Static exciter - varistors 4

[Skogsgurra]

I had this question a couple of weeks ago:

"In a 2000 kW hydro-electric generator with a static exciter, there are MOV varistors across the rectifier bridge that feeds the pole wheel. Are they there to protect against inductive overvoltage in case someone opens the excitation winding?"

- No. I said. -The rectifier bridge takes care of that. It acts as a free-wheeling diode. I think that they may be there to take care of out-of-sync situations where lots of voltage is induced in the excitation winding.

Now, this guy gets back: "No, my customer didn't buy that. They say it is for inductive kick-back".

I know there are lots of ignorants out there. But these guys work for one of Scandinavia's largest power companies and do nothing but power generation things. So I ask myself if they know something that I don't know. And that my fellows in Eng-Tips know.

What could that be?

Gunnar Englund

http://www.gke.org

 

[LionelHutz]

I take it you're meaning what happens if you turn off the supply to the rectifier? If so, as you said, the rectifier is diodes and it has a freewheeling path already.

When I use small disc MOVs I only expect them to protect against low energy transient overvoltages. I don't expect them to handle any large currents.

 

[waross]

skogsgurra
My understanding is that if the field current is interrupted the collapsing flux will induce a voltage in the opposite direction. The diode bridge will not pass this voltage. With no discharge path the voltage will rise until it exceeds the breakdown voltage of something and makes its own discharge path. Usually the diode bridge will be the first to go. I have not tried this with a generator, but I have lost the diodes in several battery chargers and/or DC supplies when I was energizing an inductive load and the plug was kicked out. Also, when field flashing, if you use a free wheeling diode to prevent an inductive kick, it is installed so as to block the applied voltage but conduct the induced voltage.
I wondered why freewheeling diodes were not used instead of movs in rotating fields and came to the theory that the movs are used instead of diodes for faster response and to prevent over-voltage in the main output when the load is reduced suddenly and the field must be rapidly weakened.
I hope davidbeach posts in. He will possibly have the definitive answer rather than theory.
respectfully

 

[Skogsgurra]

No Waross,

Draw the diagram. Current continues to flow in same direction. Not opposite. So the bridge acts as a free-wheeling diode. As you know, a free-wheeling diode is connected with its cathode to the plus.

Gunnar Englund

http://www.gke.org

 

[ScottyUK]

Hi Skogs,

In our static exciter (brush fed, no shaft rectifier, 1200V , 4700A) there is a large bank of Volt-Trap and Metrosil non-linear resistors. I'm pretty sure the Metrosil bank is across the DC field contactor and is there to limit the field voltage when the main field contactor opens. There is an early make / late break pair of poles in antiphase with the main contactor which connects a field dump resistor in circuit before the main contactor opens. When the field current is high the peak voltage across the resistor at the must be quite high too. I think the Volt-Traps are on the bridge side of the contactor and are intended to protect the bridge. It has been a while since I looked at the power circuit, and I'm away from Teesside in Poland next week so I can't get a definite answer for a while.

The field dump resistor is there to dissipate the energy stored in the field and ensure that the terminal voltage of the generator is quickly reduced to a safe level. A generator with a heavy reactive power export which suddenly decouples from the grid will find itself grossly overexcited and will experience an overvoltage unless the field is reduced rapidly. A freewheel diode would allow the current to circulate for a long time before decaying away, so the overvoltage could potentially damage the generator or GSU transformer through either overfluxing or plain overvoltage. The Metrosil is there to limit the peak voltage seen by the field dump resistor.

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I don't suffer from insanity. I enjoy it...

 

[davidbeach]

I hope davidbeach posts in. He will possibly have the definitive answer rather than theory.

But, but, but.... I'm a relay guy, not a generator guy.

On the other hand, I happen to work with a bunch of generator people and happen to have the book from one of our classes and it does have some useful information.

Excitation systems will include a means for field current to continue to circulate regardless of the operation of the rectifier. If the rectifier is only diodes, the rectifier can provide that path. If the rectifier is semi-controlled (3 SCRs and 3 diodes) there would be a free-wheeling diode in parallel to allow field current to continue during SCR commutation and when the SCRs are turned off. If the rectifier is fully controlled, the bypass device is more likely to be an SCR or an MOV to allow negative forcing. The bypass SCR would be gated on when the rectifier is off to allow field current to continue to flow. There would also be a discharge resistor in this path.

Taking rectifier protection one step further, there should be a means of allowing reverse currents to discharge without overvoltaging the rectifier. These reverse currents will occur during a pole slipping condition; the motion of the rotor field relative to the rotating field of the stator induces an AC voltage on the rotor. If that AC voltage is not shunted off through some means, it can create an overvoltage on the rectifier. This could be through an SCR that is turned on in response to the pole slip, or it could be an MOV (in theory). The problem with trying to use MOVs is the time constant of the rotor, much much longer than the typical transient event the MOVs are designed to deal with.

 

[Skogsgurra]

I have to make myself more clear. There are NO contacts that open in this circuit.

Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

...and NO slip rings...

Gunnar Englund

http://www.gke.org

 

[dpc]

I'm confused about "no slip rings". Is this a rotating brushless exciter? A static exciter would need slip rings to get field curren to the rotor, wouldn't it?

Hydro generators are subject to sustained overspeed conditions when the unit trips. Most static exciters have a crowbar circuit to help force the field current to zero to avoid overvoltage. In the old days, dc field breakers and discharge resistors were used.

Perhaps this circuit is to protect against high voltage caused by overspeed?

 

[davidbeach]

OK, rotating exciter, but the original post does say static exciter. So you have diodes, no SCRs or other controlled devices, so current in the forward (normal) direction can continue through the diodes at all times, nothing to interrupt the circuit in that direction.

But there is still the consideration of what happens when the generator slips a pole. The AC induced onto the rotor in this condition will generate a reverse current that will need to go somewhere, and an MOV seems about the only way to get that reverse current past the diode bridge other than letting the diodes break-down.

If it is truly a rotating exciter, why would they be concerned about someone opening the field winding? How would someone open the field winding in any case?

 

[Skogsgurra]

There are static exciters with slip rings and there static exciters without slip rings. The word "static" was once introduced to make a difference from shaft mounted DC generators that were used to excite the pole wheel (also called feeder machines).

I am sorry that my simple question caused so much misunderstanding. The reason I put the question here is that I think that I know all there is to know about inductive circuits, rectifiers, overvoltage protection like MOVs, absorbed energy, free wheeling diodes - with or without series resistors, even using selenium rectifiers back/back for overvoltage protection, delays caused by overvoltage protection and all that. After all, I have done this for large companies for decades. So, when I got this answer (see OP) from the energy company people, I thought that maybe there is something that I haven't learned about. I couldn't figure out what it could be so I turned to Eng-Tips for advice.

I haven't got any valid answer yet. Most of the answers were misunderstandings because I didn't explain in detail what I thought was very obvious.

I haven't been able to talk to the end customer and the guy that asked me the question has no idea why those guys couldn't accept the explanation. I am more and more concvinced that they were just repeating something they heard long time ago - without thinking about the actual situation.

I can appreciate Scotty's thinking about rapid reduction of field when the generator loses load. But since the diode bridge is connected directly to the field winding, there is no chance of getting faster reduction than the time constant of the winding (reduced by a couple of volts drop in the bridge). The MOVs never get engaged.

davidbeach describes the configuration in his words: "Excitation systems will include a means for field current to continue to circulate regardless of the operation of the rectifier. If the rectifier is only diodes, the rectifier can provide that path".

He also says "Taking rectifier protection one step further, there should be a means of allowing reverse currents to discharge without overvoltaging the rectifier. These reverse currents will occur during a pole slipping condition; the motion of the rotor field relative to the rotating field of the stator induces an AC voltage on the rotor. If that AC voltage is not shunted off through some means, it can create an overvoltage on the rectifier. This could be through an SCR that is turned on in response to the pole slip, or it could be an MOV (in theory). The problem with trying to use MOVs is the time constant of the rotor, much much longer than the typical transient event the MOVs are designed to deal with" And that's exactly what I mean with "they may be there to take care of out-of-sync situations where lots of voltage is induced in the excitation winding". We agree there. We also agree that MOVs are not up to the energy handling capacity needed for such a situation. There are six of them in parallel. Each being a heavy-duty block varistor with 140 J capacity.

For a slip situation, this is not at all sufficient (as davidbeach points out) and that's why I ask why they have been mounted at all. What is their purpose?

I felt that I needed to explain this in detail so that the valued forum members don't answer the wrong questions.



Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

davidbeach,

I was writing my epic drama when you posted your answer and your questions.

Yes, that's what I am saying. There's no way of opening the field winding in the generator. But the excitation for the rotating exciter (Sometimes called "rotating transformer" - which is also wrong. No wonder there are so many misunderstandings here) can of course go to zero or someone can open that circuit. But that doesn't cause any overvoltage in the rotor winding - only across the stationary excitation winding.

Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

And you get a PLS from me too

Gunnar Englund

http://www.gke.org

 

[davidbeach]

skogsgurra, the product line for the excitation side of the house make a distinction between static excitation, which is a DC source off the shaft and a slip ring connection to the field, and rotating exciters where the excitation system controls the stationary field of the rotary exciter, the output of which is rectified in the diode pack on the end of the shaft and fed into the main field. They are both static in reference to the use of solid state electronics, but not from the aspect of one DC source sitting unmoving (static) on the floor while the other DC source is spinning rapidly. When I read "static exciter" only one of the two comes to mind, and "rotating exciter" only brings the other to mind.

I will email this thread to someone who knows far more about what's going on inside the generator than I ever will and see what response we get. That response, though, won't happen until at least Monday, and I will spend most of Monday on the road, so don't hold your breath waiting for an answer.

 

[Skogsgurra]

Thanks davidbeach,

I have obviously used too loose a terminology. But the actual situation is as you describe it. I look forward to the answer. I will breathe normally.

