power factor when start the induction motor

[PAP]

Dear all
Anyone can suggesst me about the power factor of the motor when starting. I've known that it's about 0.2 and when try to calculate the starting power (assume starting current is 4 times of rated current for a large machine.) the starting power is just about rated power of the machine (rated pf is 0.7) . Is it correct? Thank in advance for your suggestion.

 

[jbartos]

Suggestion: The motor Circle Diagram could help. The starting motor input power versus motor rated input power is relative, depending on the motor parameters. If you happen to have the motor Circle Diagram for your specific motor, then you could derive it from there.
The output power is different. It is zero at the motor start and then it increases to the rated output (shaft) power, if the shaft has the rated load.

 

[electricpete]

You need to clarify whether you are talking about input power or output power.

I think that if you are talking about input power then you are right that power is the same order of magnitude as full power during the start, by the logic you have given. In rough numbers, a starting current 5X full load current and starting power factor of 1/5 or more of full load power factor create full load power input or more.

If you are talking about output power, then as jbartos suggests it is initially zero. If you had the torque-vs speed curve, you could create the power vs speed curve as the product of torque and speed.

This suggests that the losses created during a motor start can be equal or greater than the the full power of the motor. That is not unreasonable since a factor of 5 increase in current (from full load to starting) results in factor of 25 increase in I^2*R losses.

 

[jOmega]

JBartos & Electricpete,

Since PAP asked: "Anyone can suggesst me about the power factor of the motor when starting. " you have me confused.

In over 40 years in the motor-drive industry, I have never heard of output power factor......

Perhaps you would be so kind as to provide some elucidation on the topic of motor output power factor for the un-initiated.

Thanks

 

[electricpete]

jomega - The only person in this thread who used the term "output power factor" was you. jbartos and I made the distinction of input power vs output power.

It is a necessary clarifcation to the question
'the starting power is just about rated power of the machine (rated pf is 0.7) . Is it correct?'

Rated power can be input or output depending on whether you look at European or US standards.

I suggest in the future that you try to understand something before you criticize it.

 

[ielivaz]

Assuming you motor is unloaded when you power it, at the very begining the source will see the motor as a resistive load with no induction at all, so at the very begining the power factor will be 100%.

As the motor acelerates (unloded) it will build up inductive load moving the power factor form 100% to almost 0% (a perfect motor should have a 0% pf when unloaded).

As the motor is loaded the pf will increse until at the rated power if it were a perfect motor the pf will be 100%.

If you want to know the starting power you must measure the motor resistive load using an ohmeter.

 

[jbartos]

Suggestion: The starting power factor of 0.2 is normally linked to ordinary induction squirrel cage motors (ISCM) that have locked-rotor amps 6 times full load amps.
The starting motor power factor can be calculated from the ISCM electrical equivalent circuit, e.g.
PF=cos{arctan[(Re+Rr'/s)/(Xe+XLr')]}
where s is the variable slip.
See Figure 5.21(a) on page 376 in Reference:
1. Gordon R. Slemon, Magnetoelectric Devices, Transducers, Transformers, and Machines, John Wiley and Sons, Inc., 1966
Alternately, the starting power could be measured, for example by wattmeters.

 

[electricpete]

ielivaz - from jbartos' equation, we can see that power factor is not near 100% at initial start. It will be 10% or 20% at start, due to the relatively low value of (R1+R2/s) compared to (X1+X2) in the equivalent circuit where s close to 1 at initial start.

 

[busbar]

The default PF for induction-motor starting in the 1955 Beeman text is 0.2.

 

[jOmega]

Electricpete,
A little oversensitive, maybe ?
The quesiton wasn't criticism; but that's how you chose to
internalize to it.

It was a request for clarification which you have kindly provided. Funny, In my haste, I saw PAP's original post as being a question about Power Factor during starting, and next thing you know, the responses from you and JB are about output power, which wasn't part of the original post. Hmmmmm, must have missed something in PAP's post or is there a bit of non-sequitur in the air?

Funny, as you read thru the posts on this forum, more often than expected, someone asks a question about how to get to Topeka and the next thing you know, he's being given directions to Pensacola, including a discussion on the quality of life therein. Probably best not to mention it, though. Some people just might take offense.

Harumph!

 

[electricpete]

I'll accept the label of oversensitive if you try to hold down the sarcasm on your next clarification. This is after all a friendly forum and we shouldn't try to belittle anyone explicitly or with sacrcasm, whether we agree or not.

By the way, I've heard the Panhandle is beautiful this time of year. Gotta love that white sand on Santa Rosa Island ;-)

 

[jbartos]

Clarification/correction (I beg your pardon):
PF=cos{arccotan[(Re+Rr'/s)/(Xe+XLr')]}=
=cos{arctan[(Xe+XLr')/(Re+Rr'/s)]}

 

[peebee]

jbartos has managed to whip out yet another equation that makes my eyes spin and my head hurt. Yuck. I'm going home.

