induction motor power factor more than one

dear engineers, energy auditing time i saw one induction motor is showed a pf of 1.458 indactive is it possible. that time its current is: 4.3 voltage: 398.5 is it possible or not

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Power factor is the cosine of the phase angle. How does a cosine get above one?

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Obviously, the phase angle must be complex:=).
Bill
Bill

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--------- It is definitely not possible.

On Jun 18, 6:53 am, snipped-for-privacy@gmail.com wrote:
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No. PF is the cosine of the phase angle between the voltage and current vectors. Theoretical max PF is 1.00 when they are in phase but that will never happen as the current always lags the voltage. Minimum PF value is zero when no power is being developed, not practically possible due to losses but theoretically possible if you arrange to drive the induction motor mechanically as an induction generator so as to just match the losses! Parallel capacitors (to take leading current) are another matter but PF can still never exceed unity. Cheers, Roger

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No. PF is the cosine of the phase angle between the voltage and current vectors. Theoretical max PF is 1.00 when they are in phase but that will never happen as the current always lags the voltage. Minimum PF value is zero when no power is being developed, not practically possible due to losses but theoretically possible if you arrange to drive the induction motor mechanically as an induction generator so as to just match the losses! Parallel capacitors (to take leading current) are another matter but PF can still never exceed unity. Cheers, Roger
If the capacitor bank is large enough, a leading PF is feasible, although this is usually uneconomic, the minimum PF is not zero when no power is being developed.
Regards.

,  the minimum PF is not zero when no power is
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And how is power developed? If you mean dissipated then please tell me what is the PF of a theoreticaly perfect (lossless) reactance of either type?

| | , ?the minimum PF is not zero when no power is |> being developed. |> | And how is power developed? If you mean dissipated then please tell me | what is the PF of a theoreticaly perfect (lossless) reactance of | either type?
I'll try some explanation here.
If by lossless you mean that all power being supplied goes into delivering motive power to the rotating device, then this does not change the PF at all. It can still be no more than 1.0. You just don't have any resistive losses that would reduce efficiency (a different measurement).
If you are referring to a plain reactive device, an inductor or capacitor, with no resistive loss involved anywhere, then PF is zero and no power is lost anywhere. Of course, this is never a reality.
Power can produce work done by a motor, as well as heat. If you include the heat in the equation, you would have a slightly higher PF than if you omitted the heat. But it can still never be above unity.
Basically, unity PF means you are using all the power taken from the source. Less than unity means you are giving some back in terms of current/voltage phase angle difference. You can have a negative PF, which would mean more power given to the source than taken from it. That's what you would get with a generator feeding the system.
At zero PF, the phase angle of the current is at 90 degrees from the voltage. During half of each half-cycle, power is taken from the source, and during the other half of each half-cycle, power is given back in the same amount. The phase angle tells you the PF when everything is a perfect sine wave. The calculation is more complex when harmonics are involved.

On Jul 10, 4:41 pm, snipped-for-privacy@ipal.net wrote:
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Power factor is the ratio of KW / KVA. It is not possible for the KW to be bigger than the KVA. If it was, there would be perpetual motion!! I would suggest that if this is a three phase motor, you have a wiring problem with the instrumentation.
Best regards, Mark Empson http://www.LMPhotonics.com

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Is Phil's misconception likely to go away if one thought of incident and reflected powers instead of power factor? RF engineers do that all the time. I cannot ever remember anyone ever talking about power factor in regard to RF. It is a bit easier to tune out the effect of mismatch with reactive loads at RF than at power frequency. On the other hand, it may be silly to talk about reflected power from nearby resistive loads at power frequency.
Bill

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| |> Power factor is the ratio of KW / KVA. It is not possible for the KW |> to be bigger than the KVA. If it was, there would be perpetual |> motion!! I would suggest that if this is a three phase motor, you have |> a wiring problem with the instrumentation. | | Is Phil's misconception likely to go away if one thought of incident and | reflected powers instead of power factor? RF engineers do that all the | time. I cannot ever remember anyone ever talking about power factor in | regard to RF. It is a bit easier to tune out the effect of mismatch with | reactive loads at RF than at power frequency. On the other hand, it may | be silly to talk about reflected power from nearby resistive loads at | power frequency.
I don't understand what you see as a misconception. I already knew KW/KVA, but just didn't state it in my description.
I did describe the "negative PF" as the concept where as you reach zero you have everything out of phase by 90 degrees, and as you continue it goes on to be out of phase by 180 degrees. But that is apparently not how PF is actually figured beyond that point. I guess that comes from the fact that you can have zero PF with 90 degrees lagging or 90 degrees leading, so there are two different ways to have zero PF. It seems like we need something a bit more complex? A Smith Chart (from RF transmission lines) comes to mind. But probably never used for power engineering (though it might start to make some sense for transmission lines crossing the country).
We don't normally thing of the possibility for a "mismatch" of resistive loads in power systems, and the resultant reflective power. But once you get a transmission line long enough for there to be propogation of voltage change taking significant time, then you do get reflection because the initial propogation assumted the characteristic impedance of the transmission line because of the time difference to the load.
If you wanted to build a scale model of the electric grid, you should scale the wavelength down by the same scale factor. What frequency would that be?

