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quiz: motor Vregen higher after interrupt from no-load or...?
5

quiz: motor Vregen higher after interrupt from no-load or...?

quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
Let's say you had an instrument set up to monitor motor terminal voltage in the 20 millisecond period after opening the breaker.  In which case would you think the measured residual voltage would be highest:
A - Interrupt from no-load condition
B - Interrupt from full-load condition
C - Interrupt while accelerating during DOL start (let's say at 60% of full speed) ?

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

2
(OP)
And just to clarify, I am just talking about the simple sinusoidal decaying residual voltage at a frequency of rotor speed (would last a lot longer than 20 milliseconds), and I'm not considering any capacitive ringing.  

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

3
Intuitively, I would say C.

The reason, I think, is that we have the highest torque in such a situation and therefore the highest 'frozen' flux in the rotor.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: quiz: motor Vregen higher after interrupt from no-load or...?

I could argue that considering the regen voltage to be the back EMF, That would be highest at the highest speed or at no load.
On the other hand, after a day to think about it I may be able to argue Gunnar's position.
I really don't know but I'm looking forward to an interesting discussion.
I visualize the highest torque not as higher flux but as "stretched" flux. Consider a synchronous machine. As the torque load on the rotor increases there is a slight angular displacement. This displacement "stretches the flux across the air gap. Even a slight elongation of the air gap will require greater magnetizing current but the flux density may not increase.
I expect a similar situation in an induction machine but with more complicated calculations due to the slip frequency.
Still don't know though. Just a suggestion pending posting by the experts.  

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
Based on simulations (attached), A has highest residual voltage, B has slightly less, C has much less.  It was not intuitive to me, but I included an attempt to explain it(after the simulation 20-20 hindsight) in slide 6.  

Bill – you win the prize!

This was part of a simplistic attempt to analyse voltage across recently-interrrupted contacts under various conditions. Conclusions on slide 8... any comments on that?
 

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
In the unlikely event anyone else wants to try to validate my simulation, here are the motor parameters used:
R_1    =0.1524510
X_1    =2.6161253
R_2    =0.2047222
X_2    =2.3677244
X_M    =102.4918380
FullLoadSlip    = 0.0065265
And of course Vln = 13200/sqrt(3), sync speed = 3600rpm.

The above are the full-speed parameters and should be close enough for these simulations done at speeds 60% or more.    For completeness, I'll mention some far less important parameters: I used a deep bar correction assuming rectangular bar with assumed depth of 0.0176122m for purposes of correcting R2 and with assumed depth of 0.0109968m for purposes of correcting X2.  

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
Regarding the attempted intuitive explanation on slide 6, it could be improved by breaking viewing the transient in two different ref frames:
1 - use the sync ref frame to establish initial conditions just prior to interruption (it is easy to study the behavior in the sync ref frame because the applied voltage source is dc and solutions are well known/familiar)
2 - use the rotor ref frame for interruption and post-intteruption transient.  After interruption all that is left is the magnetizing branch and rotor circuit.  When viewed in the rotor reference frame, the speed voltage dependent source in the rotor circuit disappears and we are left with simple R/L circuit.
 

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
And of course what looks like decaying dc in the rotor ref frame R/L circuit will transform to decaying ac at rotor frequency in the stator ref frame where we view the final stator voltages.  

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?


The answer is B.

Depending on the generator's synchronous reactance the full-load excitation current can be as much as twice the no-load excitation current (the no-load excitation current is proportional to the rated generator voltage). This full-load excitation current eventually (after 3 times field time constant have elapsed) can generate about 125 to 130 percent of rated voltage, depending on the no-load saturation curve.

One thing is more important, however. As soon as the generator current is disconnected from the system, the internal generator's leakage reactance voltage drop deminishes at once, resulting in a voltage jump at the terminals about equivalent to the generator transient reactance, regardless of the voltage regulator.

Wolf
www.hydropower-consult.com   


 

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
I had in mind a squirrel cage induction motor. I think you are answering a different question?

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?


My fault. I was on the generator track. Sorry.

Wolf

RE: quiz: motor Vregen higher after interrupt from no-load or...?

The results of that simulation are not what I was expecting. I shall try and do an RL 'simulation' - that is, measure on an actual drive. I will probably not be able to do it on an MV motor. But there should be quite a lot of 400 and 690 V motors with large fans (for heavy start) around.  

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
That would be very interesting Gunnar – would love to hear your results. If you get ready to do a test I would also be interested to hear the nameplate data so I can do a simulation.

