power factor when start the induction motor
power factor when start the induction motor
(OP)
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.
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.





RE: power factor when start the induction motor
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.
RE: power factor when start the induction motor
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.
RE: power factor when start the induction motor
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
RE: power factor when start the induction motor
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.
RE: power factor when start the induction motor
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.
RE: power factor when start the induction motor
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.
RE: power factor when start the induction motor
RE: power factor when start the induction motor
The default PF for induction-motor starting in the 1955 Beeman text is 0.2.
RE: power factor when start the induction motor
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!
RE: power factor when start the induction motor
By the way, I've heard the Panhandle is beautiful this time of year. Gotta love that white sand on Santa Rosa Island
RE: power factor when start the induction motor
PF=cos{arccotan[(Re+Rr'/s)/(Xe+XLr')]}=
=cos{arctan[(Xe+XLr')/(Re+Rr'/s)]}
RE: power factor when start the induction motor
RE: power factor when start the induction motor
Best regards,
Mark Empson
http://www.lmphotonics.com
RE: power factor when start the induction motor
Minor clarification to the equation:
At start s=1, and can therefore be deleted!
RE: power factor when start the induction motor
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?
RE: power factor when start the induction motor
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)\\\
RE: power factor when start the induction motor
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.\\\
RE: power factor when start the induction motor
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.
RE: power factor when start the induction motor
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.\\\