Induction motor Inertia 1

[luis7006]

Hello,

We are currently simulating the motor starting of some large medium voltage motors.
I have some questions that I hope to receive your feedback to understand:

1) Manufacturers' catalogue state intertia. Is this value the inertia value for the motor without load, I mean, only the rotor without any mechanical load coupled?

2) Does this inertia value vary too much?
I mean, in one manufacturer it could be 100 and for the same motor characteristics, another manufacturer says 20.

3)If we do not have any data about the mechanical data, and we know it is only a fan or a pump, how can we model the load characteristic?

I thank in advance your answers/ideas.

 

[dpc]

1. Yes

2. It should be similar for same hp and speed.

3. You can't. But consider typical variable torque loads such as fans and pumps will have similar speed-torque curves.

 

[zekeman]

If you are talking a transient startup, the load you are referring to is extraordinarily complicated, especially for a fluid being pumped; it involves transient flow of the fluid in a flow field which makes the coupled inertial dynamics and electrical motor characteristics seem like a "walk in the park".

 

[sreid]

2) Make certain that the motor inertias are in the same units.

3)For a first order approximation, model the torque proportional to the square of speed ( or power as the cube of speed).

 

[luis7006]

Hello,

I thank your answers!

Just one additional question:

We modeled a load using a square torque approximation and the motor starts faster with this load that without it.
Could anybody give us a clue or tell us why?

Best Regards

 

[sreid]

A wrong sign in your equations somewhere (a - where there should have been a +)?

 

[dpc]

I agree with sreid, the result indicates a problem in your modeling.

 

[jraef]

What are you using to do the modeling? I used to use SKM Power Tools, they had a library of typical load profiles that I found to be very useful in situations like that.


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[7anoter4]

The moment of inertia may be included in the manufacturer data as:
GD^2 [metric] in to variants: in kgf.m^2 or kg.m^2 where kgf.m^2=gmetric*kg.m^2 [gmetric=9.81 m/sec^2 gravity acceleration]
Wk^2[British] also in to variants in lbf.ft^2 or lb.ft^2 where lbf.ft^2=gbritish*lb.ft^2 [gbritish=32.19 ft/sec^2 gravity acceleration]
I [or J] =GD^2/4/gmetric if GD^2 is expressed in kgf.m^2 or I= GD^2/4 if it is kg.m^2
In British units I=Wk^2[lb.ft^2] or Wk^2[lbf.ft^2]/gbritish.
Then 1kg.m^2[GD^2] corresponds to 5.933 lb.ft^2[Wk^2]
If one knows rotor Mass [kg or lb] and the rotor diameter [m or ft] then I=Mass*Dia^2/8 [kg.m^2 or lb.ft^2].
If one don't know rotor Mass [kg or lb] and the rotor diameter [m or ft] then [as a [very!] approximate mode of calculation] one could calculate as follows:

Regards