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]

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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[electricpete]

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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[electricpete]

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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[rockman7892]

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.

 

[electricpete]

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:

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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[electricpete]

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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[electricpete]

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

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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[electricpete]

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=Lt...lNDeCQ&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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