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GE electromechanical motor protection relay high-dropout flag 1

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electricpete

Electrical
May 4, 2001
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This question concerns GE electromechanical motor protection relays which have a high-dropout trip in addition to the normal time overcurrent trip and instantaneous trip.

For example IAC66M described at:


Circuit diagram page 17.

The high-dropout trip is a separate instantaneous plunger element (50-1IOCB) set at 1.1*LRC (our setting) which picks up a telephone relay (50-1/OX) with 0.1 sec delay. After 0.1 seconds the telephone relay output NO contact will close to complete a path through the high-dropout-instataneous plunger contact 50-1IOCB (IF still closed) to energize a seal-in coil (T-SI) which closes/seals-in the trip output at pin 8.

In addition to the normal two flags for normal time overcurrent and instantaneous trip (bottom of the relay), there are two flags for the high-dropout at the top of the relay. One is driven mechanically by the plunger 50-1ICOB, and the other by the T-SI sealin relay.

I would expect that for any normal start, the starting current would exceed 1.1LRC during the initial quarter cycle due to dc offset, and then decay to LRC long before 0.1 seconds. I expect the result would be a single flag at 50-1/IOCB during every start (but no trip). We never see this flag.

Can anyone explain where I have gone wrong.
 
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You made my point... there are many unkowns! Too many, in fact, to zero-in on the 50 relay's shading coil, or DC decrement.

In my experience(s), and several were provided, the breaker contacts were never engaged long enough to enable the motor to produce sufficient breakaway torque... human effort could, but electricity couldn't? And its obvious that the 50 relay takes a finite time, at least 8 ms, to react. Of course, the proof is in the current measurement.

In closing, I understand your point of view.

Good luck hunting!
 
Hmmm. Relay has a dropout adjustment range from 80%-90%.

We set it at 110% LRC.

Max dropout 90%*110%LRC = 99%LRC.

If relay doesn't reset after dc decays (but LRC remains), it will trip.

1% would not likely be enough margin to account for voltage variation, much less relay cal tolerances.

I am surprised that we haven't had more trips.
 
Electricpete - The high-dropout instantaneous that I am familiar with on the old GE IAC66K relays (and similar Westinghouse COM-5) was used differently than you have described. It was generally set 150-200% of FLA and it was used to supervise the tripping of the time overcurrent unit (51 and 50HD contacts in series). The philosophy was that the time overcurrent (set at 100-110% FLA) alone would only alarm the operator of a motor overload up to the setting of the 50HD unit. Above the setting of the 50HD unit, the combination of the 51 and 50HD units would trip the motor. The high-dropout characteristic was necessary to assure that the 50HD unit would reset after a start, since locked rotor current would most certainly pick up the unit. A normal instantaneous unit might not reset at FLA.
 
Thx jwerthman. I believe that IEEE C79.96 describes both the technique that you describe (#2 below) and the technique that we have used (#1 below). But I think we have been mistaken in the selection of 1.1*LRC as a setpoint.

IEEE 37.96-2000. page 79

#1 - "When it is necesssary to set a direct tripping IOC lower to provide adequate fault protection, it may be delayed with a short time delay (6-15 cycles) to prevent operation on assymetrical starting current as shown in Figure 38."

Fig 38 shows figure with high-dropout at 0.1sec delay combined with another instantaneous at 1.7*LRC no intentional delay. I was mistaken when I earlier said that this figures shows the HDO at 1.1LRC….. it is not labeled but by interpolation I would say it is higher. (we have 1.1*LRC in the calculation that established HDO setpoint for our motors many years ago…no-one left to ask why).

#2 - "An ICO can also be used to supervise a TOC relay (see figure 40). This scheme allows the TOC to trip for faults and serious overloads, but to alarm only for small overloads. Typically the IOC is a high dropout unit set for 1255-2005 of full load current."
 
last sentence should be "125%-200%" of full load current. ("%" without the shift accidentally turned into "5").
 
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