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Discharging PFC capacitors
2

Discharging PFC capacitors

Discharging PFC capacitors

(OP)
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 ?

RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

(OP)
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.

RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

Quote:

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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RE: Discharging PFC capacitors

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

Quote:

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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RE: Discharging PFC capacitors

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

Quote:

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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RE: Discharging PFC capacitors

Quote:

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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RE: Discharging PFC capacitors

(OP)
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.

 

RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

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

RE: Discharging PFC capacitors

(OP)
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.

RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

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

RE: Discharging PFC capacitors

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

RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

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

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RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

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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RE: Discharging PFC capacitors

I'm sorry, we don't rotate the Xm curve....we compress it vertically.  But still roughly same conclusion - doesn't seem likely to go into saturation if undercorrected at nameplate voltage (especially remembering that Eres is very far below nameplate voltage), but can't rule it out completely.  

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RE: Discharging PFC capacitors

My discussion of the shape of the curve was correct but more complicated than it needed to be.  If we subtract out Erem, the magnetizing E vs I curve is the same shape as  the B vs H curve at a given frequency.  The knee of the E vs I curve is the same as the knee of the B vs H curve... no need to call it a hump.    

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RE: Discharging PFC capacitors

Note that we can use the figure posted above also to explain how a high voltage condition exists over a range of speed.  If we are intiially highly overcorrected, then the slope of the Xc line is more horizontal and shifts further to the right with an intersection for out on the horizontal part of the Xm curve.  As we coast down, the Xc curve shifts to the left and the Xm curve shifts to the right (and slightly down to keep volts/hz constant on the horizontal part of the curve) and the operating point at the intersection shifts to the left but still on the horizontal highly-saturated part of the Xm curve over a range of speeds.  So let's say as a simplified example that speed drops 5% and we are still on the far right of the curve.... then voltage has only dropped 5% (due again to the fact that the Xm curve compresses vertically during coastsdown to maintain the same volts/per hz for saturation)

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RE: Discharging PFC capacitors


Electricpete

This is a very interesting explanation.  We have several 4.16kV motors here that are PFC corrected and it seems tham many times when we shut these motors off we blow cap fuses or the caps themselves.

At initial review it appears that the Xc of the caps are less than the Xm of the motor but there may be more to take into consideration as you talked about above.  This is something for me to keep an eye on.

 

RE: Discharging PFC capacitors

Attached in the first slide I have provided illustration that for an initial highly-overcorrected condition, the voltage remains in the saturation region even as speed coasts down.  

Second slide is a rough illustration of under-corrected condition where the voltage (at intersection of the curves) will be below nameplate.  

I believe it is always the case that if we are undercorrected at operating voltage./frequency (say 90% or less), then the voltage at intersection of these curves when disconnected but still near rated frequency will be below nameplate voltage .   That makes sense since that is the thumbrule used to prevent overvoltage.

It is tough to prove it graphically.  We need log curve to show salient points of the B vs H, but slopes have different meaning on a log curve.   

Here is my rough proof that if we capacitive vars are 90% or less of inductive (undercorrected) and reasonable values of Eres (<10% of nameplate), the voltage after disconnecting from the power system (intersection of the curves) will be less than nameplate voltage:

Quote (ElectricpeteProof):


Define:  Imnp = magnetizing current which occurs at nameplate voltage and frequency.  

ASSUME: Capacitive vars <= 90% of inductive at nameplate voltage.
=> Slope of of capacitive curve E vs I is >= 110% higher than slope of inductive E vs Im curve at nameplate voltage (need linear scale to judge slope).  
dEc/dI = Xc > 1.1* dEm/dI(Imnp))

ASSUME  - the nameplate voltage Enp corresponds to the knee of the E vs Im curve or somewhere to the left.   Therefore - for any lower voltage, the slope of the E vs Im curve is lower than the slope at nameplate (that's what defines the knee... the point of highest slope).
dEm/dI((Imnp) > dEm/dI(All I < Imnp)

Combining the above
dEc/dI = Xc > 1.1* dEm/dI((all I < Imnp))

i.e. the slope of the capacitive curve is > 110% of slope of inductive curve for all points up to nameplate magnetizing current  = up to nameplate voltage.

