86 relay current draw? 5

[eeprom]

Hello,
I'm troubleshooting a failed trip circuit for an 86 lockout relay. This is a GEI-28712E, similar to an Electroswitch. The 86 coil is tripped by one of several relays. The contacts for one of the relays has been welded closed, and I'm trying to figure out how that could have happened.

There was a standing trip on one of the relays, and the operator didn't know, and he kept trying to reset the 86. He held the 86 in the reset position long enough that it started to smoke. An observer told me he held the switch for at least 5 seconds, trying to get it to reset. I'm pretty sure this is what caused the welded relay contacts.

The current rating of the welded contacts is 30A for 0.2 second, and a continuous rating of 7A (at 125VDC). The trip voltage is 125VDC. The GE manual for the 86 says that at 125VDC, the resistance of the 86 trip coil is 24.3 ohms at 25 degrees C, and that the seal in time is 0.2 seconds. But this doesn't add up with a welded contact that can handle 30A. According to the manual, I could hold the handle to reset as long as I want and I'd never pass 5A.

I'm trying to figure out what the steady state current through that coil would be. If it's just a coil, then after the transient, it's a short circuit. But the manual says 24.3 ohms? Is there a resistor in series with the coil, or is that the impedance of the transient measured at 0.2 seconds?

Thanks for your help
EEPROM

 

[RedSnake]

I am not entirely sure I understand you question.

But how many times a relay can go on/off before it welds together is usually shown in a graf like this.

This is from another relay since you didn't say, what the model and number was.

So dependent on how big the effect is over the tung with the highest effect on the relay it will be able to pull and release a specific amount of times before it welds together which it will do sooner or later.
Breaking up DC voltage givs more arc and is harder to turn out, gives relays that handles DC voltage shorter lifespan.

BR A



“Logic will get you from A to Z; imagination will get you everywhere.“
Albert Einstein

 

[eeprom]

No, this is not addressing the right question. I have a closed relay contact (a standing trip) upstream of an 86 lock out relay. The closed contact is not changing states. The 86 is held in the reset position by an operator, making the trip circuit, which draws current through this closed contact, and through the 86. I would like to know how much current is flowing if you hold the lockout relay in the reset position.

 

[cranky108]

I don't use those but, In some lockout designs where high speed is required, the coil is rated for 24V, with a resistor inserted on a partial travel of the lockout. This is to allow targeting of older relays.
Thus, if you hold the lockout in the reset position the resistor is never inserted, and the contactor never interrupted.

You will need to refer to the manual for that lockout to be sure.

And the reason I don't use high speed lockouts, is the speed is not worth the complicated circuit. The normal speed is fast enough for me.

 

[LionelHutz]

Smoking coil = shorted coil maybe.

 

[RedSnake]

The GE manual for the 86 says that at 125VDC, the resistance of the 86 trip coil is 24.3 ohms at 25 degrees C,

Is this when it is pulled?
Because if it had welded together it doesn't mean it was welded together in a fully drawn state as it would have been if it had been pulled and alright.
The welding takes place when the relay tries to pull apart mening that between the tungs there will be a build up of metall keeping the tungs at a distans from each other so when you try to pull it again it's not sure it will be able to do so completely.

“Logic will get you from A to Z; imagination will get you everywhere.“
Albert Einstein

 

[eeprom]

I don't know what happened, that's what I'm trying to figure out. I don't believe the relay changed states. I just want to know how many amps (approximately) are drawn from an 86 relay when it's forced into the reset position with a standing trip. If that number is 10A, then I don't believe the relay is question could have been damaged or welded. But if the amp draw is 50A, and the operator held the pistol grip for 5 seconds, that's a different story.

I just need a good estimate on what the 86 will draw. I'm guessing that no one posting here knows that.

 

[thermionic1]

Is it possible the initiating relay opened its contacts faster than the 86 internal contacts in the trip circuit? When the 86 operates, these "b" contacts in the 86 should interrupt the trip current. Why didn't the initiating relay trip circuit release the trip on no fault (or similar). Is this by design? SEL relays have a setting called "Unlatch Trip" and can be assigned a number of elements (NOT Trip) for example.

I have an Electroswitch series 24 125VDC in front me and its ~28 ohms.

The attached shows a series 24 LOR with the internal contacts in series with the 86 coil that should interrupt the trip current as the LOR operates. If something is not right for these internal contacts, you can be applying a signal on a device normally rated for intermittent duty continuously.

