Power factor and VA vs Watts

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Cold Fusion

Senior Member
Location
way north
You mean to tell me the 3 questions (from post 463 - cf) (actually more like 2.5 :roll:) have no significant answer... :confused:
Well, the 2.5 Qs have answers. I wouldn't have considered them sigfinicant for this discussion. The questions were trivial simple. The concepts are easily derived from most normal power system models.

cf
 

rattus

Senior Member
Bes' statement makes perfect sense to me. There are no known models of power systems where the concept of instantaneous power having a phase angle has any use what so ever.

Listen up now. The phase angle applies to the current--not the power.

As for your reference, I have no clue as to the context nor what the authors were trying to show. I suspect if we had the reference available, we may have a different conclusion. But no way for any of us to tell.

I am countering besoeker's claim that instantaneous values have no phase angles. No other conclusion is possible.

As for finding a reference that says," Instantaneous power does not have a phase angle" - that is a little silly. Who is going to discuss a concept that has no value?
cf
Again, the claim was that instantaneous voltages and currents have no phase angles. Someone needs to provide proof of this claim. If they can't, they must be wrong!
 

rattus

Senior Member
I have seen theta and phi used both ways. However, the more common usage seems to be phi (φ) is phase angle while theta (θ) is the difference of two phase angles (φ1 ? φ2).

Smart, you miss the point. I am demonstrating that phase angles can ideed be used with instantaneous voltages and currents. Choice of symbol is immaterial.
 

K2500

Senior Member
Location
Texas
Again, the claim was that instantaneous voltages and currents have no phase angles. Someone needs to provide proof of this claim. If they can't, they must be wrong!

How can an instantaneous measurement be assigned anything but a scalar quantity?
How would it affect the outcome of an equation based on instantaneous measurements?
 

Cold Fusion

Senior Member
Location
way north
How can an instantaneous measurement be assigned anything but a scalar quantity?
How would it affect the outcome of an equation based on instantaneous measurements?
It can't.

It won't

R found an out of print text describing a concept that has no use in power systems analsys. Who knows what that particular section of the text was trying to show?

cf
 

Cold Fusion

Senior Member
Location
way north
I have seen theta and phi used both ways. However, the more common usage seems to be phi (φ) is phase angle while theta (θ) is the difference of two phase angles (φ1 ? φ2).

Smart, you miss the point. I am demonstrating that phase angles can ideed be used with instantaneous voltages and currents. Choice of symbol is immaterial.
No, Smart accurately got the important part of your point.

cf
 

Cold Fusion

Senior Member
Location
way north
Bes' statement makes perfect sense to me. There are no known models of power systems where the concept of instantaneous power having a phase angle has any use what so ever. ...

Listen up now. The phase angle applies to the current--not the power. ...

Oh right. Thanks for pointing out my mis-type.

There are no known models of power systems where the concept of instantaneous current having a phase angle has any use what so ever.

...As for your reference, I have no clue as to the context nor what the authors were trying to show. I suspect if we had the reference available, we may have a different conclusion. But no way for any of us to tell. ...

...I am countering besoeker's claim that instantaneous values have no phase angles. No other conclusion is possible. ...
Just possibly, there could be some that might disagree and also have a logical explanation


...As for finding a reference that says," Instantaneous power does not have a phase angle" - that is a little silly. Who is going to discuss a concept that has no value? ...

...Again, the claim was that instantaneous voltages and currents have no phase angles. Someone needs to provide proof of this claim. If they can't, they must be wrong!

Actually, the more I think about it, you and smart can insist that instantaneous current has a phase angle. And I won't say you are wrong. Of course that would be right up there with:
The moon has a phase angle. (Well, prove I'm wrong:roll:) Yes, equally useless.

cf
 

rattus

Senior Member
How can an instantaneous measurement be assigned anything but a scalar quantity?
How would it affect the outcome of an equation based on instantaneous measurements?

Glad you asked. Let,

v = Vm*sin(wt)
i = Im*sin(wt + phi)

p = vi = (VmIm/2)[cos(-phi) - cos(2wt + phi)]

Further identities may be used to expand the equation to 3 terms.

p is a scalar--not a phasor! phi merely shifts cos(2wt) in time.
 