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Maybe I confused you with the person who posted about getting power factors greater than one. Certainly, a negative power factor corresponds to the "load" becoming a source of power.
Bill

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remove the X to answer

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Part of the problem we are having in communication is that of engineering jargon.
If pf is defined as cos(theta) where theta is the phase difference between voltage and current phasors, then assigning positive pf to inductive loads and negative pf to capacitive loads is mathematically untenable.
That inconsistency is present even if EEs understood what is meant. I am one EE who never heard of such designation.
By the way, although I have not followed it through, power factors greater than unity would correspond to complex phase angles. In turn, What would make sense then is if there is an exponential growth (or decay) rather than a constant amplitude sine wave source. I do not wish to go there.
Bill

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Instead of the phase difference..how about the amount by which the Voltage phasor leads the Current phasor as a better definition?
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me neither...especially seeing as i would not know the way....[my maths book are coming out at this point!]

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I have no problem with such a definition. I just should not be called power factor. There is just too much history behind the definition of power factor to allow a new definition to creep in.
I do know of a definition that illustrates my point. The direction of light polarization was defined (as it turned out) to be the direction of the magnetic field. This was done before Maxwell's theory was well in place. Especially after quantum mechanics was established, it was realized that the electric field provided the strong interaction with matter. To change the emphasis onto the electric field, the old definition of polarization direction was kept, while the electric field was called the direction of vibration. Confusion still reigns.
A simpler example is that of the definition of positive and negative by Benjamin Franklin. There was no way that Franklin could have known that it was negative charge (electrons) that was predominantly the mobile charge.
Bill

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------------------ You are absolutely right - I responded without adequate thought. If I had said lagging reactive (rather than lagging pf ) is considered positive- then that would be be OK.
I stand corrected.
The definition was initially applied on the basis of generators having to be overexcited to supply inductive loads and the reactive was then treated as positive. Hence for lagging pf, the output reactive was considered positive and by association, not sense, the pf was considered +. Yes, in practice, I always indicated pf lag or lead rather than + or -.
As for complex phase angles -do you use pf = cosh(angle)?? :) Worth avoiding unless there is some real (no pun meant) reason to do so.

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It is interesting to note that many, but not all, physicists usually use an exponential variation of form exp(-i*w*t) while most engineer use exp(j*w*t) to deal with sine waves. The first form will lead to negative reactance for inductors. I have also seen misinterpretations and errors as a consequence. I remember a case when someone was sure he had found a flaw in Maxwell's equation when calculating the phase shift of waves totally internally reflected in a prism. You are almost always going to land in hot water if you take equations at face value without understanding how they were derived.
Bill

| Part of the problem we are having in communication is that of | engineering jargon. | | If pf is defined as cos(theta) where theta is the phase difference | between voltage and current phasors, then assigning positive pf to | inductive loads and negative pf to capacitive loads is mathematically | untenable.
I think what makes this hard is the notion that the "normal" case is defined as being 1 instead of defined (somehow) as being 0. One could construct a formula to make it be 0 for the non-reactive.
The problem is, it really is a circular situation. That is, the phase angle can be represented precisely as a vector that can be at any position in a circle. So really, PF is merely the real component of a complex value. So we need an additional imaginary value to account for the reactance and define the phase angle completely (in two dimensions). But for most scenarios it seems the PF is all that is needed.
I've always wondered what would happen on a shared neutral circuit if an almost pure inductive load was on one side, and an almost pure capacitive load was on the other side. Normally with resistive loads, or at least with loads of like reactance, the neutral would have only the imbalance current, which would be zero with identical loads. But in the case of inductive on one side and capacitive on the other, of equal current, the neutral is going to be getting double current (assuming single phase).

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--------- But if the pf corresponds to cos (angle) as in the situation with single frequency sinusoids. then the phase angle is known- it can be one of two values. What is not known is whether the phase angle is in the upper or lower half of the unit circles and that is where "lead" or "lag" information is needed. -------------------
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-------- Think of the worse situation that would occur if the neutral opened.