Power factor and VA vs Watts

[Smart $]

again you are wrong with 1W/VA
You have not completed what you want to proof continue your equation
cosx = Watt/VA = (I * V) / (I * V) = 1
here all unit cancels each other and nothing remains.

you are thinking it as acceleration formula = v (velocity)/t(time) = v/sec but it is not like this nor it is a constant to have a unit.




Did you not read it completely that I have proved it in previous post
V* I * cosx ( represents real power) = V* I ( represents apparent power) ; now we will write it like this

p.f = cosx = ( V * I ) / ( V * I ) = 1 -----------2 no unit
you see we got pure value without any W/V-A. That is why I am saying we can not put W/V-A in our equation like Eq.1 which is written by you.

Your still not getting it. You wrote....

cosx = Watt/VA = (I * V) / (I * V) = 1​

For one, you cannot have cos(φ) on only one side of the equation. It should be...

cosx = (I * V * cosx) / (I * V) = ???​

..and two, you are starting with a unitless ratio by itself on one side of the equation. This is a ploy to force the value on the other side of the equation to also equal a unitless ratio. It cannot be done without manipulating the terms, thus making it invalid.

More succinctly, it should be...

Power Factor = Watts/Volt-amperes = (V * I * cos(φ)) / (V * I) = cos(φ) * 1 Watt/VA​

 

[Hameedulla-Ekhlas]

Power Factor = Watts/Volt-amperes = (V * I * cos(φ)) / (V * I) = cos(φ) * 1 Watt/VA

Ok now you proved the p.f = cos(φ) * 1 Watt/VA. I am not agree only with Watt/VA which you have written.
see now your own equation result

Power Factor = Watts/Volt-amperes = (V * I * cos(φ)) / (V * I) = cos(φ). I cancelled Ampere by Ampere and voltage by voltage and result is just cos(φ). Am I wrong?

If you are insisting that you can right like that please give any book reference then I will accept you. You can find in many books just cosx = p.f without Watt/V-A

 

[Hameedulla-Ekhlas]

Your still not getting it. You wrote....
cosx = Watt/VA = (I * V) / (I * V) = 1
For one, you cannot have cos(φ) on only one side of the equation. It should be...

Here I have just considered resistive and proved that cosx = 1 or p.f = 1

cosx = (I * V * cosx) / (I * V) = ???[/

Here again if you consider it resistive you can get the result cosx = 1 or p.f = 1.

 

[Hameedulla-Ekhlas]

More succinctly, it should be...
Power Factor = Watts/Volt-amperes = (V * I * cos(φ)) / (V * I) = cos(φ) * 1 Watt/VA

no you are wrong again and again by writing power factor like this.

You have already read the wikipedia which states clearly
"The power factor of an AC electric power system is defined as the ratio of the real power flowing to the load to the apparent power,[1][2] and is a dimensionless number between 0 and 1 (frequently expressed as a percentage, e.g. 0.5 pf = 50% pf)."

Now still if you are not agree lets do it in other method to proof that. Suppose you have a simple circuit.

Apparent power * p.f = Real power
now therefore
p.f = Real power / Apparent power

Going one step further, another formula for power factor can be developed. By substituting the equations for true power and apparent power in the formula for power factor, you get:

p.f = [(I)^2 * R] / [(I)^2 * Z ]

Since current in a series circuit is the same in all parts of the circuit, current in resistor equals current in impedance. Therefore, in a series circuit,

p.f = R / Z

by contraction this you can get only and only a number or a decimal number nothing else with.

Tell me can you write p.f = cosx = R / Z *1 Watt/ V-A? Ofcourse not adding 1Watt / V-A is a wrong idea.

Hope to get agree now.

 

[Smart $]

no you are wrong again and again ...

Gee-zow-wee

...Hope to get agree now.

