Power factor and VA vs Watts

[LarryFine]

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

Watts, power.


(I know that much. )

 

[Hameedulla-Ekhlas]

Watts, power.


(I know that much. )

Yes, the formula is correct. It is watt but I just asked about the letter W. He has shown with W and W also represents the Work. It should be represented by P not W :grin:

 

[rattus]

Yes, the formula is correct. It is watt but I just asked about the letter W. He has shown with W and W also represents the Work. It should be represented by P not W :grin:

The equation is poorly written and vague.

Depending on the units of V and I, the unit of power may be a milliwatt, kilowatt, megawatt, etc.

Formulas should be written with symbols for the electrical quantities not the units of measure.

Units should be included when evaluating the formula. e.g.

P = 120V * 10A * 0.9 = 1080W = 1.08KW

 

[Hameedulla-Ekhlas]

The equation is poorly written and vague.

Depending on the units of V and I, the unit of power may be a milliwatt, kilowatt, megawatt, etc.

Formulas should be written with symbols for the electrical quantities not the units of measure.

Units should be included when evaluating the formula. e.g.

P = 120V * 10A * 0.9 = 1080W = 1.08KW

exactly, I am agree with you.

 

[Smart $]

Yes, the formula is correct. It is watt but I just asked about the letter W. He has shown with W and W also represents the Work. It should be represented by P not W :grin:

Agreed. Yet I did not write the formula. It is copied from a text.

 

[Hameedulla-Ekhlas]

Agreed. Yet I did not write the formula. It is written as such in many texts.

oh, good text

 

[Smart $]

oh, good text

American Electrician's Handbook

 

[Smart $]

...

Units should be included when evaluating the formula. e.g.

P = 120V * 10A * 0.9 = 1080W = 1.08KW

If you're going to include units, should you not also include power factor (pf or PF)?

Granted it is a unitless value in most usage forms, but in truth it does represent the ratio watts/volt-amperes and thus yields the appropriate units to the result.

 

[Smart $]

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.

...

I believe he was referring to a half-cycle of power, which would be a more accurate description, since voltage and current are, in a theoretical context, 90? out of phase with each other.

 

[Besoeker]

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.

Good man.
The instantaneous value of power is instantaneous voltage times instantaneous current.
Still with this?

 

[wasasparky]

Do threads automatically close after 400 replies?

 

[Smart $]

Good man.
The instantaneous value of power is instantaneous voltage times instantaneous current.
Still with this?

Well this is where the straw broke the camel's back....

If your question is limited to our hypothetical one watt-second produced and consumed scenario, then the answer is yes.

If not, the answer is dependent on whether any part of the instantaneous product is VAR (volt-amperes reactive). If there is no VAR present, then the answer is also yes. If there is VAR present, the answer is no.

 

[Smart $]

Do threads automatically close after 400 replies?

I don't know... but we'll find out before long

 

[Besoeker]

Well this is where the straw broke the camel's back....

If your question is limited to our hypothetical one watt-second produced and consumed scenario, then the answer is yes.

If not, the answer is dependent on whether any part of the instantaneous product is VAR (volt-amperes reactive). If there is no VAR present, then the answer is also yes. If there is VAR present, the answer is no.

My point was about the instantaneous power.
The instantaneous product of current and voltage is the power at that instant. I don't know why you would reject that. VAR is irrelevant in terms instantaneous values.

 

[rattus]

If you're going to include units, should you not also include power factor (pf or PF)?

Granted it is a unitless value in most usage forms, but in truth it does represent the ratio watts/volt-amperes and thus yields the appropriate units to the result.

One could include W/VA, although I have never seen it in texts. One could also use degrees and radians in conversion factors.

 

[Hameedulla-Ekhlas]

If you're going to include units, should you not also include power factor (pf or PF)?

Granted it is a unitless value in most usage forms, but in truth it does represent the ratio watts/volt-amperes and thus yields the appropriate units to the result.

Smart$:
Can you please make it a little bit clear saying " Yields the appropriate units to the result"

If I am not wrong, you mean it is only for unit cancellation. Am I right?

 

[Smart $]

My point was about the instantaneous power.
The instantaneous product of current and voltage is the power at that instant. I don't know why you would reject that. VAR is irrelevant in terms instantaneous values.

Says who? ...and even more so, why?

Instantaneous VAR is relevant. If you are saying it only matters over time, all we are doing over time is using a multitude of instances. Yes, volts times amperes is representive of power present at that instant, but it is of power produced. Our concern in this discussion, however, is power consumed.

If any portion of the volt-ampere product, even in an instant, represents volt-amperes reactive, not all the power at that instant is consumed. Just because we generally confine our calculations to using effective values over specified time intervals does not make the physics of the issue change.

 

[Smart $]

Smart$:
Can you please make it a little bit clear saying " Yields the appropriate units to the result"

If I am not wrong, you mean it is only for unit cancellation. Am I right?

Yes, if you include units in the formula (or calculaiton therewith), units not of the result must cancel within the calculation.

 

[Hameedulla-Ekhlas]

If you're going to include units, should you not also include power factor (pf or PF)?

Granted it is a unitless value in most usage forms, but in truth it does represent the ratio watts/volt-amperes and thus yields the appropriate units to the result.

Originally Posted by rattus
...

Units should be included when evaluating the formula. e.g.

P = 120V * 10A * 0.9 = 1080W = 1.08KW

It is ok and p.f is just a ratio and ratio is always unitless.

they also cancel each other, if you want I will do it.
we know that S = P only if p.f = 1 and here we have 0.9 not one.
it means 90% is real power and 10% is reactive power. So, since we multiply it by 0.9 it means we get pure real power without reactive. we know when there is no reactive, V-A = W.
So, 0.9 represent a ratio and I think it is correct. Am not I correct?

units not of the result must cancel within the calculation

The unit cacels each other and we get proper unit ( W, V-A, Var)

 

[Besoeker]

Yes, volts times amperes is representive of power present at that instant,

It is power at that instant. No representative about it.
You accepted that a constant 1V times 1A for one second was 1 J.
Rightly so.
You even volunteered that the instantaneous rate was 1W.
Again, rightly so.
So let's break the one second down into a million 1us segments.
It is still 1W in every segment.
Now, any one segment could be part of an ac waveform or continuous DC.
Over that sort of time interval on a system operating at power frequency the difference between ac and dc would not matter.
Still with me?