Gunnar Englund

http://www.gke.org

 

[waross]

Hi davidbeach,
I hope that i didn't put you on the spot, but I'm more than satisfied with your response. I'm also looking forward to your answer. Thank you.
Hi skogsgurra
I really misunderstood your original post. Sorry.
Now I'm wondering a few things. As I understand the action of the discharge or freewheeling diode, the current and voltage both decay to zero in about 5 time constants. I was taught that if the freewheeling diode was rated to carry normal operating voltage and current that it would function with no danger of burnout.
Why then have I consistently lost diodes when the power was interupted to a power supply or battery charger used to test an inductive load? One instance could have been overvoltage on the incoming supply rather than a power loss problem. Is it possible that I had partially damaged bridges in the other instances? If a section of a bridge was open and it was running as a half wave rectifier the discharge path would include the AC source winding.
I have a question on the action of the MOVs.
When an MOV conducts to protect an inductive coil from over voltage, what percentage of the energy is disapated as I^2R losses in the protected coil and what percentage is actually disipated in the MOV. I understand that with a freewheeling diode virtually all the energy is disipated in the coil, and I'm wondering about the MOV.
respectfully

 

[Skogsgurra]

Waross,

The 1/2 in the energy formula 0.5xLxI^2 says it all. It is the same thing with capacitors where the stored energy is 0.5xCxU^2.

I use the capacitor example because it is easier to follow. As you know, when a capacitor is charged, it has 0.5xCxU^2 joules stored. Why not 1xCxU^2? That is what you would expect if you were guessing without knowing the correct answer.

The answer is that half of the energy is consumed in the series resistance when you charge the capacitor. "What if there is no series resistor?" You ask.

There is always a series resistor. Even if you use 000 gauge wires. The current simply increases to the extent that integral(i^2xRdt) dissipates half the energy.

Knowing you, waross, I know that your next question will be "What if I cool down to absolute zero and get superconducting circuits?" That's a very good question. The answer is that the Q of the circuit gets very high and instead of an RC circuit, you have more of an LC circuit where the closure of the switch starts heavy ringing so that the current oscillates. An oscillator emits energy, so the energy lost in R at room temperature is now lost as radiated energy instead. The resulting energy is still 0.5xCxU^2.

The L and C duality (Vs/A = As/V, I = U, Short = Open) makes the proof valid for inductors as well. Example, charge a capacitor and short it with a piece of wire. A bang is heard and a spark is seen (if enough energy involved). Next make a current flow in an inductor (current instead of voltage, see?). Now open the circuit (open instead of short) and see the spark. Spark is now a result of high voltage (it was high current when shorting the capacitor). The energy dissipated in the sparks are the same (0.5xAxB^2, where A is L or C and B is I or U).

Now, let the inductive current circulate in a diode bridge. The total energy is - again - the same. So, it doesn't matter how you dissipate energy. It is always the same amount you have to get rid of. If you do it in a MOV, a zener, a resistor or a diode influences the time it takes to dissipate the energy, but not the energy itself. A high voltage MOV reduces energy faster and a single free-wheeling diode does it very slowly, sometimes used to delay drop out of relays and valves up to several hundred milliseconds.

To the question about what percentage is dissipated in the coil and the MOV: It depends, a high voltage MOV (say breaking down at five times coil rated voltage) dissipates a much higher percentage than the simple diode where most of the energy, as you say, is dissipated in the coil resistance. It is not too complicated to solve for that analytically if you simplify the MOV and diode to have a constant voltage drop while they conduct.

Gunnar Englund

http://www.gke.org

 

[waross]

Hello skogsgurra
Thank you for your time and trouble to provide the explanation to me. It is a proof of your thorough understanding of the subject that you are able to explain it with a minimum of technical words and mathematic formulas. I am sure that there are a lot of technicians striving to increase their understanding of things electrical by quietly following these fora.
Actually, I wasn't thinking superconducting yet.
May I re-state the issue in my own words to verify my understanding?
With a field voltage of 100 volts discharging through a diode;
The voltage drop of the diode is less than a volt, so that about 99% of the energy will be disipated in the winding.
The voltage drop across the MOV in your example was 5 times rated voltage, so about 1/6th of the energy will be disipated in the winding and about 5/6ths of the energy will be disipated in the MOV.
If I am still on track, if the current is 1 amp, the peak energy disipated by the MOV will be 5 x 100 volts x 1 amp, or 500 watts.
Am I anywhere near the ballpark, skogsgurra
respectfully

 

[Skogsgurra]

Yes, but 500 W is the peak power. Current decreases almost linearly as long as the MOV voltage is constant, so power is almost zero at the end of the process. For energy, you need to factor in the time as well. And, as said before, the total energy dissipated is 0.5xLxI^2.

Gunnar Englund

http://www.gke.org

 

[amps21]

Skogsgurra,
Your explaining of energy storage in LC circuits brought my classroom days back to me. Ur teaching skills are praisable.