Power factor and VA vs Watts

[Cold Fusion]

No, not wasted. It is just sent back to the generator. The part that is wasted is the losses in the transmission of the .293W, shuttling it from the gen to the load and banc to the gen. The .293 is still available to use again, but the extra power it took to get it out and back are gone.

cf

You're making it sound like the system on the whole is a finely tuned RCL oscillator. It is not. You'd have to have some major PFC caps hiding [and connected] somewhere.

You are right. The system is ot a finely tuned oscillator. However, the reactive power shuttles back and forth exactly as I outlined.

Let's reduce the system to one generator, one transmission line, one motor.

Looking at the motor first:
Real power is either coming out the motor shaft, given off as heat in the windings, or windage and friction. In addition, the motor requires reactive power for excitation. That is stored in the stator mag field for part of the cycle and given up in other parts of the cycle.

Now lets look at the generator:
The generator driver must put out enough torque/rpm to the gen shaft to match the real power the motor is using/putting out. As you have said, that power is sent to the motor as current in-phase with the voltage. However, in order to supply the reactive power, the generator is over-excited, which makes it produce vars. Again as has been discussed, this reactive power is sent to the motor as current lagging the voltage.

That reactive power is shuttled - gen the motor and back - as the motor mag field builds and collapses. This power is not lost, the generator produces it and takes it back twice a cycle.

Now lets look at the transmission line and to a certain extent the gen and motor I^2R losses. The gen has to send the current it takes for the real power - no way out of that. So any transmission/motor/gen I^2R losses associated with that current are tough luck - have to have it.

The reactive current also produces I^2R losses. So this reactive current that is not producing power, requires real power from the gen/driver to make up for the losses.

So, no "finely tuned RCL oscillator" or "major PFC caps hiding". Just the way the laws of God and physics work out.

cf

 

[Smart $]

see this graph.

Your graphs are valid.

For an RCL circuit with a lagging current (i.e. 0? < φ < -90?), show the apparent, reactive, and real power curves (Hint: real is R only, reactive is CL only, and apparent is all three combined).

P = P + Pcos2wt......(1) ???

Is this an equivalent of S = P + Q?

 

[jghrist]

S = P + Q, where S = complex power ( or apparent power or VA), P = real power (Watts), Q = Reactive power (VARs). All three quanities are vectors, with magnitude and phase angle.

Complex power is P + jQ. P and Q are not vectors, they are scalar values.
P = |V|?|I|?cos? where ? is the angle between V and I.
Q = |V|?|I|?sin?

See Elements of Power System Analysis, W.D. Stevenson, Jr., 1962, p. 137.

 

[Smart $]

You are right. ...

So, no "finely tuned RCL oscillator" or "major PFC caps hiding". Just the way the laws of God and physics work out.

Very nice explanation. Another aspect which you did not include is "shifting" of the current (negative power) from one load to another without going back to the generator (though I do understand you used a simplistic perspective of one generator and one load).

Yet if we open the matter to beyond the electrical system itself we would have to know the efficiency of the generator. If it takes "1.0 watt of fuel" to generate 0.7 watt of real power on the consumer end we are already in the hole without yet accounting for transmission losses.

With all that said and understood, let me just say my use of the word "wasted" was simply a poor choice of wording. Just seemed appropriate at the time.

 

[Smart $]

Complex power is P + jQ. P and Q are not vectors, they are scalar values.
P = |V|?|I|?cos? where ? is the angle between V and I.
Q = |V|?|I|?sin?

See Elements of Power System Analysis, W.D. Stevenson, Jr., 1962, p. 137.

Don't have the book on hand. Does he provide a reason for using the moduli of the vectors?

That also seems to contradict other publications (no citation and doubt I will make any) in that they are as depicted below...

...though I personally prefer the VAR at the other end of W.

 

[Besoeker]

Not in complete disagrement with you, but the issue we are discussing is a little more complex than this.

I was referring to your post #149:

FWIW, instantaneous power represents apparent power

1V at the same instant as 1A IS 1W.

 

[Smart $]

I was referring to your post #149:

1V at the same instant as 1A IS 1W.

In a discussion of apparent, real, and reactive power it is not. It is 1 VA, not 1W. In simple terms, it is not 1W until there is a power [usage] factor. With no usage it is transferred potential energy, such as with the ideal capacitor or inductor.

 

[Hameedulla-Ekhlas]

Your graphs are valid.

For an RCL circuit with a lagging current (i.e. 0? < φ < -90?), show the apparent, reactive, and real power curves (Hint: real is R only, reactive is CL only, and apparent is all three combined).

P = P + Pcos2wt......(1) ???

Is this an equivalent of S = P + Q?

no,

S = P + Q

What kind of formula is this. Have you forgot the Sqaure root of each element. I have never seen and used such a formula in all five years engineering. Can you give me a book reference to contain this formula

Besoeker
1V at the same instant as 1A IS 1W.

wrong:

Another power quantity that is commonly used to define the power rating of electrical equipment is the apparent power S.
Apparent power S is defined as

S = V*I

The unit for the apparent power in the international system is VA(volt-Ampere). This power is usually understood to represent the maximum reachable active power at unity power factor.