It is not this big of an issue, Ham'. Simply put, in electrical terms, the defined relationship is:

Watts = Volts ? Amperes ? Power Factor​

or transposed

Power Factor = Watts ? (Volts ? Amperes)​

You should know by now, no matter how you try to put across that Power Factor is a unitless ratio, I can use the definition by equation to prove it is in units of W/VA. There comes a point where one accepts the definition by equation as a universal truth and moves on...

If you want to consider this an agreement, then by all means please do so

 

[Hameedulla-Ekhlas]

Gee-zow-wee


It is not this big of an issue, Ham'. Simply put, in electrical terms, the defined relationship is:

Watts = Volts ? Amperes ? Power Factor​

or transposed

Power Factor = Watts ? (Volts ? Amperes)​

You should know by now, no matter how you try to put across that Power Factor is a unitless ratio, I can use the definition by equation to prove it is in units of W/VA. There comes a point where one accepts the definition by equation as a universal truth and moves on...

If you want to consider this an agreement, then by all means please do so

Ok anyway. But tell me when this thread is going to finish. I think Never

 

[Smart $]

Ok anyway. But tell me when this thread is going to finish. I think Never

Well, never is a pretty deep subject... so let's not digress

If nothing else, this thread certainly has provided me with a good bit of entertainment :roll:

 

[weressl]

Gee-zow-wee


It is not this big of an issue, Ham'. Simply put, in electrical terms, the defined relationship is:

Watts = Volts ? Amperes ? Power Factor
​

or transposed

Power Factor = Watts ? (Volts ? Amperes)
​

You should know by now, no matter how you try to put across that Power Factor is a unitless ratio, I can use the definition by equation to prove it is in units of W/VA. There comes a point where one accepts the definition by equation as a universal truth and moves on...

If you want to consider this an agreement, then by all means please do so

Yep, we are ALL in agreement that you're wrong.

 

[jghrist]

For a little light reading...

http://en.wikipedia.org/wiki/AC_power

From that source:

If the load is purely reactive, then the voltage and current are 90? degrees out of phase. For half of each cycle, the product of voltage and current is positive, but on the other half of the cycle, the product is negative, indicating that on average, exactly as much energy flows toward the load as flows back. There is no net energy flow over one cycle. In this case, only reactive energy flows?there is no net transfer of energy to the load.

There is energy flow with a reactive load. It just averages to zero over a full cycle.

This goes along with a definition of real energy being the average of the instantaneous power. For reactive load, the instantaneous power is positive for 1/2 cycle and negative for 1/2 cycle.

Just for fun, graph I??R for a purely resistive circuit. Compare this to the previous graphs of real power.

 

[Hameedulla-Ekhlas]

Gee-zow-wee


It is not this big of an issue, Ham'. Simply put, in electrical terms, the defined relationship is:

Watts = Volts ? Amperes ? Power Factor​

or transposed

Power Factor = Watts ? (Volts ? Amperes)​

You should know by now, no matter how you try to put across that Power Factor is a unitless ratio, . There comes a point where one accepts the definition by equation as a universal truth and moves on...

If you want to consider this an agreement, then by all means please do so

I am agree with you to accept every defination by equation too.

Power Factor = Watts ? (Volts ? Amperes) = any number from 0 to 1

I can use the definition by equation to prove it is in units of W/VA

See below if I expand your power factor equation what it becomes and even by equation you will not accept it.

Originally Posted by Smart $

More succinctly, it should be...
Power Factor = Watts/Volt-amperes = (V * I * cos(φ)) / (V * I) = cos(φ) * 1 Watt/VA

Now you proved by equation and see I will expand it what it becomes then tell me will you accept it or not.
power factor = cos(φ) * 1 Watt/VA = cos(φ) * [ V * I * cos(φ) ] / [ V * I ] ; after voltage and current cacellation it becomes
power factor = cos(φ) ^ 2 ; Will you accept through equation deffination. When something is wrong, we even can not define it through equation too.



Ok anyway. But tell me when this thread is going to finish. I think Never

 

[ggunn]

Ok anyway. But tell me when this thread is going to finish. I think Never

Pass the popcorn, please.

