breaking inductive currents

[saltukhan]

as far as i know, it is more difficult to break lagging currents than break resistive currents. is it true? if it is true why? does it matter the lag of the current,?

 

[Skogsgurra]

Yes, true. Many, many will tell you that. And in many different words. The basic reason is that you have to take the stored energy in the magnetic field into account. Another, and more common way of saying that, is that the voltage developped across the breaker is the sum of instantaneous voltage plus the L*di/dt voltage. Others will tell you that it is analogous to stopping a heavy car. The heavier, the more force you need. And the quicker you want it stopped, the more force you need. And then, some will tell you that current 'wants to contnue flowing in inductors'. That is also true, but I do not think that current has a will of its own.

Anyhow, it is true. It is also true that using capital letters in the beginning of a sentence is a good habit.

Gunnar Englund

http://www.gke.org

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

 

[davidbeach]

It is also true that using capital letters in the beginning of a sentence is a good habit.

So very true.

 

[saltukhan]

Yes you are right dear friends, thank you ...

 

[electricpete]

I agree with Gunnar's response. He also mentioned there are many considerations.

Another consideration is phase relationship between voltage and current as it relates to interrupting at a current zero. There are more details to that subject than I know. Without looking it up in a book to refresh my memory (an ahead-of-time excuse for not being exact), it is easier to interrupt at the natural current zero when the voltage is also zero at that time (resistive current) since in that case no voltage will appear accross the arc/contacts when they reach open circuit condition. That is why it can be more difficult to interrupt a capacitive ac current than a resistive ac current... even though there is no L*di/dt voltage in the capacitive current. The lack of natural current zero's is also the reason that it is tougher to interrupt a dc inductive current than an ac inductive current. I'm sure other forum members can add more if they care to.


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[TurbineGen]

Saltukhan - It is more difficult to interrupt any reactive circuit than a primarily resistive one. Current is interrupted at a Zero crossing of the current. If the load is stricly reactive whether lag or leading, the voltage as you know is offset from this current at 90 degrees (theoretically). Thus the voltage is actually at peak. A resistive circuit's voltage would be at zero when current is zero. Thus interrupting reactive circuits is much more difficult.

With capacitive circuits (leading), this can in fact be even worse. The capacitors will hold some charge in them from the time of trip and slowly deteriorate over time. However this charge can cause restrike in your breaker since the voltage differential across the interrupter is higher than nominal.

One more thought, NEVER try to interrupt a DC circuit with an AC breaker. There is no zero crossing and thus it is likely not to clear the circuit.

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If it is broken, fix it. If it isn't broken, I'll soon fix that.

 

[Compositepro]

I think that the leading or lagging issue is not at all helpful in understanding the difficulties in switching inductive loads. As Gunnar said, inductors store energy in their magnetic fields. There is a physical relationship between the magnetic field and the current through a coil (inductor). If a switch tries to interrupt an inductive current the energy in the magnetic field will keep the current going until there is no more energy in the magnetic field. If precautions aren't taken the energy is dissipated in the switch. If the energy in the inductor is sufficient, the switch will be damaged or destroyed.

Understanding how ignition coils and magnetos work would be very helpful to understanding the concept.

 

[marks1080]

We have cap breakers on our 230kV system that I have been told are different than 'normal' breakers. I am thinking specifically of SF6 type. Visually they look just like any old SF6 breaker. Is this (the topic of this thread) the difference that people are referring to? Is there is physical difference in the structure of the interrupting head or is it a difference in the control system (at what voltage level the breaker opens) or maybe a combination of both?

Any other comments on HV cap breakers would be appreciated as well.

Thanks,
Mark

 

[Marmite]

Capacitor switching is onerous for circuit breakers because of the magnitude of the inrush current, and also its frequency. The rate of rise of current di/dt can exceed the rate of rise of contact pressure, which leads to burning of the contacts. Switchgear for switching capacitors has to be specifically type tested for that application and assigned a rating. Back to back capacitor bank switching is even more onerous and circuit breakers have to be additionally type tested for that specific duty. A circuit breaker which has been type tested for single bank switching may not necessarily have an assigned back to back rating.
Regards
Marmite

 

[Skogsgurra]

To avoid confusion, I think it would be better starting a new thread on the specific problems associated with capacitive switching. This thread is, as its title suggests, about breaking inductive current.

Also, when you start that other thread, please do consider the difference between returning voltage in HV switching and the inrush current when closing low inductance and low resistance capacitive circuits. The two problems have very little with each other to do. So, two threads are recommended.