Further, if Xc >  1.1* dEm/dI((for all I < Imnp)),   then it is also greater than the average slope of Em on that interval.  i.e.
Xc >  1.1* dEm/dI((average))
dEm/dI((average)) < Xc/1.1  [Equation 1]
where  dEm/dI((average)) is average on the interval I = 0...Imnp

Now find another expression for slope of curve dEm/di/(Average)

Em(Imnp) = Eres + Imnp * dEm/dI(average) = Enp
dEm/di/(Average) = (Enp - Eres)/Imnp
ASSUME - Residual voltage is less than 10% of nameplate voltage  Eres < 0.1 * Enp
dEm/di/(Average) > (Enp - 0.1*Enp)/Imnp
dEm/di/(Average) > 0.9*Enp/Imnp   [EQUATION 2]

Compare equation 1 to equation 2
0.9*Enp/Imnp  < dEm/di/(Average)   < Xc/1.1
0.9 * 1.1  Enp < Xc * Imnp
Enp < Xc * Imnp [EQUATION 3]

If we we draw a vertical line at Inp as shown in slide 2, it intersects the magnetizing curve at Enp (by definition), and it intersects the Xc curve above the magnetizing curve by equation 3.

Based on the Xc curve above the magnetizing curve at Inp, it is clear that the intersection of these curves will occur to the left (at some voltage less than nameplate).  

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RE: Discharging PFC capacitors

Whoops.  The above "proof" was based on the assumption that inductance at operating voltage is the instantaneous slope of the curve.  That is wrong.  Back to the drawing board for a proof.  Anyway, the slides I think give a good picture of the overcorrected and undercorrected conditions.

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RE: Discharging PFC capacitors

Here is corrected proof which ends up being simpler.  We only need to look at the average slope, not the instantaneous slope.

Quote (CorrectedProof):


Define:  Imnp = magnetizing current which occurs at nameplate voltage and frequency.  

ASSUME: Capacitive vars <= 90% of inductive at nameplate voltage.
Slope of of capacitive curve E vs I is >= ~110% higher than the effective inductance at 60hz.  The effective inductance at 60hz is the average slope of inductive Em vs Im curve over the interval from I = 0 to Imnp

dEc/dI = Xc > 1.1* dEm/dI(average))
where dEm/dI(average) is the average slope of magnetizing curve over the interval I=0 and Imnp

dEm/dI((average)) < Xc/1.1  [Equation 1]

Now find another expression for slope of curve dEm/di/(Average)
Use the equation of a line startin at (I,E) = (0,Eres) up to (Imnp,Enp):
Enp  = Eres + Imnp * dEm/dI(average)
dEm/di/(Average) = (Enp - Eres)/Imnp
ASSUME - Residual voltage is less than 10% of nameplate voltage  Eres < 0.1 * Enp
dEm/di/(Average) > (Enp - 0.1*Enp)/Imnp
dEm/di/(Average) > 0.9*Enp/Imnp   [EQUATION 2]

Compare equation 1 to equation 2
0.9*Enp/Imnp  < dEm/di/(Average)   < Xc/1.1
0.9 * 1.1  Enp < Xc * Imnp
Enp < Xc * Imnp [EQUATION 3]

If we we draw a vertical line at Inp as shown in slide 2, it intersects the magnetizing curve at Enp (by definition), and it intersects the Xc curve above the magnetizing curve by equation 3.

Based on the Xc curve above the magnetizing curve at Inp, it is clear that the intersection of these curves will occur to the left (at some voltage less than nameplate).

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RE: Discharging PFC capacitors

A few followups to support the above proof:

Two references suggest that the max residual flux density in steel parts is on the order of 50 Gauss:
http://www.trifield.com/gauss_meter.htm
http://books.google.com/books?id=LtlFcoeIP-UC&pg=PA277&amp;lpg=PA277&dq=motor+%22residual+magnetism%22+gauss&;source=bl&ots=a1_iOzFzhQ&amp;sig=Ej0k8NZUBSoYo42FYis-gI_3QCg&hl=en&ei=vvbQSbX-AZ_tlQeOlNDeCQ&sa=X&oi=book_result&resnum=2&;ct=result

50 gauss is 0.005 T.   The airgap flux cannot be more than the iron flux.  Airgap flux when energized to nameplate voltage is typically 0.5 T or more.  So the assumption Eres < 0.1 Enp is very conservartive... should more likely be Eres < 0.01 Enp which provides a stronger proof of the above conclusion.

Also for completeness, I should have defined residual voltage.  It would be the stator terminal voltage generated by rotor residual magnetism when the rotor is rotating at the frequency of interest (usually at or just below line frequency) and nothing externally connected to stator winding (no caps, no supply, etc) and no significant mechanical load.
 

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