P=V*Icos0 incase cos0 = 0 then P = S

But without mentioning the unity of power factor it is wrong statement.

 

[Hameedulla-Ekhlas]

Your graphs are valid.



P = P + Pcos2wt......(1) ???

Is this an equivalent of S = P + Q?

no
What kind of formula is this. Have you forgot the Sqaure of each element. I have never seen and used such a formula in all five years engineering. Can you give me a book reference to contain this formula


Sorry in my previous isntead of sqaur, I have mentioned sqaure root. It is just sqaure

 

[steve66]

Don't have the book on hand. Does he provide a reason for using the moduli of the vectors?

That also seems to contradict other publications (no citation and doubt I will make any) in that they are as depicted below...

...though I personally prefer the VAR at the other end of W.

The reason is because vectors have direction. But power and VAR's don't really have a direction. They are technically scalars.

However, because the apparent power (complex power) has a phase angle, it is useful to use vectors to represent complex power so we can keep track of the power factor (and phase angle).

Although Watts and VA's aren't true vectors, linear algebra (and things like the fact that independent components are orthogonal) tell us that we can use vectors and get the right answer.

Steve

 

[Smart $]

no,



What kind of formula is this. Have you forgot the Sqaure root of each element. I have never seen and used such a formula in all five years engineering. Can you give me a book reference to contain this formula


wrong:

Another power quantity that is commonly used to define the power rating of electrical equipment is the apparent power S.
Apparent power S is defined as

S = V*I

The unit for the apparent power in the international system is VA(volt-Ampere). This power is usually understood to represent the maximum reachable active power at unity power factor.

P=V*Icos0 incase cos0 = 0 then P = S

But without mentioning the unity of power factor it is wrong statement.

Actually you say no... but it is... and you even confirm this with your post unknowingly. You've already stated S in your post.

S is the vector for apparent power​

Now look at the image you posted.

P is the vector for real power
Q is the vector for reactive power​

BTW, what book did that image come from... looks like a google book search, being the word power is highlighted.

Also, my last chart essentially depicts the same waveforms overlaid upon each other rather than in two separate charts.

 

[Smart $]

The reason is because vectors have direction. But power and VAR's don't really have a direction. They are technically scalars.

However, because the apparent power (complex power) has a phase angle, it is useful to use vectors to represent complex power so we can keep track of the power factor (and phase angle).

Although Watts and VA's aren't true vectors, linear algebra (and things like the fact that independent components are orthogonal) tell us that we can use vectors and get the right answer.

Steve

I can live with your explanation.

So in technical summary, depending on which train of thought I have at the moment, I am agreeing, yet I am also not

This leaves me wondering whether Descartes can think himself back into existence

 

[Hameedulla-Ekhlas]

Actually you say no... but it is... and you even confirm this with your post unknowingly. You've already stated S in your post.

S is the vector for apparent power​

Now look at the image you posted.

P is the vector for real power
Q is the vector for reactive power​

BTW, what book did that image come from... looks like a google book search, being the word power is highlighted.

Also, my last chart essentially depicts the same waveforms overlaid upon each other rather than in two separate charts.

Yes, I said no for S = P + Q because you have not put j with Q. S is a complex power and every power and circuit books like this

S = P + jQ = Vrms*Irms cos0 + jVrms*Irms sin(0)

That 's why I told you. You had written wrongly the equation and you will not find any book to write the complex power like you.

 

[Cold Fusion]

Wow -
I never expected "complex power expressed as vectors" to stir up such a hornet's next. The best I can say is the model makes perfect sense to me, is calculable, and consistantly defined. Complex power needs a phase angle to be useful to me.

Complex power is P + jQ. P and Q are not vectors, they are scalar values. ...

...See Elements of Power System Analysis, W.D. Stevenson, Jr., 1962, p. 137.

Yes, I sure in that context that definition is exactly right.

...What kind of formula is this. Have you forgot the Sqaure root of each element. I have never seen and used such a formula in all five years engineering. Can you give me a book reference to contain this formula ...

No, I didn't forget. In the context I was speaking, S, P, Q are vectors.

...Apparent power S is defined as

S = V*I ...

It sounds like you are saying "Apparent power S" is a directionless scalar. I am using "S" to denote Complex power - which is a vector,

...Apparent power S is defined as

S = V*I

And I would say:
S = V*I(conjugant)
where S, V, I(conjugant) are ll vectors

...Sorry in my previous isntead of sqaur, I have mentioned sqaure root. It is just sqaure

I knew what you meant. You are adding the Scalar quantities of P and Q with a known orthoganal relation.

Don't have the book on hand. Does he provide a reason for using the moduli of the vectors?

That also seems to contradict other publications (no citation and doubt I will make any) in that they are as depicted below...

I'l agree with smart on this one. And I also doubt I'll spend time finding a reference. either.

Steve -
These are all good comments. And undoubtly textbook correct - at least for the context of the particular textbook.

... But power and VAR's don't really have a direction. They are technically scalars. ...

