electric motor rotor moment of inertia

[wilans]

what is the effect when i replace an electric motor with "rotor moment of inertia" from 8 kgm2 to 16 kgm2. the load is centrifugal compressor ( original motor has rotor moment of inertia 8 kgm2 )> any limitation that te ration should never exeed certain limit???? thanks in advance for the response.

 

[rcwilson]

Depends on what else is different in the newer motor and characteristics of the laod.

If the motor has the same torque characterisitcs it will take about twice as long to get up to full speed when operated uncouopled since it has twice the inertia to accelerate.

If the compressor load is much larger than the rotor inertia, it may not make much of a difference.

Higher inertia will also lead to a larger flywheel effect that will damp some of the speed fluctuations.

 

[wilans]

first of all thank you for te reply, but i have looked at the motor data sheet that the starting time is almost the same and the torque caracteristic is also almost the same. Please advice te meaning of motor rotor inertia is 16 kgm2 ( what does it actually mean? is it defining the rotor inertia itself or the maximum permissible load inertia to the rotor? )

 

[Besoeker]

first of all thank you for te reply, but i have looked at the motor data sheet that the starting time is almost the same and the torque caracteristic is also almost the same. Please advice te meaning of motor rotor inertia is 16 kgm2 ( what does it actually mean? is it defining the rotor inertia itself or the maximum permissible load inertia to the rotor? )

The inertia is a flywheel effect as rcwilson has already noted
The greater the inertia, the harder it is to accelerate it.
I have never seen rotor inertia used to describe anything other than that of the motor rotor itself.
Surprising that the two motors have such different inertias if they are the same in other respects - speed, power, voltage etc.

In theory at least, there is no upper limit to load inertia. In practical terms, acceptable acceleration times might be a factor in rating the motor for the application.
We do a bit of work in paper mills and some of the loads are very high inertia. In some cases a motor that can can accelerate to speed uncoupled in about two seconds, takes two minutes when coupled to the load.

 

[wilans]

thank you can understand but any rule or code about load to motor rotor ratio e. g compressor rotor inertia/motor rotor inertia ??

 

[Besoeker]

thank you can understand but any rule or code about load to motor rotor ratio e. g compressor rotor inertia/motor rotor inertia ??

None that I'm aware of.
Practical considerations may well dictate the maximum load inertia.
If it's too much, the run up time could be too long and exceed the thermal capacity of the motor for a direct-on-line start. The motor would burn out.
In the example you gave, I don't think there's much danger of that.

 

[kingpb]

Moment of Inertia and Rotational inertia are the same thing, and is a resistance to change of rotation; and related to angular acceleration.

The more rotational inertia an object has, the less it responds to being spun. The outside parts of a spinning object have to move much faster than inside parts near the axis.

Torque and Angular Acceleration is Newton?s Second Law for Rotations

 

[Besoeker]

Moment of Inertia and Rotational inertia are the same thing, and is a resistance to change of rotation; and related to angular acceleration.

Sort of. They are related but rather the other way round.
In the case being discussed, the inertia of rotor is a property it possess whether or not it is accelerating.

 

[kingpb]

Torque, angular acceleration and the moment of inertia are related by :



and;



Therefore I'd think they are directly related

 

[Besoeker]

Torque, angular acceleration and the moment of inertia are related by :

View attachment 6137

and;

View attachment 6140

Therefore I'd think they are directly related

This I know. And very much simpler in SI units.

My point was that rotor inertia is a value. It is what it is. Regardless of acceleration.
It is not directly related to acceleration.
It's the other way round. Changing the rate of acceleration does not change inertia.

 

[GeorgeB]

My point was that rotor inertia is a value. It is what it is.

And (another thread in mind<g>) does it depend on unstable operation of the motor?

NO, that is not my question ... just teasing.

I'm having a discussion with another (totally non electrical) engineering friend that some of this would contribute to understanding ...

A "typical" hydraulic pressure compensated piston pump requires approximately (per the manufacturer) 3 revolutions to reach stability. I am trying to determine how long a "typical standard NEMA or IEC" unloaded squirrel motor requires to accelerate from 0 to "nominal" speed started across the line. I'm not really concerned with whether the speed is 0.9PU or 0.999PU. For the sake of discussion (if it matters), let's assume a motor in the 7.5kW (10HP) to 75kW (100HP) range, 4 pole.

The moment of inertia is published for these pumps, and is usually (from visual guestimation) significantly less than that of the motor rotor ...