Cold Fusion

Senior Member
Location
way north
Glad you asked. Let,

v = Vm*sin(wt)
i = Im*sin(wt + phi)

p = vi = (VmIm/2)[cos(-phi) - cos(2wt + phi)]

Further identities may be used to expand the equation to 3 terms.

p is a scalar--not a phasor! phi merely shifts cos(2wt) in time.

Wow that is interesting. So could one also say:
"i is a scalar -- not a phasor! phi merely shifts sin(wt) in time."

cf
 

Smart $

Esteemed Member
Location
Ohio
Smart, you miss the point. I am demonstrating that phase angles can ideed be used with instantaneous voltages and currents. Choice of symbol is immaterial.
I was not commenting on your demonstration (which is fine in my book...errrr that would be "head" for you bookworms :roll:)




I was commenting on...
The theta (more usually termed as phi) is a phase displacement. For a given frequency that is a time displacement between current and voltage.
For instantaneous quantities there is no time element thus no time displacement.
Your statement does not make sense. The argument for the trig functions must be in radians, not seconds. e.g., the units of "wt" are radians.

Theta is the lead or lag as the case may be between the voltage across and the current through a load. e.g., the general form for i(t) is,

i(t) = Im*sin(wt +
theta)

Theta is described as a phase angle in,

[Kerchner and Corcoran, (Alternating-Current Circuits, 3rd edition, John Wiley & Sons, 1951)]

If you think otherwise, give us a solid reference.
The slightly less general form of i(t) is:
i(t) = Im*cos(ωt ? φ + θ)

where:

φ is the associated voltage phase angular delay from reference (omitted if the voltage phase angle is the reference angle)
θ is the displacement amount of the current to voltage phase angles (leading is positive, lagging is negative)
...and the others you know...
...and I also used the cosine function instead of sine... and you know why.​

That said, your usage of theta is the most common form.
 

Smart $

Esteemed Member
Location
Ohio
Glad you asked. Let,

v = Vm*sin(wt)
i = Im*sin(wt + phi)

p = vi = (VmIm/2)[cos(-phi) - cos(2wt + phi)]

Further identities may be used to expand the equation to 3 terms.

p is a scalar--not a phasor! phi merely shifts cos(2wt) in time.
The dot product of two vectors being scalar is a vector math limitation. In the physics world, AC power is a sinusoidal waveform when both voltage and current are sinusoidal waveforms. It therefore can be modeled as a vector or phasor if one desires.
 

Smart $

Esteemed Member
Location
Ohio
Well, the 2.5 Qs have answers. I wouldn't have considered them sigfinicant for this discussion. The questions were trivial simple. The concepts are easily derived from most normal power system models.

cf
Please indulge me and answer the questions...
 

rattus

Senior Member
Wow that is interesting. So could one also say:
"i is a scalar -- not a phasor! phi merely shifts sin(wt) in time."

cf

I should say that a postive phi advances i(t), and a negative phi delays i(t). v(t) is our reference with an angle of zero. The fully expanded equation contains no phase shift terms at all.

All the terms are scalars.
 

Besoeker

Senior Member
Location
UK
But there is. The equation in question was derived from,

v(t) = Vm*sin(wt)
i(t) = Im*sin(wt + theta)

The term "cos(theta)" turns out to be the power factor.
I said there was no displacement between instantaneous values.
Let me rephrase that.

There is no phase displacement between the current and the voltage occurring at the same instant.
The power at that instant is just current times voltage.
Real and reactive components have no meaning in this context.
 

Smart $

Esteemed Member
Location
Ohio
I said there was no displacement between instantaneous values.
Let me rephrase that.

There is no phase displacement between the current and the voltage occurring at the same instant.
The power at that instant is just current times voltage.
Real and reactive components have no meaning in this context.
IMO, they actually have more meaning than total power at that instant when considering reactive loads (...but this opinion may be heavily influenced by my present disposition ;)).

However, if our objective is to plot the data to get a "picture" of relationships, we don't really want a set of values at one and only one particular instant. Rather we want to amass a multitude of instant values over a range of no less than one full voltage cyle. I prefer two full voltage cycles. Yet, in the process, the capability of determining a set of values at one particular instant is made available to us.
 
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