Personally the intuitive view that I presented above seems fairly reasonable to me (maybe that's not surprising ... if there is an error buried in my thinking then I guess I'd be the last to see it). But to recap: looking at currents in the rotor and magnetizing branches... they are slowly-varying dc in the reference frames of interest (sync ref frame before interruption and rotor ref frame after interruption) and it is clear that one of them will have to reverse direction upon opening of contacts (*)

I was thinking a little more about comparing magnetizing branch with low current high inductance to rotor branch with lower inductance and high current (during start) and which would "win" (one of them loses and reverses and the other one wins and continues to flow in same direction).   One parameter to look at to get a preliminary intuitive feel on which would win (and whether it is a close battle) would be the time required to bring an initial current I down to a final current of zero with some high constant voltage V applied (such as the high voltage that might occur during contact opening.)
V = L * di/dt.   
dt = L/V * di
t = L/V * I
For a given V, the time parameter is proportional to the initial product L*I.  I consider the same V for both branches because those inductances are roughly in parallel (assuming motor resistances are small compared to the resistance associated with the opening contact).

In my case  X_2=2.3677244 and X_M = 102.4918380 and the ratio LM/L2 = 43.

So I think the magnetizing branch "wins" the battle as long as initial rotor current is not more than 43 times higher than initial magnetizing branch current.   Locked rotor current is 1500+, magnetizing branch current during start is less than 70, so that ratio is less than 20.   It seems that the magnetizing branch wins the battle as the simulation shows.  And even if it didn't win, it would still be a very close battle and the opposing currents strongly weaken each other to result in very small remaining current after the interruption and low residual voltage.

*  Exactly what happens during the moments of interruption is a tricky part to model and probably the weakest part of the whole analysis, but I think what I have done should be pretty close.

So far I've been talking about an equivalent circuit suggesting that (as long as initial magnetizing branch is strong enough to win) the initial rotor current will have to reverse direction between pre-interruption and post-interruption conditions. Can we say it in more physical terms?  The best I can come up with at the moment: During initial conditions before interruption, the magnetizing current and associated flux are driving the rotor current.  After interruption, the rotor current is driving the airgap flux.   That's not really precise terminology... I'll have to think some more... maybe someone can polish it up.

I'd be interested to hear any comments or thoughts and definitely looking forward to any test results.
 

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

I mentioned this to a winder shop with good testing facilities and they invited me to do tests in their test bench. We will have a 30 kW 400 V motor available and can brake it fully. The supplying grid is from a 1 MW generator with wide frequency and voltage range. We will use our ARCUS recorder (ten channels, twelve bits, around 200 us sampling).

Any more tests (other than the EMF measurements) that come to mind?

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
I assume you will have a circuit breaker or contactor to deenergize the motor.

For the purposes of validating my simulation, you could monitor voltage for each phase referenced to a single reference point such as ground.  Or for that matter phase-to-phase would tell the story pretty well as far as voltage magnitude.

For the purposes of identifying the challenge to breaker operation, you could measure transient recover voltage, which would be the voltage  directly accross each breaker contacts (line-side-to-load-side).  It could include some L/C ringing after interruption.  That TRV (peak and rate of rise) is what challenges the breaker contacts.   It wouldn't match my simulatnion because our motors/systems would be different and I didn't model any shunt capacitive effects on either side of breaker or series inductive effects on the supply side. But it would be interesting to see if it's readily available.

Either way, I definitely appreciate your efforts. It is rare for someone to spend time to actually do some measurements related to forum questions, but would be very informative and interesting.

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
Of course if you just measure each phase of motor to ground and each phase or power supply to ground, then all other voltages of interest can be computed from that.

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

This is a very interesting discussion and I am looking forward to test results.

Thanks Gunnar and Pete!

RE: quiz: motor Vregen higher after interrupt from no-load or...?

OK Pete. We may both be somewhat 'unusual'. I tend to observe reality and you tend to simulate/calculate. And, to tell the grim truth, I think we are both fascinated with that. I do not see it as spending time to do measurements, I see it as an opportunity to learn more.

Some day, I may come out with a collection of measurements with comments. What about AC DC ABC BY GKE?

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: quiz: motor Vregen higher after interrupt from no-load or...?

I'd buy it, Gunnar!

Good on ya,

Goober Dave

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
I'd buy it too.
How about ac-dc flick-of-the-switch?
Whoops already taken:
http://www.acdc.com/us/music/flick-switch

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

Back from test with induction motor and interrupted start.

I'm afraid you win Pete. Preparing a report that soon will appear on a computer near you.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
If it was a competition, then I think Bill won because he's the only one that guessed it... I was caught by surprise when I saw it on my simulation and had to double-check it a few times before I believed it.

But we all win by getting the chance to look at some measurements.   I'm definitely looking forward to that.

Thanks again.

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
That is very interesting and good explanations. The interruption period is definitely more complicated than I thought.  I guess the current lags the voltage by just under 90 degrees, so the glitches that occur just prior to the peaks of the voltage waveform (3.215 to 3.225 sec in the interrupted start) correspond to the times of current zero-crossing and perhaps represent failed initial interruption attempt at current zero when the contacts are just beginning to part... which successfuly interrupted a cycle or so later.