To furnish, perhaps, TMI (too much information), I am interested in determining whether starting load current would be lower with the pump outlet vented (low outlet pressure at full flow) or blocked (displacement will go from full to "zero" during that 3 revolutions) with the compensator vented (stable pressure reached after, WE THINK, those 3 revolutions, but that pressure will be on the order of 5x the vented pressure. OK, real numbers as example ... 100cc pump, vented at 3 bar, compensated at 15 bar, say 1500 synchronous RPM. To expand, this pump MIGHT operate at 220 bar, 140 liter/minute and consume 60kW. Oh, hard numbers ... for this 100cc pump, J is on the order of 0.02 kgm^2.

What I'd like is some thought of what the 1st second current might look like ... NO PUMP LOAD.

I've looked at things like this before with a CT into a signal conditioner, but accepting the time constant built in all I've used, things are so smoothed that I don't trust the data; it is great in operating modes, though.

To put this in perspective ... 50Hz, motor turning 25 revolutions/second synchronous ... it is shorter than the first 0.2 seconds where my HYDRAULIC loads are of interest.

My friend's employer has actually commissioned a local university to do a research project to evaluate energy savings possible via VFDs, but their equipment (and inherent slow acceleration via the VFD) won't show any more than I can do, and probably less, in startup.

 

[Besoeker]

And (another thread in mind<g>) does it depend on unstable operation of the motor?

NO, that is not my question ... just teasing.

And nothing wrong with that....

I'm having a discussion with another (totally non electrical) engineering friend that some of this would contribute to understanding ...

A "typical" hydraulic pressure compensated piston pump requires approximately (per the manufacturer) 3 revolutions to reach stability. I am trying to determine how long a "typical standard NEMA or IEC" unloaded squirrel motor requires to accelerate from 0 to "nominal" speed started across the line. I'm not really concerned with whether the speed is 0.9PU or 0.999PU. For the sake of discussion (if it matters), let's assume a motor in the 7.5kW (10HP) to 75kW (100HP) range, 4 pole.

I've always guessed, or maybe I was told at some point in the dim and distant past, that the starting time uncoupled was in the region of one second.

The following is for a rather larger motor that I happen to have the inertia for:

I assumed constant rated torque for the run up period. I know that the torque isn't constant but looking at the curve, it looks like it would average out somewhere around there.
So one second seems to be in the right ball park.

The moment of inertia is published for these pumps, and is usually (from visual guestimation) significantly less than that of the motor rotor ...

To furnish, perhaps, TMI (too much information), I am interested in determining whether starting load current would be lower with the pump outlet vented (low outlet pressure at full flow) or blocked (displacement will go from full to "zero" during that 3 revolutions) with the compensator vented (stable pressure reached after, WE THINK, those 3 revolutions, but that pressure will be on the order of 5x the vented pressure. OK, real numbers as example ... 100cc pump, vented at 3 bar, compensated at 15 bar, say 1500 synchronous RPM. To expand, this pump MIGHT operate at 220 bar, 140 liter/minute and consume 60kW. Oh, hard numbers ... for this 100cc pump, J is on the order of 0.02 kgm^2.

What I'd like is some thought of what the 1st second current might look like ... NO PUMP LOAD.

Although it might sound counter intuitive, the loading doesn't affect the magnitude of the run up current, just the duration. Here's one I posted well....you know where........

What you will notice is that the current is high and close locked rotor current value until it gets near to the normal operating range. Take, for example, 0.6pu speed. At that point, the current is around 6.8 times rated current and that's what it is at 0.6pu speed regardless of what's hanging on the shaft - or nothing.

I've looked at things like this before with a CT into a signal conditioner, but accepting the time constant built in all I've used, things are so smoothed that I don't trust the data; it is great in operating modes, though.

I usually use a a clip on CT with a resistive burden and look at it with an oscilloscope. It takes out any intermediate signal conditioning.

To put this in perspective ... 50Hz, motor turning 25 revolutions/second synchronous ... it is shorter than the first 0.2 seconds where my HYDRAULIC loads are of interest.

My friend's employer has actually commissioned a local university to do a research project to evaluate energy savings possible via VFDs, but their equipment (and inherent slow acceleration via the VFD) won't show any more than I can do, and probably less, in startup.

Unless the motor and load need to be variable speed there is not a lot of point of considering a VFD.
Do you have any more detail of what they are trying to achieve?

 

[GeorgeB]

Although it might sound counter intuitive, the loading doesn't affect the magnitude of the run up current, just the duration. Here's one I posted well....you know where........

What you will notice is that the current is high and close locked rotor current value until it gets near to the normal operating range. Take, for example, 0.6pu speed. At that point, the current is around 6.8 times rated current and that's what it is at 0.6pu speed regardless of what's hanging on the shaft - or nothing.

