Discharging PFC capacitors 2

[edison123]

In my motor repair shop, I use PFC capacitors across motors when they are tested at no-load at 415 V. These capacitors have discharge resistors right at their terminals. Yet, if I switch off the circuit (the main incoming breaker) at 415 V, the capacitors recharges the outgoing bus (where my voltmeter and the caps are connected) to as high as 700 V.

Should I use additional discharge at the outgoing bus to ground (with probably a resistor thrown in series) to reduce this back feed voltage from the caps ?

 

[electricpete]

Does the bus charge to 700volt ac or dc and does it go away as motor coasts down?

If it is ac and goes away - one possibility is that you have over-corrected the motor.
i,e At power frequency Xc > Xm
1/(j*w*C) > j*w*Lm
where Xc and C are associated with cap and Xm and Lm are associated with motor magnetizing current.

Then when you disconnect as w decreases Xc increases and Xm increases and you'll reach a point where they are equal (resonance) and motor residual magnetism can cause high voltage.

Or on the other hand, if it is DC, then perhaps you are right about needing discharge - would be in parallel to discharge any dc off the cap.


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[edison123]

Thanks pete. It is ac and dies down very slowly. The cap correction is for about 90% of no-load current only. I don't think resonance is the issue here since the voltage decay is quite long for over 3 minutes.

 

[electricpete]

If you have ac voltage present after opening the input breaker.... it must be coming from the motor right? A capacitor cannot generate ac. If residual capacitor charge were the problem it woudl be dc.

I think reducing the Xc to keep below Xm will eliminate the problem.

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[electricpete]

I think reducing the Xc to keep below Xm will eliminate the problem.

Whoops. I meant you have to keep Xc above Xm (which means capacitive vars from cap below inductive vars demanded by motor)

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[electricpete]

I see I had it backwards up above also. Should have been:
...

one possibility is that you have over-corrected the motor.
i,e At power frequency Xc < Xm
1/(j*w*C) < j*w*Lm
where Xc and C are associated with cap and Xm and Lm are associated with motor magnetizing current.

Then when you disconnect as w decreases Xc increases and Xm increases and you'll reach a point where they are equal (resonance) and motor residual magnetism can cause high voltage.

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[electricpete]

Also of course as w decreases Xc increases and Xm decreases. So there is yet another correction required:

one possibility is that you have over-corrected the motor.
i,e in an overcorrected condition at power frequency Xc < Xm
1/(j*w*C) < j*w*Lm
where Xc and C are associated with cap and Xm and Lm are associated with motor magnetizing current.

Then when you disconnect as w decreases Xc increases and Xm decreases and you'll reach a point where they are equal (resonance) and motor residual magnetism can cause high voltage.

I have read this one twice, I think I have the explanation of overcorrected right now...


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[electricpete]

Thanks pete. It is ac and dies down very slowly. The cap correction is for about 90% of no-load current only. I don't think resonance is the issue here since the voltage decay is quite long for over 3 minutes.

If it truly is not over-corrected, then what is the source of the problem and why would discharge resistor help? I don't understand what else would cause high AC voltage other than resonance excited by motor residual voltage.

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[edison123]

May be I should explain my test system better.

415 V supply to the test board =>Mechanical disconnect switch no. 1 =>3 ph contactor (A) controlled by the tester=>3 ph variac input=>3 phase variac output (where the voltmeter is connected)=>Cables direct to motor + Capacitor bus with a mechanical disconnect switch no. 2.

When I trip the 3 ph contactor (A) in the test board, the voltage boosts to nearly 700 V. If I open the disconnect switch no. 2 of the capacitor bus, the voltage immediately drops to zero (and the motor is spinning at this time at probably 80 to 90%).

Now I am thinking of introducing a 3 ph contactor in between the 3 ph variac output and the capacitor bus, so that the bus is physically cut out.

 

[electricpete]

There is no cause I can think of other than resonanance of the capacitance and motor inductance. I didn't fully understand your system, but if you can isolate the caps from the motor preferably before cutting the power or at same time or less preferably immediately after, I think it would solve the problem. (I think this is what you were talking about with the added disconnect).

I agree with you that I wouldn't have thunk it would last for three minutes. But resonance involving an iron-core component (ferroresonance) is more complicated than simple resonance because when the voltage increases, the core goes into saturation and the effective inductance changes. Perhaps that extends the speed (frequency) range over which resonance occurs (?).

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[electricpete]

By the way, I am not positive whether it is safe to deenergize capacitors using a disconnect. I do remember there are special requirements for switching capacitors just like inductors. It's not as easy to understand as the inductive kick that we are familiar with for inductors.

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[waross]

I think that both of you have been looking at the induction and magnetizing current of the motor only. As I understand the connection, the auto transformers are also in the circuit.
The Variacs may be complicating the issue.

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

 

[edison123]

pete

I will try to post a single line diagram of the set up here.

The mechanical disconnect switch 2 of caps bank (which is a fused switch) is switched on before the test starts and is not switched off while the test is on. I think the introduction of an additional contactor before this switch and tripping at along with the main incoming contactor should solve the problem of bus voltage rise.

But then I wonder what happens to the caps voltage itself at the time of switching off. I need to put a voltmeter across the caps and see what happens.