Gunnar Englund

http://www.gke.org

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

 

[m3ntosan]

marks1080,

If CB has single pole operation, maybe there's just a "Controlled point-on-wave switching device" that makes the difference.

For inductive load switching some CB use resistors to damp out transient overvoltages caused when switching transformer magnetising or shunt reactor currents.

For capacitive current switching the problem is to build up quickly sufficient dielectric strength between the contacts to withstand twice normal voltage one half cycle after interruption. Switching of capacitors or open-ended lines is especially susceptible to switching surges. When switching a capacitive element, the CB could end up with opposite polarity voltages on either side of the switch. This could lead to twice nominal voltage across the circuit breaker. The voltage that appears across the open switching device is called the recovery voltage.
The higher the recovery voltage, the greater the likelihood of a restrike occurring.
When the circuit breaker interrupts the capacitive current the voltage is at a maximum value. One half cycle after the CB opens, the capacitor is still charged to a minimum value but the system voltage has reached a maximum. The circuit breaker now has twice system voltage across it. The probability of a restrike has increased.

When the current is zero the voltage is at a maximum in a capacitive circuit.

 

[electricpete]

Deja vu !

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[electricpete]

I agree inrush is a whole different subject that does not belong in this thread. The reason for bringing up up
capacitor deenergization was that it demonstrates di/dt voltage is not the whole story by itself.... phase angle
between voltage and current plays a role separate from di/dt voltage. That is why deenergization of an ac
capacitive circuit poses greater challenges than a ac resistive circuit of same current.

http://ecmweb.com/powerquality/electric_trouble_capacitors_part/

De-energizing a capacitive load poses even more challenges. Because the current waveform leads
the voltage waveform by 90°, the current is interrupted very close to its zero crossing when the
voltage is at its maximum absolute value....
A capacitive switching device must be designed to endure the thermal stresses caused by the
re-ignitions and restrikes. Some circuit breakers fail to meet this level of performance. This is
why switching devices used for capacitor switching must be designed specifically for that application.
In many cases, such devices have a higher transient recovery voltage rating than general-use circuit
breakers

"Transmission and Distribution Electrical Engineering", 3rd Ed, by Bayliss and Hardy

http://books.google.com/books?id=6m...o interrupt than resistive circuits."&f=false

The characteristic waveform of the recovery voltage is shown in Fig. 13.2. A
high frequency voltage oscillation, known as the 'transient recovery voltage'
(TRV), fluctuates about the power frequency recovery voltage waveform. Its
behaviour is determined by the circuit parameters and the associated rapid redistribution
of energy between the network component electric and magnetic fields.
If the power factor of the faulted circuit is high (i.e. resistance is a significant proportion
of the total fault impedance) then the circuit or power source voltage at
current zero will be low. At low power factors (predominantly inductive or capacitive
circuits) the circuit voltage at current zero will be high and result in a
tendency for the arc to re-strike. This is the basic reason why inductive and
capacitive circuits are more difficult to interrupt than resistive circuits. The circuit
breaker must, therefore, be designed to withstand the transient recovery voltage.
Whether or not the arc extinguishes after the first current zero depends upon
establishing adequate dielectric strength across the circuit breaker contacts faster
than the rate of rise of TRV and the peak TRV involved.

Scroll up a little bit to figure 13.2. You can see that at the moment of current zero, the recovery voltage jumps
to the normal system voltage (which is maximum at natural current zero for inductive and capacitive
circuits). There is also an L-C ringing which causes transient overshoot as Marmite has alluded.
All you have to do is read the bolded sentence above to see that this author singles out as the "basic reason"
for the challenge of interrupting ac inductive (and capacitive) currents is the phase relationship between
voltage and current and it's effect on the voltage at the time of current zero. There is very little about L*di/dt
voltage in this discussion.

V = L*di/dt is the most important factor for dc circuits.

For ac circuits the angle between voltage and current at the time of natural current zero also plays a very big role as described above.

I think that the leading or lagging issue is not at all helpful in understanding the difficulties in switching inductive loads. As Gunnar said, inductors store energy in their magnetic fields. There is a physical relationship between the magnetic field and the current through a coil (inductor). If a switch tries to interrupt an inductive current the energy in the magnetic field will keep the current going until there is no more energy in the magnetic field. If precautions aren't taken the energy is dissipated in the switch. If the energy in the inductor is sufficient, the switch will be damaged or destroyed.

Understanding how ignition coils and magnetos work would be very helpful to understanding the concept.

Lead and lag are very relevant to the question as shown above. DC circuits such as ignition coils are not representative of the principles of ac circuit interruption because DC circuits do not have a natural current zero.

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