GE electromechanical motor protection relay high-dropout flag 1

[electricpete]

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:

http://www.geindustrial.com/products/manuals/iac/gek86105.pdf

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.

 

[electricpete]

Thinking about my approach for power factor, I can see it may have been an error to consider only stator and rotor resistance. Other losses (stator core losses, stray losses) also represent energy losses which should perhaps be considered. And possibly even a rotor core loss element that doesn't show up in the full power losses (since field frequency with respect to rotor is much larger during start). That would push the R even higher and the decay
even faster. I did confirm we measured 0.342 ohms on all three phases during dc winding resistance check.

I realize this is very approximate approach. Actual measurements of conditions during starting would be preferable. We are planning on instrumenting one of these motors during its next start.

 

[DanDel]

electricpete, I had the same idea and am looking at some monitoring info I have about some 5kV 2500HP motors at start. The only problem is, these motors have a reactor start, which makes me think, what kind of starter are you using?

 

[electricpete]

Both motors are DOL start 13.2kv, horizontal sleeve bearing motors.

The first one with the problem was 7000hp, 3600rpm.

The one I am troubleshooting now is 2500hp, 1200rpm.

 

[DanDel]

The initial PF on the graphs I have from the monitoring drops down to 0.2 after start. I'm not sure if the response is quick enough in this PF graph. But the disturbance graph showing the voltage and current sine waves at start indicate a DC offset which dissipates within three cycles.

 

[Shortstub]

Elecricpete,

I presume that there is an instantaneous element for each phase? If so, are the unwarranted trips occurring in different phases? If no, something is wrong with the element. If yes, I suggest that you set the instantaneous pickup to 2x the rms value of starting current inrush for the condition of max expected supply volts to the motor.

The theoretical DC offset is twice the rms value. But the value at 1/2 cycle is somewhat less the DC decrement. No setting rule says it has to be sqrt3! And, 2x always worked for me.

 

[Shortstub]

To all,

In situations like this when there aren't enough H-eyes to watch the targets and hardware, I used a video-cam. Set it for 10-20x normal frame-rate.

 

[1x1x1x]

Electricpete,

You were right of the duration of transient inrush current, when I made a quick simulation just using X and R of the circuit at pf = 0.1 the DC component was practically gone after 6 cycles. The subtransient peak current was 3343 A which is 1.73 times of the 1567 A, but the 1.73 is the peak current not an RMS current.
The theoretical value of the maximum DC offset is two, therefore the instantaneous value of the subtransient current is sqrt (2)*2*RMS = this 2.82 times of the rms value.
I think the reason for the generally used 1.73 multiplier is that at pf = 0.1-0.2 X/R ratio of system gives you a DC offset factor about 1.73, and the majority of the motors X/R ratio falls in this region. the formula to calculate “k” the DC offset factor is

k = 1.02+0.98*e^-3*R/X

 

[Shortstub]

Electricpete,

I'm afraid you've experienced math-runaway.

I too, was wrong regarding the maximum instantaneous peak value. It is not twice the LRCrms value, but istead, 2.83 times. Following is the calculation.

Assuming maximum DC offset, then the peak asymmetrical current magnitude, at t=0, is equal to:

Ipk = 2 x LRCpk = 2 x sqrt2 x LRCrms = 2.83 x LRCrms.

Of course, if the time constant equaled zero, then the magnitude of the first-cycle peak (at t = 8.333 msec) will also be 2.83 times LRCrms. For any a time-constant greater than zero, the magnitude will have have decayed to a smaller multiplier. Maybe, 1.7, 2.3, etc, times LRCrms. It can certainly be calculated if you know the time-constant.

My point is, that none of the above has anything to do with the response of the instantaneous relay element. I'm sure GE can provide the information if you really want it. However, the bottom line is that it should be set to "ignore" whatever the maximum value is, is (excuse me, I couldn't resist)! And, I reiterate my earlier suggestion, that you set the pu to 2 x LRCrmc.

Now, the real problem is to determine if the relays are actually tripping! I'm sure the camcorder will be helpful!

 

[1x1x1x]

Sorry for my English and let try again

The peak (subtransient) value of the transient AC current can be calculated as follows:

Ipeak = Irms*sqrt(2)*k

Where the Irms = steady state rms current

k=1.02+0.98^-3*R/X

The R is the resistance of the circuit and X is the inductive reactance of the circuit.

 

[1x1x1x]

Electricpete,

To find out what is going on use a storage oscilloscope with three DC current probes, pre-trigger the scope on the relay tripping and record about 500ms before the relay trips. It will reveal the transient condition and the duration of the current transient.

 

[jbartos]

Suggestion to electricpete (Electrical) Feb 5, 2003 marked ///\\I would expect that for any normal start, the starting current would exceed 1.1LRC during the initial quarter cycle due to dc offset,
///This expectation is in the right direction. Actually, the dc offset will vary depending on the sin(alfa-theta) value. See my previous posting.\\ and then decay to LRC long before 0.1 seconds.
///Again, this time varies according to dc offset decay speed.\\ I expect the result would be a single flag at 50-1/IOCB during every start (but no trip). We never see this flag.
///Traditionally, protective relay flag on an actuation of the protective function. The protective function will take place if the dc decay is prolonged, e.g. by a short.\\Can anyone explain where I have gone wrong.
///By not reading good books on protective relaying, e.g.
A. R. van C. Warrington, "Protective Relays Their Theory and Practice," Chapman & Hall Ltd, 1971, Section 4.1.6 Instantaneous Overcurrent Relays on pages 146-147 including dc offsets and decays.\\

 

[electricpete]

Shortstub - I will disagree on your comment of math runaway. I have not overlooked the role of the sqrt(2).

