Induction motor modelling. 1

[gokulkrish2]

Hi everyone.

I need some help with modelling an induction motor. All i want to do is to get a model of an induction motor which can simulate the starting time(Time it takes to get to the full speed) with no load attached to it. I tried to use matlab simulink but there is a problem. I have the motor parameters like R and X from the manufacturer. They have given me 2 sets of data(Data Attached). I dont know why they have given me 2 sets for full load and Locked rotor. But when i try these values i get two different starting times. I believe its giving the deep bar effect of the motor. But i want to have an idea from you guys. please help.

gokul

 

[magoo2]

I only see 1 set of data. It covers full load and starting.

 

[gokulkrish2]

ya i was saying two sets for Full load and starting only.

My question is R and X should be one value right.

I dont know if i am wrong. But i think that how can the same winding can have different values while starting and at full load.

 

[electricpete]

I would say mabye R1 changed due to temperature (assume higher temperature running than starting).

R2/s of course changed due to change in slip.

Varying levels of sSaturation of certain areas may affect X1, X2, Xm.

Deep bar effect may play some role as well for R2 and maybe X2.

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[gokulkrish2]

Ya I agree that slip changes. But when we calculate with the corresponding slip value too R2 is differing.

Major doubts i have is that i dont know if this motor has a rotor which accounts for deep bar effect.

And suppose it provides deep bar effect then it should have a higher resistance and reactance at the starting right. But for my agony it is in the other way around. Please give me some reason for that.

thanks
gokul

 

[electricpete]

This does in fact show higher rotor resistance at starting as expected by deep bar effect if you account for slip.

Locked rotor:
R2/s = 0.0769
s=1
R2 = R2/s * s = 0.0769

Running
R2/s = 2.8292
s = (1800-1785)/1800 = 0.01
R2 = s* R2/s = 0.0283

R2 higher during locked rotor due to deep bar effect.


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[gokulkrish2]

ok. Thank you for that clarification. Lets take that that motor accounts for deep bar effect.

Now my question is why is then the reactances x1 adn x2 are greater when they are at full load? They are also kind of resistances and hence should be high at starting and less at running correct.

This is where this issue confuses me

 

[electricpete]

As I said the reactances are a little trickier.

Here is my take fwiw.

The leakage reactance paths saturate during start due to high current. For example in closed rotor slots that would be the bridge section. In open or semi-open slots, the tooth and tooth ear corners. This is the iron part where flux the flux that encircles the conductor (leakage flux) IS constricted (high flux density). Saturation of leakage reactance during start means the effective leakage inductances are lower during start.

On the other hand, the airgap flux is higher during run than start and the toothtops tend to move more into saturation. That will make running magnetizing reactance slightly higher.


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[electricpete]

Two corrections in bold below.

The leakage reactance paths saturate during start due to high current. For example in closed rotor slots that would be the bridge section. In open or semi-open slots, the tooth ear corners. This is the iron part where flux the flux that encircles the conductor (leakage flux) IS constricted (high flux density). Saturation of leakage reactance during start means the effective leakage inductances are lower during start.

On the other hand, the airgap flux is higher during run than start and the toothtops tend to move more into saturation. That will make running magnetizing reactance slightly lower.

The two corrections account for the following:

1 - The area that saturates for semi-open or open slots is tooth ear corners (not teeth in general).
2 - Whichever one saturates gets lower effective reactance. So the running magnetizing reactance is lower (since it is more saturated.


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[gokulkrish2]

Thank you so much for that heads up. Do you have some specific literature explaining these things online so that i can go through them and get known. Thanks.

gokul

 

[cherry2000]

Gokul..Typically to calculate the starting time, the driving torque & driven torque are needed. Do you have the Moment of Inertia of the motor only?

 

[burnt2x]

cherry2000,
Please take a close look at the attachement:
Rotor Inertia = 87 lbs- ft²

 

[cherry2000]

One more data required for calculating the starting time is the Motor speed torque characteristic. The no load curve also will be required. If both can be plotted together,the starting time can be calculated by deriving the plotted values on an excel sheet and using some fundamental formulae

 

[gokulkrish2]

i do have the speed torque characteristics curve and bunch of other curves as given in the attachment. Can you please tell me now how should i approach on the issue. Thank you very much

gokul

 

[safiabd73]

the one set of data in locked rotor when s= 1
r2 vary with s in locked rotor or at start r2 is minimum
the secend set is in full load when s=sr=ns-nr/ns=min
r2 =max

 

[electricpete]

You want time to start an unloaded motor... all it requires is torque speed curve and inertia.

Start from first principles:
dw/dt = (Telec-Tmech)/J
dt = [J/(Telec-Tmech)] dw
time = [J/(Telec-Tmech)] dw, w=0..wfinal

For N in rpm, J in lbm-ft^2, T in ft-lbf, time in seconds, we can express the above equation as follows :
time = [J/<(Telec-Tmech)*307>] dN, N=0..Nfinal
The constant 307 = 60*32.2/(2Pi)
Of course in your case Tmech=0.

The integration can be carried out numerically. A long time ago I created a a spreadsheet that does the job. Requires reading off of the torque speed curve. I have plugged in coarsely-spaced data from your torque speed curve along with inertia = 87 lbm=ft^2 mentioned above. I get 0.54 seconds acceleration time at full voltage and 0.846 seconds at 80% voltage.

You will certainly want to check the calculation carefully for yourself to make sure I have correctly represented your problem and make sure you agree with the calcualtion approach and make sure I didn't make any mistakes.

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[electricpete]

I left out the integral sign in expression for time - I think you can figure out where it goes.

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[electricpete]

The endpoint of the integration deserves a little more discussion. You actually cannot integrate all the way to the point of 0 accelerating torque since it would require smaller and smaller step size and the time to reach 0 accelerating torque is theoretically infinite (integral 1/x, x=0..x0 is infinite) even though for all measurable and practical purposes the speed becomes steady.

To get around this, we need to select an endpoint which is somewhere short of the 0-accelerating torque point (for example within some tolerance). In the case of no-load motor full load start, assuming we are looking to size relays, a convenient place to stop the integration might be the speed corresponding to full-load torque, since current will drop to FLA at this point. So to be more precise I should have divided the interval beyond breakdown torque into more slices and stopped at the speed corresponding to full load and stated it was the time to accelerate to full load speed. If you perform that excercise, I don't think you will see much change in the result.


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[electricpete]

Attached is a revised calculation - time to accelerate the unloaded motor up to the point that current drops to/below full load current.

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[electricpete]

Attached is a corrected version of the acceleration time calculation (corrected the torque at 1740 rpm)

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