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What Is Normal No Load Idle Current? 2

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OhioAviator

Electrical
Sep 8, 2003
123
Hi All,

Someone told me today that as a general rule of thumb, motor idle current will increase as the motor nameplate speed decreases. In other words, a 3600 rpm (2 pole) motor might have 10-15% FLA at no load, an 1800 rpm (4 pole) motor might have 25-30% FLA at no load, and a 600 rpm (12 pole) motor might have 40-50% FLA at no load. Is this true?

The reason I ask is that I have a 4,500 HP Siemens-Allis 12-pole (600 RPM) squirrel-cage induction motor operating an automobile shredder. With the motor completely uncoupled from the shredder, the motor idles at about 250 amps, which is about 48% FLA. Does that sound normal to you guys? If not, what should the no load FLA be?

Thanks!
 
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Yes, it is true that no-load current typically increases as speed decreases and the values you mention sound reasonable (except 10% sounds a little on the low-side to me even for 2-pole motors).

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Here is a simplistic way to think about it.

If you have two motors both 100hp, both 480vac with one 2-poles and one with 12-poles:

They will both have roughly the same full-load current since that is related primarily to Hp and voltage (and lesser extent power factor and efficiency).

The low-speed motor will be a larger motor with a larger airgap and will therefore require higher magnetizing current. This means that for the low spee motor, the no-load current will be higher and the full load power factor will be lower (in general)

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I agree with Electripete, your figures look right except for 2 poles which could be around 20 to 30%. The 250 amps, current which is about 48% FLA for a 12 poles motor looks right.
No load currents are an option the motor designers will handle to comply the torque, speed, efficiency, temperature rise, cost, etc. No standard regulate that parameter and could change from motor to motor and manufacturer.
 
Thank you, electricpete and aolalde. I sure do appreciate your help. It all makes sense and you also confirmed what other 'motor people' have told me. Since my experience has been largely confined to 1800 RPM motors, I guess I never payed much attention, or given much thought, to that particualar facet of motor operating characteristics before.

Again, many thanks!
 
Sounds kinda nasty since I bet an auto shredder has a high idle percentage.
 
The current is high, but the power factor is low. So, unless the auto shredder is paying power factor penalties, there probably isn't much difference in the power consumed between the 2 pole and the 12 pole motor while unloaded.
 
electricpete, Sorry to say but on low speed motors the air gap decreases. Typical for 2 pole at 500 hp would be around 0.12", where an 8 pole would only be around 0.05".

 
That is true Mendit; the air gap reduces with the increase of poles on induction motors. However the magnetizing current still increases with the increase of poles. I tested no load a large vertical motor 1250HP, 28 poles, 60 HZ, 255 rpm, 4 kV, 255 amperes at full load. The no load current tested 197.5 Amperes or 77.2% as compared to full load current.
 
Sorry to day mendit, but you have misunderstood my intent.

"larger motor with a larger airgap"

Let me clarify. The airgap has a length (around the circumference), a height, and a depth (across the airgap). By larger I was referring to the length and the height (not depth). It should be obvious that length and/or height must increase when we increase the physical size of the motor.

The general form for inductance associated with an airgap is N^2 * depth / (A *mu0). For our cylindrical airgap, A is airgap length times core height.

As airgap length or height increases, A increases, magnetizing reactance decreases, magnetizing current increases.

The increase in length and height of airgap for slow speed motors is unavoidable and tends to cause increase in magnetizing current.

There are no inherent constraints associated with the speed of the motor that affect the depth of the airgap. However, for slow speed motors, manufacturers will tend to try to make the airgap smaller to compensate for the lower power factor caused by longer/taller airgap. There are limitations of how small a depth airgap you can make without unreasonable risk of pullover and rub, so this decrease in depth generally cannot compensate for increase in length and height.

Bottom line, the factor which causes decrease in power factor for slow speed motors is increase in length and height of the airgap.

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I don't think the inductance formula I provided is directly applicable but it's just intended to illustrate the point that longer and higher airgap require more exciting current.

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A better illustration would have been Flux = B * A.

Field B needs to stay constant. That means as A increases we have more flux. We need more exciting current (NI) to achieve that higher flux.

Mabye someone can provide a better explanation.

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I was mistaken, when you apply more than 1 dimension to the airgap. Typical in a 2 pole there are 2 A phase 2 B phase and 2 C Phase groups of coils. Thus each coil group has a circumfrencial airgap of 60 degrees
on an 8 pole this is only 15 degrees. The magnetic pull is around and not just radial. Thus making the magnetic pull arc shorter. Thanks for the clarity. Just being mechanical I was only concerned about the radial airgap.
 
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