 

[Besoeker]

The concept that instantaneous power is either reactive or real does not fit any known math model currently in use.

I did not say either/or. I said one, the other, or a combination thereof.

But it isn't. And you can't seem to grasp that.
One Volt and one Amp occurring at the same instant is one Watt.
Just that. One Watt. Real or reactive is totally meaningless in this context.

 

[rattus]

My book says:

My book says:

But it isn't. And you can't seem to grasp that.
One Volt and one Amp occurring at the same instant is one Watt.
Just that. One Watt. Real or reactive is totally meaningless in this context.

My book defines the vi product simply as power, that is:

p(t) in watts = v(t) in volts * i(t) in amps

 

[Cold Fusion]

...One Volt and one Amp occurring at the same instant is one Watt. Just that. One Watt. Real or reactive is totally meaningless in this context.

...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.

These two statements highlight the difference in understanding.

...It is really quite simple, and you would have gotten the gist if you would quit trying to define it in mathematical terms rather than what it is. ...

...You guys really need to get a grip on the english language... but I guess you have a more universal language and consider english but a mutterance. ...

...That is correct. I'm not referencing any book. I write what is in my head. ...

And these three highlight why this difference is occuring.

There is nothing to debate.

No there is not.

 

[ggunn]

These two statements highlight the difference in understanding.







And these three highlight why this difference is occurring.


No there is not.

Good luck on your quest there, Mr. Quixote.

I think that the Humpty Dumpty quote was the most incisive contribution to this thread to date.

 

[rattus]

Either way:

Either way:

Some authorities acknowledge the use of VAs and VARs as units for the vi product, but I think WATT is preferred by most. Apparent (average) power is useful in load calculation and transformer ratings, but I see no advantage in its use with p(t).

I think the key to this discussion is the fact that on average, no work is done by reactive power, but at any instant, unless p(t) is zero, the source is moving energy to and from the load. That is, the source is giving and taking it back. Looks like work to me.

Furthermore, I don't like the idea of adding VARs and WATTs algebraically.

 

[Smart $]

Some authorities acknowledge the use of VAs and VARs as units for the vi product, but I think WATT is preferred by most. Apparent (average) power is useful in load calculation and transformer ratings, but I see no advantage in its use with p(t).

No advantage other than understanding the production, transfer, and consumption of energy, and most especially why production and consumption are different. That certainly is not a disadvantage.

I think the key to this discussion is the fact that on average, no work is done by reactive power, but at any instant, unless p(t) is zero, the source is moving energy to and from the load. That is, the source is giving and taking it back. Looks like work to me.

You're contradicting yourself. First you say no work is done by reactive power, then you turn around and say the source is "moving" energy to and from the load and then calling it work.

Furthermore, I don't like the idea of adding VARs and WATTs algebraically.

If you're going to make a statement as such, please explain.

 

[Besoeker]

NYou're contradicting yourself. First you say no work is done by reactive power,

No. What he said was:

on average, no work is done by reactive power,

 

[rattus]

Seems to me that at any instant when the source is delivering energy, that is positive work. When the source is absorbing energy, that is negative work. Therefore it seems appropriate to use the WATT as the unit of power.

Real and reactive powers are added vectorially (although they are not vectors) in the power triangle. Doesn't seem right to add them algebraically. And too, the usage of the VA and VAR is with steady state (average) values. I see no benefit in separating out reactive power in the p(t) formula. That being said, you may use any units you wish if it makes things clearer for you.

 

[jghrist]

Just for fun, graph I??R for a purely resistive circuit. Compare this to the previous graphs of real power.

Actually, I should have said for a circuit with resistance an reactance. You will find that I??R peaks at the same time as current, not at the same time as voltage as on Post #229. If you insist on treating real power as a variable with time, then I think plotting I??R would be the thing to plot.

Personally, I consider real power to be the average of instantaneous power, so it really has no meaning except when considered for a particular period of time like a full cycle. I think plotting it as a function of time has no utility.