 

[Marke]

The start power factor is typically between 0.1 and 0.2 when the motor is at rest. As the speed of the motor increases, the power factor also increases. When the shaft is stationary, there is no shaft power, only shaft torque. Shaft power is a function of rotational speed and torque. There is input power to the motor however, and that is dissipated in the stator and the rotor. Typically, the starting current (LRC) for large machines is between 550% and 900% of the rated current. Modern high efficiency motors will typically have a higher starting current and a lower starting torque.
Best regards, Mark Empson

http://www.lmphotonics.com

 

[Shortstub]

Jbartos,

Minor clarification to the equation:

At start s=1, and can therefore be deleted!

 

[charlierod]

I found this one during and advanced search and would like to add a comentary:

Starting PF=0.2
=cos(arctan(Xeq/Req))
=cos(arctan(377*Leq/Req)), at 60 Hz

Time constant would be calculated as follows:

arctan(377*Leq/Req) = 78.46 degrees
377*(Leq/Req) = 5

So time constant = (Leq/Req) = (5/377) = 13 ms.

Does this mean that the initial high inrush (not locked rotor) lasts 40 ms approximately (3 times 13 ms)? By the way, does the value of power factor given apply to initial inrush or locked rotor?

 

[jbartos]

Comment on Shortstub (Electrical) Feb 22, 2003 marked ///\\Jbartos,
Minor clarification to the equation:
At start s=1, and can therefore be deleted!
///Agree. I just stated that the starting power factor can be calculated from the equation. There will be other data needed to evaluate the power factor, e.g. Re, Xe, Rr', XLr', (these are more difficult to obtain)\\\

 

[jbartos]

Suggestion to charlierod (Electrical) May 22, 2004 marked ///\\I found this one during and advanced search and would like to add a comentary:
Starting PF=0.2
=cos(arctan(Xeq/Req))
=cos(arctan(377*Leq/Req)), at 60 Hz
///Xeq and Req are actually (Xe+XLr')/(Re+Rr'), to be more accurate)\\Time constant would be calculated as follows:
arctan(377*Leq/Req) = 78.46 degrees
377*(Leq/Req) = 5
So time constant = (Leq/Req) = (5/377) = 13 ms.
Does this mean that the initial high inrush (not locked rotor) lasts 40 ms approximately (3 times 13 ms)?
///The time constant is actually obtained via tangent at t=0 sec. The high inrush current transition to the full load current takes longer, e.g. 40 ms or more, e.g. 0.5 second, etc.\\ By the way, does the value of power factor given apply to initial inrush or locked rotor?
///To the power factor actually applies to the inrush current and voltage motor condition, not to the locked-rotor condition, since the inrush condition causes the motor terminal voltage to be lower (it may be as low as 75% depending on the rated motor transient starting voltage) and consequently the current to be lower (therefore the current is called inrush current rather than the locked-rotor current.\\\

 

[charlierod]

jbartos:

I don't understand why you're saying that the transition from initial inrush to full load cutrrent lasts only 40 to 50 ms. As far as i'm concerned transition to full load amperes takes several seconds depending on voltage and load.

The differences between inrush and locked rotor current has been discussed in this forum before. The conclusion is that despite of motor voltage inrush current (motivated by RL circuit transient) is higher than steady state locked rotor current (which defines KVA code of a motor). This fact is taken into account to adjust overcurrent relay settings.

 

[jbartos]

Comment on the previous posting marked ///\\jbartos:
I don't understand why you're saying that the transition from initial inrush to full load current lasts only 40 to 50 ms. As far as i'm concerned transition to full load amperes takes several seconds depending on voltage and load.
///Agree. In fact, I mentioned "..40ms or more, e.g. 0.5sec, etc. The larger motors tend to have a longer transition over the inrush current.\\The differences between inrush and locked rotor current has been discussed in this forum before. The conclusion is that despite of motor voltage inrush current (motivated by RL circuit transient) is higher than steady state locked rotor current (which defines KVA code of a motor). This fact is taken into account to adjust overcurrent relay settings.
///The inrush current definition from IEEE Std 100-2000: "The rapid change of current with respect to time upon the motor energization. Inrush current is dependent on the voltage impressed across the motor terminals and the motor inductance by the relationship V=Ldi/dt."
The instantaneous values around inrush current peak can be very high, i.e. the higher than rms value of the locked rotor current; however, if one compares apples to apples, i.e. rms of the locked-rotor current to rms of the inrush current (up to its diminished value to the motor FLA), then the inrush current rms will be smaller than the locked-rotor current since the motor terminal voltage will be lower than motor terminal rated voltage.\\