LRC by my equations is defined as the rms value of locked rotor current. The recommendation is to set the instantaneous current to 1.73*LRC. That means it is calibrated with a sinusoidal current whose rms value is 1.73*LRC. (1.73 is ratio of the rms's or ratio of the peaks, take your pick). The rms content of that sinusoidal test current is equal to the rms content of the fully offset dc waveform which goes between zero to 2*sqrt(2)*LRC.

I agree with you the peak value of the offset dc waveform is 2.8*LRC. In my discussion above I have identified that the coincidence of a factor of exactly 1.73 in itself suggests that this author viewed rms and not peak as the quantity of interest in tripping of instantaneous relays on non-sinusoidal currents. Certainly we normally think of an ideal instantaneous relay as responding to peak of a waveform, but real electromechanical relay may be more complex. That is what I am exploring.

I agree there is reason to set below it's peak of 2.0*LRC (again this represents the rms of a sinusoidal test current as for all relay settings) due to first half-cycle decay, but exactly 1.73 seems too much a coincidence.

 

[electricpete]

Delete the words "it's peak of" from my last sentence. I intended the word peak in this case to mean with worst possible dc offset and excluding the effects of decay, but peak was a poor choice of words.

 

[Shortstub]

Electricpete,

Have you confirmed that an instntaneous element is indeed causing the trip?

 

[electricpete]

I have history of an old trip on a similar motor (both horizontal sleeve bearing motors) where we did record the high-dropout flag dropped and no others. In that case we found the motor was binding. It had been sitting idle for several months. After breaking it free it started fine.

The motor I am working on now has tripped three times over 2 years. I have conflicting information on what flags were tripped. We have done quite a bit of investigation (after the fact) of circumstances at the time of those trips, but it is too much to describe here. Within a month we will have the motor instrumented during it's next scheduled start.

 

[Shortstub]

eletricpete,

I don't understand the reason for your reluctance to raise the setpoint, but that's your perogative.

Getting back to the inst element. If it's not the cause of the unwarranted trips, then it sounds like a problem I had with an 11,000Hp, 11kV, DOL, some years ago.

Is the breaker air, reduced-oil, or vacuuum type? The breaker I had the problem with was manufactured in Europe, and I believe, under license from GE. Any way, when I locate my records I'll supply additional info!

 

[electricpete]

It is a GE air breaker. 13.2kv system.

Reasons for being reluctant to change the setpoint:
5 other sister motors have not tripped
This motor never tripped before 2 years ago
This tells us maybe something else is wrong that needs fixing.
As I mentioned I don't know 100% sure which flags tripped (I have some conflicting info). So at this point changing setpoints would be a shot in the dark. It is an option we will keep open when we learn more.

 

[Shortstub]

Electricpete,

Like most, I enjoy a good joke, and using my 48-45-6-10 rule of life, I thought I saw and heard everything! For example:

I saw a large motor MV motor fail to start because one breaker pole failed to travel far enough to make "contact". Thus the motor single-phased.

I saw an end-winding movement demo of large MV motor where the end-bells were removed and the rotor was "blocked" in the stator with wooden wedges. Full voltage was repeatedly applied. After some 10's of cycles, the wedges loosened, and the motor started to spin, without the benefit of bearings. Upon reaching full speed unbalanced axial forces shot the rotor out of its stator like an shell from an Iowa class 16" gun.

Even had a 22,000 Hp [11kV, DOL] motor trip with a 50 at initial start. You better believe the setting was increased so that it would'nt happen a second time.

And in my last post, I alluded to a situation similar to yours, where the 50's were thought to be causing unwarranted trips. Instead it turned out to be a problem in the trip-free circuitry.

Now you're telling your readers that a large machine, intentionally so, in order to produce large torque... using the ROT rule, about 16,000 lb(f)-ft [1,600 kg(f)-mt or 21,000 N-mt]... couldn't roll because it was binding? Yet, a few brutes (I accept that they weighed a couple thousand pounds apiece) were able to spin it by hanging on a strap-wrench? And, the investigation never made the books!

Now, I know it isn't April 1st, so maybe DanDel's right... perhaps you're over analyzing the problem!

 

[jbartos]

Suggestions: Visit

http://www.ee.sc.edu/classes/Fall02/elct751/SHORT_1.pdf

for DC offsets

http://www.beckwithelectric.com/powerlines/articles/i38/comtrade.htm

for COMTRADE software analytic tools

http://www.rocoil.cwc.net/Pr7o.pdf

for Rogowski Coils instead of CTs consideration

http://www.ee.sc.edu/classes/Fall02/elct751/Fundamentals_1.pdf

for DC offset derivation
etc.

 

[electricpete]

Hi shortstub. You are entitled to your opinion. As I stated excessive force was required to break it free. Also note that one can apply many times more force than one's weight in a short-duration jerking force AND that torque is the product of force times distance. You and I know nothing about how the force was applied other than the the writeup stated excessive force was required to break it free (cheater bar, hammer on cheater bar, come-along?). You may be comfortable to jump to judgement without knowing the circumstances, but I am not.