Reclosure Circuit Holding Coil

[UtilityGy]

Hello Everbody,

My first post on this website. I am sure my diemna will be resolved. I spent all Friday evening to figure this out but my logic did not make sense. May be my physics skills are not the best to understand concept of holding and closing coil.

http://www.gedigitalenergy.com/products/brochures/hga18.pdf

Could you kindly explain how this single shot reclosing circuit works and what is the role of capacitors. I am trying to picture how this can be part of a circuit breaker. Could you kindly advise how a blocking reclosure option be used in the circuit above.

Thanks in anticipation.

 

[rhatcher]

It took a while to figure this out but I think that I have it right. The manual for the relay is located at the link below but it does not offer much more detail.

http://www.gedigitalenergy.com/products/manuals/hga18/gek49819.pdf

For my explanation the notation such as resistor 79/RES 9-10 indicates the 79 resistor located between terminals 9 and 10.

The simple explanation is that both the operating coil and the holding coil require full supply voltage (say 125VDC) to close. The use of the two separate operating and holding coils are necessary to provide the one-shot action of the relay in conjunction with the capacitor 79/CAP. The operating coil 79/OC closes the relay from the discarge of the capacitor 79/CAP. The holding coil seals in the relay using the supply voltage to hold the relay closed until the breaker closes even if the capacitor 79/CAP has already discharged. The capacitor 79/CAP is used to provide the lockout if it is not fully recharged and to provide the time delay for auto resetting based on the time to fully charge when in series with the charging resistor 79/RES 9-10. If the breaker opens again before the reset time, the capacitor 79/CAP will not be fully charged to the supply voltage and the operating coil 79/OC will not see full voltage from the supply because it is in series with the charging resistor 79/RES 9-10. For a step by step explanation of the operation, read what follows.

The breaker control switch (52/CS) is normally in a neutral position and is spring loaded to return to neutral after being operated by hand. When the switch is momentarily turned to the close position, the contact 52/CS 4-4C momentarily closes, causing the breaker to close. The contact 52/CS 4-4C opens when the switch is released and it returns to the neutral position. The contact 52/CS 5-5C also closes but it remains closed until the switch 52/CS is turned to the trip position. For clarification on this, see the contact condition chart for the 16SB1B9 switch. The condition for the contact 5-5C is normally closed after a closing operation (NOR AFT CLOSE). The contact 4-4C is open.

Once the switch 52/CS has been operated and released, the breaker is closed, the contact 52/CS 5-5C is closed, the contact 52/CS 4-4C is open, and both of the contacts 52/B are open. The capacitor 79/CAP begins to charge with a charge time that is determined by the resistor 79/RES 9-10. This charge time is the relay reset time. As the capacitor charges, the current flowing through it reduces, thereby reducing the voltage drop across the resistor 79/RES 9-10 and eventually allowing the capacitor 79/CAP to fully charge to the supply voltage.

If the breaker trips open, both contacts 52/B close. This allows the fully charged capacitor 79/CAP to discharge through the contact 52/B 8-7, applying full supply voltage to the operating coil 79/OC. The operating coil 79/OC pulls in the relay, closing the contacts 79 1-2 and 79 5-4. This allows the supply voltage to pass through the contact 52B 8-5 and the contact 79 5-4 to operate the holding coil 79/HC and seal in the relay until the breaker closes. This also allows the supply voltage to pass through the contact 79 1-2 to the breaker closing circuit, therefore reclosing the breaker.

If the breaker trips open again before the capacitor 79/CAP is fully charged, the operating coil 79/OC will not operate because it does not see the full supply voltage since it is in series with the charging resistor 79/RES 9-10. Without the operating coil pulling in, the rest of the sequence is interrupted.

Therefore this is a one-shot relay. To reset the relay, the breaker must be manually closed with the closing switch 52/CS and must remain closed long enough for the capacitor 79/CAP to fully charge to the supply voltage of 125VDC.

I hope this helps you.

 

[UtilityGy]

Rhatcher,

I am going to take a printout and study it. I sincerely appreciate your writing such a big response. I will come back tomorrow with questions. I can see this website surely has dedicated people to land a helping hand.

Thanks

 

[rhatcher]

No problem, that is what we are here for. Good luck.

 

[UtilityGy]

Hi Rhatcher,

I got it. Thanks to your explanation. I found out that my relay is HGA18 A2A. I got its manual from GE archive. I have attached it for your kind consideration. Fig. 9 shows the schematic. To my surprise when I got to the substation, it was not quite the same as I found in the manual .

http://files.engineering.com/getfil...73-4fa7-9637-f7d31519b686&file=GEI-11325E.pdf

Please see the link below for the schematic:

http://files.engineering.com/getfil...c76-4349-82f2-0ed15609c741&file=Reclosure.jpg

If you kindly look in the middle of the schematic, just below 6R-7R and 5, there is a normally closed contact drawn with pencil. Actually, this normally closed contact was introduced by substation guys later on to block the reclosure scheme, if they want to.
But there are some strings attached to this scheme. When there is a fault and the breaker is open and crew is working on line, and that control room guys disable the reclosure and the normally closed contact discused above is opened. Now if the control room guys put the recloser scheme back in circuit by closing the normally closed contact, the breaker that was opened, it reclosed. It was surprising but I figured it out why that happens. It was kind a strange to hear how come a breaker open on fault with a disable reclosure scheme will reclose as soon as a the reclosure scheme is enabled and the breaker is open.

I think it is happening because the capacitor which is charged because 3 and 5 CS contacts are still closed and the capacitr is still charged which recloses the breaker discharging through the holding coil.

I want to seek opinion on the solution.

The way I have decided to implement this solution is by creating a using the normally open contact(Form C) between 9 & 6 in parallel with capacitor, this contact would be closed and short the capacitor in addition to the normally closed contact between 7R and 5. this contact would be closed when they When the control station guys, unblock the reclosure scheme when the breaker is open, the capacitor would discharge through the shorted contacts and not reclose the breaker.

This contact shots the capacitor momentarily discharges it and then becomes open in parallel with capacitor.

I am not sure, if I phrased it right. My concern is if we will shorting a capacitor can cause any problem. There is 179R in series to limit the current.

Kindly advise.

Thanks

 

[stevenal]

Probably don't want to discharge the cap with dead short, since the initial current might be very high. How about relocating your reclose disable switch so that it is series with the 179/R resistor?

Has the recloser always performed this way?

 

[UtilityGy]

Appreciate the response Stevenal.

Please see the link below:

http://www.gedigitalenergy.com/products/manuals/hga18/gek65602.pdf

This seems to the direct replacement of old model. In this doc on Page 12 they have shown 100 ohm res. in series with disabling contact of the reclosure scheme.
Para. 3 on Page 4 talks about using a resistor to discharge capacitor.

Questions:

1)What would happen, if I short the capacitor without the resistance.

2) Any time delay that might crop in by inserting a resistance in series with normally open disabling contact that might trigger reclose operation.


Thanks

 

[UtilityGy]

One more addition to my question, the existing resistance 79/RES 9-10 is 5 mega Ohms. My supply voltage is 125 V DC.

This will give a current in mAmps. How a holding coil will pick up on this current.

am I getting this right ?

 

[stevenal]

Looks like they did it with one contact. I was concerned that the two contact solution you proposed could be a problem because contacts never operate exactly together.

1) After the capacitor charges to 125 V, the capacitor becomes a 125 V source. 125V/0 ohms is a really big current.

2) There's always a time constant associated with RC circuits, but I don't see that causing 179/OC to energize.

3) Sounds too big to me.