I understand the current at 0.6pu==current at 0.6pu, so I'm likely confused by the axis markings. Is pu speed horizontal? If so, it appears this plot, at 0.6pu speed (and further assuming full vertical is 7pu rated current), current would be about 0.3pu. What am I missing? Green is current and blue torque? Current peak occurs at about 0.93pu speed on that graph? Torque is substantially constant around 3pu below 0.6pu speed?

In any event, that revelation of ABOUT 1 second changes lots of my questions as the hydraulic system will be at equilibrium in well under that time.

I usually use a a clip on CT with a resistive burden and look at it with an oscilloscope. It takes out any intermediate signal conditioning.

I recently upgraded my data acquisition hardware such that I can look at the AC waveform ... previously I was limited to 60 samples/sec at best which resulted in "poor" data ... I'll have to hook across my burden resistor and investigate it.

Unless the motor and load need to be variable speed there is not a lot of point of considering a VFD.
Do you have any more detail of what they are trying to achieve?

One goal of the organization is to offer energy savings to customers via various hydraulic improvements they can sell. We "know" that a pressure compensated pump SYSTEM with multiple loads which require various flows and pressures, some of them coincident, is not very efficient. We "know" that optimum hydraulic efficiency would be obtained by individual sources per load supplying the flow required capable of the pressure required. I "KNOW" that individual pumps on individual motors with flow control either by speed or displacement control is not cost efficient.

Their goal as I understand it is to determine if there is any advantage (well really sufficient advantage to justify the very minor performance issues) by just reducing the speed of the motor-pump when flow is not needed. As I see it, only friction will be reduced, but there is MUCH confusion among the mechanical engineers as to the significance of power factor and current. They feel that the VFD will reduce POWER by the line power factor improvement. After some of the thoughts here with 0.95 pf into the drive but 0.05 pf to the idling motor ... there MAY be some demand reduction ... but these salesmen are going to sell snake oil that the current went down by 75% so the power went down by 75%.

Thanks for your insight.

George

 

[Besoeker]

I understand the current at 0.6pu==current at 0.6pu, so I'm likely confused by the axis markings. Is pu speed horizontal? If so, it appears this plot, at 0.6pu speed (and further assuming full vertical is 7pu rated current), current would be about 0.3pu. What am I missing? Green is current and blue torque? Current peak occurs at about 0.93pu speed on that graph? Torque is substantially constant around 3pu below 0.6pu speed?

My apologies for the lack of annotation.
Yes, pu speed is horizontal. But green is torque, not current.
The current remains substantially constant up to about 0.8pu speed.
Torque has the familiar hump shape peaking at about 0.93pu speed.

In any event, that revelation of ABOUT 1 second changes lots of my questions as the hydraulic system will be at equilibrium in well under that time.

But maybe not from standstill if the motor is still accelerating?

One goal of the organization is to offer energy savings to customers via various hydraulic improvements they can sell. We "know" that a pressure compensated pump SYSTEM with multiple loads which require various flows and pressures, some of them coincident, is not very efficient. We "know" that optimum hydraulic efficiency would be obtained by individual sources per load supplying the flow required capable of the pressure required. I "KNOW" that individual pumps on individual motors with flow control either by speed or displacement control is not cost efficient.

Their goal as I understand it is to determine if there is any advantage (well really sufficient advantage to justify the very minor performance issues) by just reducing the speed of the motor-pump when flow is not needed. As I see it, only friction will be reduced, but there is MUCH confusion among the mechanical engineers as to the significance of power factor and current. They feel that the VFD will reduce POWER by the line power factor improvement. After some of the thoughts here with 0.95 pf into the drive but 0.05 pf to the idling motor ... there MAY be some demand reduction ... but these salesmen are going to sell snake oil that the current went down by 75% so the power went down by 75%.
Thanks for your insight.

Yep, that magic box.



George[/QUOTE]

 

[GeorgeB]

But maybe not from standstill if the motor is still accelerating?

If I accept that Eaton-Vickers, Bosch-Rexroth, and Parker know of what they speak and that PUMP equilibrium is reached within 3 revolutions, there is little doubt that OUR SYSTEMS will complete 3 revolutions in less than 1 second. To take an extreme case, assuming 50Hz power and an 8 pole motor (slowest "ever" used, 2, 4, and 6 pole are way over 99% of applications), synchronous speed is 12.5 rev/sec. ASSUME linear acceleration for 1 second, 3 rotations in about 0.25 seconds. Allow for (I've NEVER seen it that slow across the line) a 4 second acceleration, still under a second.

All that to say that your curve covers my questions ... THANKS and Merry Christmas!

George