Motor Inrush Current

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mull982

Senior Member
When a motor is started there is what people commonly refer to as "insush current" This inrush current consists of the magnetising inrush current to create the magnetic field in the windings and the motor starting current which is a function of the motors Locked Rotor Current. The magnetising current lasts briefly only for maybe a 1/10th of a second before it is diminished to the value of the starting current. The magnetising inrush current is higher than the starting current, however my question is by what factor? I have heard that a typical multiplier for this inrush current is 1.6 times the Locked rotor current of the motor? Does anyone know if there is a multiplier for determining this current?

Also I believe the LRC is a function of motor slip with the greatest LRC occuring when slip=1 or when the rotor is at rest. However if the rotor is already spinning (draft on a fan) when power is applied to the motor would the slip at start be something less than 1 and therefore make the LRC less than the LRC that the motor would have at rest? If this is the case then would the magnetising inrush current for a motor be less on a motor that had a rotor already spinning prior to applying power? Would this diminished magnetising inrush current value be a multiplier (as mentioned above) times the new LRC of the motor that is already spinning?
 

mikeames

Senior Member
Location
Germantown MD
I think your over thinking it. Although I agree with 95% of what you wrote.

LRC is just that a rotor in a stalled state, anything else is a loaded motor (unless your are talking about a few RPM) When that is the case think of the motor as what it is at that point, a giant short circuit. (Thus the resistance reading you get on a motor, low to zero depending) If the rotor is moving its not really locked. A spinning rotor would always produce a smaller inrush vs a stationay rotor. Although if you are talking a few rpm or so you may as well consider it locked. I do not know about any multiple maybe there is?
 

Jraef

Moderator
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
You can't make a catch-all generalization for magnetizing (inrush) current. It totally depends upon the winding resistance, phase angle that the line s at when the contactor closes, air gap, number of turns, magnetic permeability of the core material etc. etc... In the "olden days" maybe you could have boiled it down to something as a gross generalization, although I have never heard the 1.6X LRC before.That sounds suspiciously low. However, in the past 15 years virtually every motor mfr has been tweaking their designs to squeeze more efficiency out of them, and one common side effect is that Magnetizing Current has actually been going up, because reducing wnding resistance reduces the copper losses when running. As a result, it is now not uncommon to see Magnetizing Currents as high as 20X FLC!

So let's see, if LRC = 600% FLC, then 1.6 X LRC = 9.6X FLC. The "old" Magnetizing Current rule of thumb used to be 10-13X FLC, so that sort o fits. But I'd say that is outdated. If it's an Energy Efficient motor, I would assume 13-17X FLC, maybe more.

This, by the way, is what was behind the changes in the NEC, Article 430 (.52?) that now allows for higher maximum instantaneous trip settings of MCCBs when it is demonstrated that lower levels will not hold in. It used to say 1000 - 1300%, now I think it says up to 1700% (?).
 

Cold Fusion

Senior Member
Location
way north
mull -
I can give you a statistical sample of one.

I put a recorder on a 40hp, 480V reliance, energy efficient per MG 1 standards. I looked up the recordings, so the numbers are close.
FLA = 47.7
LRC = 325A (500A peak)
Inrush - 450A for 20 ms (800A peak)

Ran this three times. nearly identical each time. Really surprised me. I never expected that 125A, 20ms bump above the LRC right at the start.

cf
 

mull982

Senior Member
You can't make a catch-all generalization for magnetizing (inrush) current. It totally depends upon the winding resistance, phase angle that the line s at when the contactor closes, air gap, number of turns, magnetic permeability of the core material etc. etc... In the "olden days" maybe you could have boiled it down to something as a gross generalization, although I have never heard the 1.6X LRC before.That sounds suspiciously low. However, in the past 15 years virtually every motor mfr has been tweaking their designs to squeeze more efficiency out of them, and one common side effect is that Magnetizing Current has actually been going up, because reducing wnding resistance reduces the copper losses when running. As a result, it is now not uncommon to see Magnetizing Currents as high as 20X FLC!

So let's see, if LRC = 600% FLC, then 1.6 X LRC = 9.6X FLC. The "old" Magnetizing Current rule of thumb used to be 10-13X FLC, so that sort o fits. But I'd say that is outdated. If it's an Energy Efficient motor, I would assume 13-17X FLC, maybe more.

This, by the way, is what was behind the changes in the NEC, Article 430 (.52?) that now allows for higher maximum instantaneous trip settings of MCCBs when it is demonstrated that lower levels will not hold in. It used to say 1000 - 1300%, now I think it says up to 1700% (?).
Good point about the Magnetizing Current being a function of several variables.