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
Out of curiosity,  what was the sample rate?

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

The sample rate is limited by the data transfer rate of the USB connection and is around 4400 samples/second.

Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

RE: quiz: motor Vregen higher after interrupt from no-load or...?

I'm not too surprised at the results since the regen voltage would be or or less proportional to the back emf and as Bill already posted, the back emf will be the highest at full-speed no load.

However, I believe this thread was partly a follow-up due to a motor control section failure electricpete is involved with. That failure involved a 13.8kV motor with an air break contactor which failed during motor acceleration due to the breaker attempting to stop the motor. In that particular system, I would expect the arc current and transient voltages to be higher when breaking a higher current. That system could also have relatively high cable capacitance and possibly other caps nearby (surge or power factor correction) which might cause extra system voltage ringing. If I assume the true question really involves what conditions would cause the highest transient breaking voltages then I'd expect the highest breaking transient voltages to occur when the current is highest. If my assumption is wrong then you can just ignore this post.

 

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)

Quote:

I'm not too surprised at the results since the regen voltage would be or or less proportional to the back emf and as Bill already posted, the back emf will be the highest at full-speed no load.

The post interruption period has a very straightforward physical interpretation: we have a a decaying dc current in the rotor (dc in the rotor reference frame) and the induced stator voltage by Faraday's law is proportional to that rotor current (times the rotor speed times a mutual inductance).

So we know the very simple relationship:
1. Higher post-interruption rotor current -> higher post-interruption stator voltage

But the examples simulated and measured showed that:
2. lower pre-interruption rotor current -> higher post-interruption stator voltage.

The only way to reconcile 1 and 2 above is:
3. lower pre-interruption rotor current -> higher post-interruption rotor current.

This last statement (3) is completely true (when comparing the 3 cases discussed in original post).  But I would submit that it may not be obvious to the casual observer.   But I'll be the first to admit that "obvious" (like beauty) is in the eye of the beholder winky smile

Yes I am certainly interested in understanding the challenge of the various scenarios to breaker interruption as discussed in thread thread238-286947: Is interrupting motor start current more challenging to a bkr ?
 

=====================================
(2B)+(2B)'  ?

RE: quiz: motor Vregen higher after interrupt from no-load or...?

Skogsgurra & e-pete,
  Thank you. This is the kinda stuff that makes 'Eng Tips Forum' the best forum on the web. I've learned more practical knowledge from you guys and the other contributors on this site than I ever learned in any classroom.
 

RE: quiz: motor Vregen higher after interrupt from no-load or...?

(OP)
There was discussion that the post-interruption voltage can be predicted based on the pre-interruption "back-emf", which I presume would be envisioned to be proportional to pre-interruption voltage across the magnetizing branch of the steady state equivalent circuit.

I think it is a reasonably good explanation depending on your purpose, but I'd like to explore a few aspects further:

1 – We know the stator current obviously changes dramatically during interruption and we may suspect rotor current also changes dramatically during interruption.  If emf is created by flux which is created by currents, then  why would we expect the "back emf" to stay roughly the same when currents change dramatically ?  The answer I would suggest (it may already be obvious to some) is that the magnetizing branch paramters (current, inductance, and associated flux linkage) tend to dominate the picture and  therefore rotor current will change in such a manner to keep the magnetizing current and flux roughly constant before/after the transition (in ref frames discussed above)... i.e. rotor current changes in a direction to cancel change in stator current and keep magnetizing current and flux the same.

2 – BUT, looking at pre-interruption back-emf alone does not explain everything.   This is illustrated in two simulations attached: :   
[*]Simulation 2A Unloaded start, interrupt at 92% of full speed, with motor parameters as above.
[*]Simulation 2B - Unloaded start, interrupt at 92% of full speed, with motor parameters modified so that L2 is 25% lower than above.

Looking only at pre-interruption back-emf, we would expect simulation 2B to have lower post-interruption voltage because by voltage divider, we expect lower voltage across magnetizing branch in scenario 2B.

However, that is not what the simulation shows.  The simulation shows that scenario 2B has the higher post-interruption voltage as can be seen by comparing slides 1 and 2.  (Additionally when the two simulations were repeated at other speeds such as 50%, again it shows that L2 caused lower post-interruption voltage.)    So for these scenarios we cannot explain post-interruption voltage soley by considering pre-interuption back-emf.  But it is explained by looking at the changes that occur between pre and post interruption conditions as discussed above.  Even though pre-interruption magnetizing flux was lower in 2B (because magnetizing branch voltage was lower), the reduction in magnetizing flux from pre-to-post-interruption was also lower (because the rotor branch inductance L2 is lower.... so in the battle of Lm*Im vs L2*I2, the magnetizing branch current Im does not lose as much of its initial value in the process of reversing the rotor current I2).
 

=====================================
(2B)+(2B)'  ?

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