converted to another form of energy, or does work [otherwise known as real power])?

Even you said energy is power times time. So even by your definition of power, energy and power are directly related where time is involved. And while my question may not be worded technically correct, I feel like you are using this to ignore the gist of the question...

It is one Watt.
Let me try yet again.....
Suppose you have one Volt DC and one Amp dc you would agree, would you not, that this is one Watt?
Does the duration of that 1V and 1A make it anything other than 1W?

A circuit being DC does not make any difference. It is still one volt-ampere if no work is done or no energy is converted to another form.

 

[Besoeker]

A circuit being DC does not make any difference. It is still one volt-ampere if no work is done or no energy is converted to another form.

OK. Care to give an example of a DC circuit where no work is done when you have V Volts and I Amps yet no Watts?

 

[Smart $]

OK. Care to give an example of a DC circuit where no work is done when you have V Volts and I Amps yet no Watts?

Charging, discharging a capacitor.

 

[Besoeker]

Charging, discharging a capacitor.

Work is done to charge it.
Where do you think the 0.5CV^2 comes from?

 

[Smart $]

Work is done to charge it.
Where do you think the 0.5CV^2 comes from?

Circuit-wise, it's a transfer of energy. Yes, there is work done, but it is by the source, not the circuit or the capacitor itself (load). Typically for our discussions here of circuits and their loads, we consider the source as a "black box". We generally don't consider the work done by the POCO generators... do we?

 

[Besoeker]

Circuit-wise, it's a transfer of energy. Yes, there is work done, but it is by the source, not the circuit or the capacitor itself (load).

Good but why the but?
Work is done by the source.
Now suppose the source is 1Vdc and the current is 1Adc and it flows for 1 second that would be one Joule of energy.
With that so far?

 

[Smart $]

Good but why the but?
Work is done by the source.
Now suppose the source is 1Vdc and the current is 1Adc and it flows for 1 second that would be one Joule of energy.
With that so far?

Yes. That is one watt of power and one watt-second of energy produced (converted). Provided the 1A is constant for the duration is indicative of one watt-second also being consumed (likely as a resistive circuit which converts the electrical energy to heat).

What is the point? Our concern in this theoretical discussion is not how much energy is produced, but how much is consumed. In other words, our concern is what the kwh meter on the service registers.

 

[Besoeker]

Yes. That is one watt of power and one watt-second of energy produced (converted). Provided the 1A is constant for the duration is indicative of one watt-second also being consumed (likely as a resistive circuit which converts the electrical energy to heat).

It could be a number of kinds of circuit...a motor converting it to mechanical, a battery charger converting it to chemical, plating, welding....etc.
The point is that the rate of conversion is 1W.
Still with this?

 

[Smart $]

It could be a number of kinds of circuit...a motor converting it to mechanical, a battery charger converting it to chemical, plating, welding....etc.
The point is that the rate of conversion is 1W.
Still with this?

Yes, sir, I am: 1 watt-second produced; 1 watt-second consumed.

I suppose I'll have to follow your paradigm. It seems any context I bring up outside of it you ignore.

 

[Besoeker]

Yes, sir, I am: 1 watt-second produced; 1 watt-second consumed.

Do you then agree that the rate of conversion is one Watt?

 

[LarryFine]

Do you then agree that the rate of conversion is one Watt?

Wouldn't a rate require specifying "one watt per (time interval)"?

 

[Besoeker]

Wouldn't a rate require specifying "one watt per (time interval)"?

The Watt is a rate.
Joules per second.

 

[jghrist]

Circuit-wise, it's a transfer of energy. Yes, there is work done, but it is by the source, not the circuit or the capacitor itself (load). Typically for our discussions here of circuits and their loads, we consider the source as a "black box". We generally don't consider the work done by the POCO generators... do we?

OK, let's take that capacitor that was charged up with no work done. Now, discharge it through a resistor. It makes heat, right? Where did the heat come from?