In my OP I refered to a lower LRC when rotor was already spinning, but I realize I used the wrong terminology due to the fact that if the rotor is spinning, this is no longer a locked rotor.

Was what I mentioned in my OP about the starting current being lower with spinnnig rotor true due to the fact that slip is less than 1 (higher rotor resistance) and that we are already further to the right on the motors speed vs torque curve with the current decreasing as we move right.

I thought I heard this 1.65 multiplication ratio somwhere but I cant remember where. Does it still hold true however that the magnetising current will be a function of starting current, so if the rotor is already turning when power is applied then magnetising current will be proportionately lower?
 

GeorgeB

ElectroHydraulics engineer (retired)
Location
Greenville SC
Occupation
Retired
A spinning rotor would always produce a smaller inrush vs a stationary rotor.
An engineer with a firm making large VFDs told me that isn't true. The more slip that exists, the shallower in the rotor shorting material is in play which is usually thinner and higher resistance. He told me (this was 8 or so years back when we were trying to use a smaller VFD and accelerate the motor with a smaller motor) that the instantaneous peak current would be much higher at 95% of sync speed than at 0%.

The peak wouldn't last long, however ... just long enough to blow the semiconductors that protect the semiconductor fuses <g>.

I have no other support of this, but after the explanation, it made sense.
 

templdl

Senior Member
Location
Wisconsin
LRA is commonly around 7x FLA.
Magnetizing currect is much higher than that. Wen a motor is first energized there is no counter EMF to oppose current flow for the first 1/2 cycle. The spike of current is basicaly dependent upon the resistance of the wire in the stator which motor have always had and quickly decays to the LRA.
When energy efficint motors evolved the motors were made more efficient be optmizing the windings which reduced thier resistance. And, yes, you guessed it, by reducung the winding resistance the magnetizing inrush current increased dramatically. If you followed the NEC 430 during the 1980s the setting point which you were allowed to set the pickup of an MCP was 13x of the FLA of the motor. The HE motors often times ran in far in excess of 13x and the MCPs nuisance tripped. Art 430-52 was ammended to allow settings of up to 17x.
 

mikeames

Senior Member
Location
Germantown MD
An engineer with a firm making large VFDs told me that isn't true. The more slip that exists, the shallower in the rotor shorting material is in play which is usually thinner and higher resistance. He told me (this was 8 or so years back when we were trying to use a smaller VFD and accelerate the motor with a smaller motor) that the instantaneous peak current would be much higher at 95% of sync speed than at 0%.

The peak wouldn't last long, however ... just long enough to blow the semiconductors that protect the semiconductor fuses <g>.

I have no other support of this, but after the explanation, it made sense.
I would believe this for a very small portion of time. However after that micro second of time I would have to believe the current required from then on would be lower. I can understand the importance when dealing with VFD's.
 

Besoeker

Senior Member
Location
UK
An engineer with a firm making large VFDs told me that isn't true. The more slip that exists, the shallower in the rotor shorting material is in play which is usually thinner and higher resistance. He told me (this was 8 or so years back when we were trying to use a smaller VFD and accelerate the motor with a smaller motor) that the instantaneous peak current would be much higher at 95% of sync speed than at 0%.
In general you wouldn't want to run a VFD at much above the slip that results in full load torque and full load current. For the VFDs motors I have dealt with, maximum slip is generally in the region of 1% or 99% 0f synchronous speed.
That said, we mostly deal with drives in the few 100 kW to the few MW.
 

Rick Christopherson

Senior Member
FLA = 47.7
LRC = 325A (500A peak)
Inrush - 450A for 20 ms (800A peak)

Ran this three times. nearly identical each time. Really surprised me. I never expected that 125A, 20ms bump above the LRC right at the start.
Well, you've corrected me before I could even make a posting, so for that I am grateful. :grin: However, that doesn't eliminate my confusion with these numbers.

A locked rotor motor is "supposedly" nothing more than an induction coil with an iron core, and I had been taught that current through an inductor cannot change instantaneously. Building the magnetic field takes energy and creates back emf, so I would have thought that the instantaneous current in the coil just after time=0 would be lower than the steady state current with the rotor still locked.

Hmmm, as I was about to ask someone to explain this, it occurred to me that maybe my model was wrong, in that this is not an induction coil, but actually a transformer with the secondary (rotor bars) shorted. I guess there must be a delay in coupling the primary and secondary coils, so maybe the instantaneous model is still an induction coil, but the steady state model is a shorted transformer. I am still not clear on how this results in the higher instantaneous current, but I would like to learn it.

