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

Reply to
raggetzu
Loading thread data ...

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

Reply to
Bob Eld

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

Reply to
Don Kelly

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

Bill

Bill

Reply to
Salmon Egg

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

Reply to
Engineer

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.

Reply to
John McLean

, =A0the minimum PF is not zero when no power is

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?

Reply to
Danny

| On Jul 10, 6:06?am, "John McLean" wrote: | | , ?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.

Reply to
phil-news-nospam

e:

f you post to =A0|

ASAP. =A0 =A0 =A0 =A0|

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

formatting link

Reply to
marke

--------- The load on the motor plus losses determines the power drawn. The above seems to imply that you can shove power from the system into the motor to get work done- you can't. The supply simply tries to satisfy the needs of the motor and load.

-------------

-------- You do this at any power factor. Note that the power of concern is average power per cycle. That is what a wattmeter measures. This should not be confused with instantaneous power (which can be negative during part of the cycle if the power factor is not unity). Reactive power, KVA and pf are also based on average values as is rms voltage and current (which don't actually exist) specific averages. Unity pf simply means that there is no "reactive" load. --------

---------- This is not true. The sign of the "real" or average power depends on whether the machine is generating or motoring but the sign of the power factor depends on the direction of the "reactive power" which is not determined by the direction of the real power. An induction machine whether motoring or generating, will have a lagging power factor, sucking vars from the source. A synchronous motor can have its pf changed from + to - by excitation control without any change in the magnitude or direction of the power delivered to the load. Similarly, it can feed power to the system with independent control of the power and the power factor. By North American convention (British too, I believe), lagging pf is considered +ve. Leading pf is -ve. European convention is the opposite.

--------------

Reply to
Don Kelly

e:

no need. thx for making the effort though. It was intended as a slightly sarcastic rhetorical question. perfect reactance theoretically has 0pf. And the power is developed at the generator (or in fact converted). it is dissipated at the load.

Reply to
Danny

8 20:22:54 -0700 (PDT) Danny wrote:

and anyway i am not the OP who posed the question

Reply to
Danny

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

Reply to
Salmon Egg

On Mon, 14 Jul 2008 23:28:40 -0700 Salmon Egg wrote: | In article | , | marke wrote: | |> 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?

Reply to
phil-news-nospam

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

Reply to
Salmon Egg

Sorry, I only partially agree. By North American convention, complex power is given by E(Iconjugate) so that lagging pf is considered positive at a load (and has to be generated ) and leading (or capacitive) pf is negative. Consider a synchronous motor tied to a grid. As a motor it draws power from the grid but changing the excitation doesn't change this (except for some changes in losses) but can swing the reactive and pf from + (underexcited) to -(overexcited) yet the load doesn't become a real power source.

If you consider an induction machine, as it always draws reactive, then the pf can be considered as - when it generates. There I agree as it is a sink for reactive.

So, the sign of the pf becomes rather confusing when you consider the possibility of four quadrant It is better to think of reactive "flow" as being +/- and real flow as being +/- and keep to one reference for + flow to either motor or generator conditions.

Now if you are in continental Europe the convention is different and complex power is given by I(Econjugate) reversing the sign of reactive- adding to the confusion.

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

Reply to
Don Kelly

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

Reply to
Salmon Egg
8X------

Instead of the phase difference..how about the amount by which the Voltage phasor leads the Current phasor as a better definition?

me neither...especially seeing as i would not know the way....[my maths book are coming out at this point!]

Reply to
Danny

------------------ 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.

Reply to
Don Kelly

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

Reply to
Salmon Egg

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.