 

[electricpete]

waross - you are right the variable autotransformer may be participating in the circuit. But wouldn't you agree that it should add more inductance in parallel which would decreases total Xm (increase magnetizing vars) and move further away being overcorrected. I think the solution remains isolating the caps. Although there is still the nagging question of why we had resonance to begin with if we met the required condition to avoid overcorrection: Xc > total Xm (the capacitive vars was lower than the inductive vars).

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[Skogsgurra]

Three minutes. Around 600 seconds. And f = 60 Hz? That would mean that the Q of the resonance circuit would be somewhere between 100 and 400. Not possible in a circuit containing iron and discharge resistors.

What happens is that motor's back EMF generates a voltage that decays as motor coasts to stillstand. Capacitors aid keeping excitation up. Happens all the time if you do not disconnect capacitors.

Gunnar Englund

http://www.gke.org

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[Skogsgurra]

Made a wrong mental calculation there. Still, if the voltage goes down as the motors coasts down, it could be a self-excitation thing.

But, there are other possibilities.

An AC voltmeter also reacts to DC. It often shows a higher value when connected to DC (has to do with calibration for sine). Opening the three-phase contactors will make the voltage increase to what happens to be across the capacitor when your contactor opens. That could easily be a high voltage (sine peak voltage).

Still do not know if your capacitors are disconnected or not. That one-line diagram would help.


Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

I agree that a self-excited (by capacitors) induction generator at no-load as suggested by Gunnar is a better framework for viewing the system.

There is discussion of that in the Induction Motor Handbook and the Variable Speed Generator Handbbok (both by Ian Boldea) – the solution of voltage for a variable speed generator at constant speed (similar to coasting down slowly) and no-load is given by considering a simple circuit of residual voltage source, capacitance and inductance. The tricky part is that the inductance changes with voltage. The net effect imo is that the resonance range is in fact extended beyond what we would see for simple linear system.

Attached is an excerpt from the above text which outlines the analysis to determine operating point (voltage) for induction generator at no-load (I added the red annotations)

As shown on the left side of the figure, there are two branches both connected in parallel to E1 = terminal voltage. One branch is the capacitor, the other is series combination of Erem (residual magnetism) and the magnetizing inductance (which changes)

On the right side the two curves are plotted on a plot of E1 vs I and the operating point will be the intersection.

The capacitor branch is a straight line with slope |Zc| = |1/wC|.

The inductor branch has an offset voltage of Erem, and the slope is the value of Xm at that particular voltage.

In this particular set of parameters plotted, the slopes are the same when the current is near zero... would indicate a resonant condition for the system linearized about that low-excitation point. The voltage increases until the Lm goes into saturation at the operating point.

The slopes are also the same in the middle of the curve, would indicate a resonance at that linearized point as well.

We can certainly see that if we took this system and increased the margin to overexcitation by increasing Xc, we would rotate that Xc line counterclockwise and reduce the voltage. That qualitative feature of the linear/resonant model remains intact.

But it is fundamentally a non-linear system and talking about resonance doesn't really tell the whole picture. There are in fact a wide range of parameters that will lead the system to saturation.

It is not too far from what I said originally – the non-linearity of the system extends the range of resonance. But I do think I was wrong to say that we had to be overcorrected (based on calculations using no-load current at rated voltage/frequency) in order to generate these high voltages near saturation. It is not too hard to imagine that nameplate votlage/frequency condition might correspond to a point toward the right of the curve where the slope of the magnetizing branch (Xm) is less than the slope of the capacitive branch (Xc), i.e. inductive vars more than capacitive vars (undercorrected), and yet the system would still go to the saturated operating point at the far right of the curve.


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[electricpete]

And I think everyone that chimed in still agrees the caps need to be isolated.

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[electricpete]

By the way I don't think it's dc based on my understanding of the stated symptoms that the motor and capacitors are connected in this period... any dc would disappear quickly.

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[electricpete]

Actually to be more precise, the effective Xm would be closer to the slope from the point (I,E)=(0,Erem) up to the point on the curve (not the instantaneous slope of the curve)... rather than instantaneous slope of the curve as I was imagining.

Also the operating conditions at nameplate would likely be on the hump of the curve or to the left of that hump. (I call the hump the point where the Xm curve deviates the most from the Xc curve... also corresopnds roughly to highest effective permeability and the "knee" of the B vs H curve.

Considering thsoe two points, it now seems less likely (but still possible) that we would end up at the final fully saturated operating point if we were not initially overexcited.

For one thing, those points over to the right of the hump likely corresdpond to a point above the nameplate voltage.

For another, everything at the hump or to the left has higher effective Xm than the initial Xm at 0. i.e. if we pick any point at or to the left of the hump for nameplate voltage on this particular curve it has a higher slope than then the capacitor curve which corresponds to an overcorrected condition.

There are certainly a lot of variables and it's not a simple problem, so I don't rule out that there can be undercorrected motors that will go to a fully saturated operating point such as the one shown. But it no longer seems like a probable occurence to me. If you are undercorrected than the slope of the curve at nameplate conditions (slightly to left of the hump) is less than the slope of the Xc curve. Rotate the Xm curve to make that happen and you have an operating point below rated voltage.

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