Can anyone explain this for me?
 

gar

Senior Member
090815-2005 EST

Rick:

If you have a coil with a ferromagnetic core that is left in a state near saturation, then apply a step voltage of a polarity that forces the core more into saturation the inductance will drop even more and the rate of rise of current will be much greater than if you had started from zero residual flux in the core.

In the case of AC excitation this means you will have a much larger peak current on the first half cycle.

See my scope waveforms P6-P8 for a transformer at:
http://beta-a2.com/EE-photos.html
This is an unloaded transformer and therefore the secondary plays no part in the current waveforms.

.
 

Cold Fusion

Senior Member
Location
way north
Well, you've corrected me before I could even make a posting, so for that I am grateful. ---
You're welcome

--- However, that doesn't eliminate my confusion with these numbers. ---
mine neither

---Can anyone explain this for me?
I wish I could. But I can't.

The one I listed is the only one I ever put a power monitor on. I had the instantaneous trip set up 11x and it was tripping occasionally. I thought it might be spinning backwards when it was starting. I put up a post asking if any one had experience, but never got anything i could use. Finally put a power monitor on it and there was the cutest little 20ms bump above the LRC right at start up. Measureable, verifiable, repeatable. The phase with the highest current would change, but the bump was always there.

But this is a statistical sample of one. I know a whole lot about that particular motor, and it's sister, so I can likely generalize to this model of motor, but I wouldn't be comfortable generalizing to all energy efficient motors.

I'll eventually get another chance and will do some more. But until then ...

cf
 

wjock

New member
If a have a pony motor to run up a larger induction motor and them near rated speed I energize the motor, what could be the inrush current?
Thanks
 

GeorgeB

ElectroHydraulics engineer (retired)
Location
Greenville SC
Occupation
Retired
If a have a pony motor to run up a larger induction motor and them near rated speed I energize the motor, what could be the inrush current?
Thanks
I had a drive manufacturer tell me it is much higher. The slip, and slot shape, make 60 Hz slip and the depth in the rotor create a (relatively) high impedance magnetic circuit. When the motor is near sync speed, there is little slip, and the magnetic circuit is far more of a short circuit.

He said that the duration of inrush type current is actually longer as well as being higher. The total time from energizing to operating speed is shorter, however.

In this case, I was working on a 1000 hp 4160 motor and just a softstart.
 
The span of the windings maybe part of it too.

Span is the slots covered by the winding and number of slots it covers.

Not sure, but it would seem that some pole winding "almost" overlap would be big on the class B motors for energy efficiency.
 

winnie

Senior Member
Building the magnetic field takes energy and creates back emf, so I would have thought that the instantaneous current in the coil just after time=0 would be lower than the steady state current with the rotor still locked.
Rick,

This is in answer to the subtopic about the 'inrush' current associated with applying AC voltage to an inductor (any inductor, including a motor or a transformer primary).

You are quit correct that it takes time for the current to build, and that the instantaneous current at time T=0+epsilon would be very close to zero.

The real key to understanding 'inrush' current is to ask what happens at about T=1/2 cycle after you've applied the voltage.

When the inductor is operating in steady state with AC that has been applied for 'a long time', then the magnetizing current flow and magnetic flux will lag the applied AC voltage by 90 degrees. Right at the zero cross of the AC voltage cycle you have _peak_ current flow and _peak_ flux.

Over the course of an applied voltage half cycle, the flux and current changes from one peak to the opposite peak. In the steady state, each AC half cycle causes a flux change equal to _twice_ the peak flux.

Now consider the transient conditions at T=0. The flux in the inductor is (ideally) zero. Presume for a moment that you got 'lucky' and managed to switch on the AC right at the zero cross. The volt-second product for that half cycle is the same as in the steady state, and thus the _change_ in flux is trying to be exactly the same as in the steady state case. However now instead of going from -peak flux to +peak flux (or vice versa), you are going from zero to 2xnormal peak flux.

This saturates the inductor and leads to tremendous current flow. As the asymmetry dies away, the current flow then falls to the steady state value. The forces that push the inductor to the steady state case are things such as magnetization losses and winding resistance; if you had a perfect inductor and AC voltage source then the startup transient would never die away; depending upon starting conditions the current waveform could end up anywhere from (0 to 2xpeak) to (-peak to +peak) to (-2xpeak to 0).

-Jon
 
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