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 ?
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
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
RE: Discharging PFC capacitors
I think reducing the Xc to keep below Xm will eliminate the problem.
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RE: Discharging PFC capacitors
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RE: Discharging PFC capacitors
...
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RE: Discharging PFC capacitors
I have read this one twice, I think I have the explanation of overcorrected right now...
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RE: Discharging PFC capacitors
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RE: Discharging PFC capacitors
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
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
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RE: Discharging PFC capacitors
The Variacs may be complicating the issue.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Discharging PFC capacitors
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
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RE: Discharging PFC capacitors
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
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Discharging PFC capacitors
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
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Discharging PFC capacitors
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
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RE: Discharging PFC capacitors
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
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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
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:
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RE: Discharging PFC capacitors
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RE: Discharging PFC capacitors
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://boo
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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