Let's look at vars from an inductive reactance:
Z = jwL

vars = V/Z
Where V is a vector, V = |V| +j0.

vars = (|V| + j0/jwL = 0 -j(|V|/wL)

and that makes vars look like a vector to me. Please be aware, I am not insisting I am clasically correct. I am saying this is the model I am using and it works really well. I have not seen it break down yet.

...However, because the apparent power (complex power) has a phase angle, it is useful to use vectors to represent complex power so we can keep track of the power factor (and phase angle). ...

Yes, yes, yes

...Although Watts and VA's aren't true vectors, linear algebra (and things like the fact that independent components are orthogonal) tell us that we can use vectors and get the right answer. ...

Wow, that sure is lucky for me. Otherwise my model would be worthless:roll:


cf

 

[Hameedulla-Ekhlas]

Wow -
I never expected "complex power expressed as vectors" to stir up such a hornet's next. The best I can say is the model makes perfect sense to me, is calculable, and consistantly defined. Complex power needs a phase angle to be useful to me.


Yes, I sure in that context that definition is exactly right.


No, I didn't forget. In the context I was speaking, S, P, Q are vectors.


It sounds like you are saying "Apparent power S" is a directionless scalar. I am using "S" to denote Complex power - which is a vector,


And I would say:
S = V*I(conjugant)
where S, V, I(conjugant) are ll vectors


I knew what you meant. You are adding the Scalar quantities of P and Q with a known orthoganal relation.


I'l agree with smart on this one. And I also doubt I'll spend time finding a reference. either.

Steve -
These are all good comments. And undoubtly textbook correct - at least for the context of the particular textbook.


Let's look at vars from an inductive reactance:
Z = jwL

vars = V/Z
Where V is a vector, V = |V| +j0.

vars = (|V| + j0/jwL = 0 -j(|V|/wL)

and that makes vars look like a vector to me. Please be aware, I am not insisting I am clasically correct. I am saying this is the model I am using and it works really well. I have not seen it break down yet.

Yes, yes, yes


Wow, that sure is lucky for me. Otherwise my model would be worthless:roll:


cf

Is the red hilighted equation correct?

Vars = V sq / z; z = jwl

 

[Cold Fusion]

Is the red hilighted equation correct?

Vars = V sq / z; z = jwl

no it's not I'm asleep. Thank you for catching this. I'm hurrying trying to get back to work and did a really poor job. Left out most of it - yuk - I try not to be that sloppy. But sometimes I am

Z = jwl
I = V/Z = 0 - j|V|/wL
The circuit is pure inductive, so vars = VI
= (|V| + j0)*(0-j|V|/wL) = 0 - j (|V|^2)/wL

and that looks like a vector.

Hopefully that sounds a lot better.

 

[steve66]

I can live with your explanation.

So in technical summary, depending on which train of thought I have at the moment, I am agreeing, yet I am also not

This leaves me wondering whether Descartes can think himself back into existence

Personally, I don't care if we call power vectors or scalars. I was just explaining why I think some authors use scalars, and then just write S=P+jQ. I think they are math purists.

Engineers tend to be the other way. We don't care if we are mathmatically exact or not, as long as we get the write answer, we're happy:grin:

I tend to use what ever seems approporate at the time. I'll even mix vectors, scalars and phasors in the same problem. (This once sparked a very long thread about something I'm not going to mention here - this thread is already going enough different directions.)

Heck, I'll even write "The integral of x dx from x=0 to x=100" much to the annoyance of my college calculus teacher.

 

[Besoeker]

In a discussion of apparent, real, and reactive power it is not. It is 1 VA, not 1W. In simple terms, it is not 1W until there is a power [usage] factor. With no usage it is transferred potential energy, such as with the ideal capacitor or inductor.

Your comment:

FWIW, instantaneous power represents apparent power

was about instantaneous power.
One Volt and one Ampere at the same instant is one Watt.
That is not a debatable point.

 

[Cold Fusion]

Engineers tend to be the other way. We don't care if we are mathmatically exact or not, as long as we get the write answer, we're happy ...

I try to keep my model consistent. Is that a pun on engineers lacking righting skills? If so, good one.

...I tend to use what ever seems approporate at the time. I'll even mix vectors, scalars and phasors in the same problem. (This once sparked a very long thread about something I'm not going to mention here - this thread is already going enough different directions.)...

I recall one thread a few years back (when I was lurking) that ended in a bitter, demeaning, fight over vectors vs phasors. I recall thinking, "And we care about this why?"

... Heck, I'll even write "The integral of x dx" much to the annoyance of my college calculus teacher.

You would get a raised eyebrow out of me on that one. I'd be wondering, "What is that defined as?"

Generally speaking, I can translate what someone is trying to get at - although I have commented a few times (mostly to management), " That violates a couple of the major laws of God and Physics. Let me work on this a bit and see if I can find another way."

cf

 

[Hameedulla-Ekhlas]

Information:


The magnitude of complex power is referred to as apparent power.


|S| = sqrt( (sqrt)P + (sqrt)Q )

Apparent power, like complex power is measured in volt ampere. The apparent power unit is volt-ampere