 

[eeprom]

The pistol grip was physically held in the reset position, that's why the b contacts don't matter. This is a case of an operator not being trained and over-riding the system.

But...now let's talk about that 28 ohms on the coil. 28 ohms is a lot of resistance for a coil. Is that DC ohms or impedance? The resistance is just wire when it's a DC circuit. The coil has an inductive transient, but after that it's a short. If the coil is 28 ohms at steady state, then I could hold the pistol grip in reset all day long and I'd never draw 5 amps. If that's true, then the contacts in question were not damaged by the operator holding the switch in reset. I'm suspecting the coil is drawing more than 5A. A lot more.

That's what I'm trying to figure out. I have an electrician today who's going to put an ammeter on the coil and hold the pistol grip to reset for 1 second and see the amp draw.

 

[MKFPE]

eeprom,

I think thermionic1 is on to something, but it is still a bit puzzling. Older lockout relays can become "sticky" after sitting unoperated for years and if the fault was intermittent the initiating relay might have dropped out before the 86 could operate. I have seen this, but the initiating protective relay contact burned open, not closed. Can you provide a photo of the welded contact?

Also, the smoke is not hard to explain. Lockout relay coils are designed for short energized time. If the voltage is 125 volts and the coil resistance is 24 ohms the power dissipated in the coil is 125x125/24 = 651 watts. The current is 5.2 amps. At over 600 watts dissipation the coil heats up quickly.

 

[eeprom]

I don't have a photo of the welded contact, and I haven't yet confirmed it's welded. I just know it won't change states.

I don't seem to be making my point. The 86 relay trips everything just fine. It's wired right and has been in service for 30 years. The problem occurred when a human being went to reset the lockout while there was a standing trip on one of the protective relays. This action would, of course, result in another trip. Now if the guy holds the pistol grip in the reset position, he's preventing the 86 from tripping. This is allowing current flow. How much is the question.

If it's really 5A, then most likely the relay contact did not get welded and has failed for other reasons. I'm trying to figure out how much current would flow if you held the pistol grip in the reset position. I am not trying to troubleshoot the lockout relay.

 

[MKFPE]

eeprom,

Relax and use Ohm's law, pretty basic stuff.

If you know the applied voltage and the resistance in the circuit the current equals the voltage divided by the resistance. Again, if the coil resistance is about 24 ohms as you said in your original post the current is 125/24 = 5.2 Amps. If your 125 volt system floats at about 132-133 volts as most do the current is about 133/24 = 5.5 amps.

 

[eeprom]

So, the steady state current draw is 5.5 amps. That's what I wanted to verify. I wish I had known of ohm's law before now.

 

[MKFPE]

eeprom,

3 more things:

1) Your lockout relay has been abused. I would be concerned about its future reliability.

2) I would get some training for your operator, understanding of protective device function appears to be needed. LOR tripping is a "call in the experts" situation, not something to try to override.

3) If you need to upgrade your protective relays take a good look at the SEL relays with their patented high interrupting output contacts that can interrupt 10A DC at 125VDC.

Good luck,

 

[eeprom]

MKFPE,
I work at many small plants, and a lack of operator training is a common problem. Very common.

At this point, my only concern is to find out what's wrong with the trip circuit, get it repaired, and to explain to the owner what happened. I don't yet know why the contact failed. Those other items you mentioned have been addressed numerous times in the past.

 

[waross]

It may be a failure of the trip coil. A failure of the economizer circuit or a mechanically jammed trip solenoid.
If the economizer circuit does not operate, the current will be very high. Hold it in and something will smoke.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Fischstabchen]

Buy a new lockout. The relay is not meant to see DC for long periods for through the coil. Whatever process you have is more expensive than than another relay. If smoke comes out of anything, you replace it.

 

[RRaghunath]

eeprom,
I am almost certain the contact welding is not due to the current rather it is due to arcing. As it was held in Reset position for only 5s, I doubt the current magnitude can cause that much heat to cause contact welding. Mind you, these contacts of 86 are tough and do not melt that easily.
When the operator was holding the pistol switch against the force of the operating coil, I guess there was arcing as the contact tried to open.

 

[LionelHutz]

Have you checked that the coil on the 86 is still OK or not? It's pretty easy to do an initial check with an ohmmeter because it should be 24ohms or maybe a bit more.

 

[eeprom]

Gentlemen,
This post has nothing to do with the lockout relay, other than trying to determine the steady state current flow. The lockout and the obvious need for operator training are separate issues and I assure you they will be dealt with. Thank you so much for all of your suggestions. I will re-post once I have figured out what happened.