Cold Fusion
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Yes, I read it. Didn't see anything significant for comment. Sometimes I can't tell which part is the trivial rambling.Post #463.
cf
Yes, I read it. Didn't see anything significant for comment. Sometimes I can't tell which part is the trivial rambling.Post #463.
You mean to tell me the 3 questions (actually more like 2.5 :roll have no significant answer...Yes, I read it. Didn't see anything significant for comment. Sometimes I can't tell which part is the trivial rambling.
cf
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.You mean to tell me the 3 questions (from post 463 - cf) (actually more like 2.5 :roll have no significant answer...
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.
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.
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!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
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).
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!
It can't.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?
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).
No, Smart accurately got the important part of your point.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.
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. ...
Just possibly, there could be some that might disagree and also have a logical explanation...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? ...
...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?
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.
I was not commenting on your demonstration (which is fine in my book...errrr that would be "head" for you bookworms :rollSmart, you miss the point. I am demonstrating that phase angles can ideed be used with instantaneous voltages and currents. Choice of symbol is immaterial.
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 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.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.
...
That said, your usage of theta is the most common form.
...
i = Im*sin(wt + phi)
...
Just possibly, there could be some that might disagree and also have a logical explanation
cf
Please indulge me and answer the questions...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
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 said there was no displacement between instantaneous values.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.
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 ).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.
Not even total power. Just power. It really is that simple.IMO, they actually have more meaning than total power