With an ac source connected to the capacitor, work is done by the source, and potential energy in the capacitor is increased during one half cycle, and then during the other half cycle, the potential energy of the capacitor is used to power the source.

I've been semi-following this thread and it seems to me that the main bone of contention is whether or not it is proper to consider real and reactive power to be sinusoidal. It seems that it is possible to consider it this way, where the real power is a sinusoid with a minimum peak at zero. I see no advantage to this construct, however, other than to make interesting graphs. Real power is the average value of the instantaneous power V?I. It is basically a scalar quantity and I see no advantage of treating it otherwise.

 

[Smart $]

Do you then agree that the rate of conversion is one Watt?

Yes, sir. A constant one watt produced for one second and a consistent one watt consumed for one second. The produced power is one watt. The consumed power is one watt. These are instantaneous values as well as average values.

 

[Smart $]

OK, let's take that capacitor that was charged up with no work done. Now, discharge it through a resistor. It makes heat, right? Where did the heat come from?

The heat comes from enrgy conversion. In the case where the capacitor is the source of energy (already charged), it does no work and converts no energy. The energy is converted by the resistor, not the capacitor. Of course we are discussing a theoretical (aka ideal) capacitor where there are no losses involved.

With an ac source connected to the capacitor, work is done by the source, and potential energy in the capacitor is increased during one half cycle, and then during the other half cycle, the potential energy of the capacitor is used to power the source.

Quite correct. But when the capacitor "powers" the source, it is only returning energy produced by the source. It does no work or conversion to perform this function.

I've been semi-following this thread and it seems to me that the main bone of contention is whether or not it is proper to consider real and reactive power to be sinusoidal. It seems that it is possible to consider it this way, where the real power is a sinusoid with a minimum peak at zero. I see no advantage to this construct, however, other than to make interesting graphs. Real power is the average value of the instantaneous power V?I. It is basically a scalar quantity and I see no advantage of treating it otherwise.

I believe the real advantage is in the understanding of the concept of apparent, real, and reactive power (also the terminology and units of measure). IMO, that is what the OP asked some 370 plus posts ago.

It is common to see the formula W = E ? I ? pf. Yet very few common texts provide a treatise on the subject. With all the rhetoric flying around in this thread, it is no wonder

 

[rattus]

For what its worth:

For what its worth:

OK, let's take that capacitor that was charged up with no work done. Now, discharge it through a resistor. It makes heat, right? Where did the heat come from?

With an ac source connected to the capacitor, work is done by the source, and potential energy in the capacitor is increased during one half cycle, and then during the other half cycle, the potential energy of the capacitor is used to power the source.

I've been semi-following this thread and it seems to me that the main bone of contention is whether or not it is proper to consider real and reactive power to be sinusoidal. It seems that it is possible to consider it this way, where the real power is a sinusoid with a minimum peak at zero. I see no advantage to this construct, however, other than to make interesting graphs. Real power is the average value of the instantaneous power V?I. It is basically a scalar quantity and I see no advantage of treating it otherwise.

Must nit-pick a bit. Energy flows into the cap for a quarter cycle, then energy flows back to the source in the next quarter cycle. That is the exchange occurs at the rate of 2wt and is sinusoidal.

Although we aren't normally concerned with instantaneous power, I believe that understanding the math helps one understand average power.

 

[Hameedulla-Ekhlas]

The heat comes from enrgy conversion. In the case where the capacitor is the source of energy (already charged), it does no work and converts no energy. The energy is converted by the resistor, not the capacitor. Of course we are discussing a theoretical (aka ideal) capacitor where there are no losses involved.


Quite correct. But when the capacitor "powers" the source, it is only returning energy produced by the source. It does no work or conversion to perform this function.


I believe the real advantage is in the understanding of the concept of apparent, real, and reactive power (also the terminology and units of measure). IMO, that is what the OP asked some 370 plus posts ago.

It is common to see the formula W = E ? I ? pf. Yet very few common texts provide a treatise on the subject. With all the rhetoric flying around in this thread, it is no wonder

Smart$,
what does the red letter represent in above formula?