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GE electromechanical motor protection relay high-dropout flag

GE electromechanical motor protection relay high-dropout flag

GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
I should clarify we have 3 relays, one per phase. We never see flags on any of them during normal start.

RE: GE electromechanical motor protection relay high-dropout flag

I'm not sure if this is what you're asking, but non-operation of a trip or target involving current flow in excess of the pickup setting for only 1/4 cycle(0.0042s) doesn't sound odd. I know typical Inst trip times for C/Bs usually take 0.07s or so, though this also includes opening times. The only device I know which responds in the first half-cycle is a current-limiting fuse.

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
Thanks for the info Dan.

I should have said 1/2 cycle... that is where highest peak can occur.  

I should mention that the whole point of the high-dropout scheme to provide another element which can be set closer to LRC than the normal instantaneous without tripping on the DC component.

We set the high dropout at 1.1*LRC for 0.1 sec.
Instanenous at 1.7*LRC.
This is consistent with GE and other guidance for electromechanical relays.

IF the DC offset component had no effect on an instaneous relay, then the scheme makes no sense... we might as well get rid of the 0.1 sec delay, or equivalently set the regular instantaneous to 1.1*LRC.  

I don't agree with that. By my thinking 1.1*LRC should result in trip due to dc component... that's why we increase the normal instantaneous setting up to 1.7x.

I'm not trying to argue... just trying to explore your comments.

RE: GE electromechanical motor protection relay high-dropout flag

Maybe I wasn't clear in my previous response. From your initial question, you seem to state that there are four targets, one for O/C, one for standard Inst, and two for the High-Dropout Inst, one of which is activated by the plunger at 50-1ICOB and one which is activated after the 0.1s delay when the telephone relay is activated.
I thought you were originally asking why you didn't get a target(not a trip) on the '50-1ICOB' from the plunger action which would occur without the 0.1s delay for the telephone relay. My response was an attempt to provide a reasoning why this wouldn't happen in such a short time(1/4 or 1/2 cycle).
Are you sure that there are two targets for the High-Dropout Inst? I see the four coils in the picture, but it seems that a target for pickup below 0.1s would not be in anyone's interest for this function, and could be very confusing. We don't have the original book for this relay(and we have almost all of them). Apparently it is a very rare relay, since my main relay engineer has never seen one. He did mention that just because the coil is there, it doesn't necessarily have to have a target. I'm not doubting you; you have the relay there, just checking. If you push up on the contact below the coil, this should drop the target, if it is there.

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
Thx Dan.  I agree with most of what you have said. We have plenty of these relays installed.  

You have now proposed two possible scenario's
#1 - The peak at 1/2 cycle is not enough to trip an instantaneous relay.
#2 - Perhaps the relay does not have a target associated with the 50-1IOCB high-dropout plunger.

#1 doesn't make any sense to me. The reason I brought up the 1.1 and 1.7*LRC settings is because they are all based on the premise that the dc component resulting in peak at first 1/2 cycle CAN trip the instantaneous relay.

#2 seems reasonable.  The relay looks like the picture in the manual. You can see the 4 shutter-type target windows, the upper right one associated with the 50-1IOCB high-dropout plunger. But I agree it seems likely that the target/flag feature associated with the 50-1IOCB high-dropout plunger may have somehow been disabled.  Certainly a flag (without trip) during every start would cause concern for many folks... so it makes sense that GE would e disabled it.  If that's the case I can see also potential for some confusion. Let's say you have a trip accompanied by a normal instantaneous flag but don't see any high-dropout flag (which you expect since it has a lower setpont and a window that looks identical to a target).... that would lead us toward suspecting calibration error when none existed.

I can't manually actuate the plunger on our installed relays without generating trip (I don't have confidence I can do it for <0.1sec). I will try it next time we have one out.

Thx

RE: GE electromechanical motor protection relay high-dropout flag

I wasn't saying that the DC component wouldn't trip the target and/or relay, I just said that I thought that
nothing would happen within the first 1/4 to 1/2 cycle; I believe the operating time of any type of electro-mechanical coil actuator would take longer than that.
I believe your initial question was 'why didn't the target for high-dropout-instantaneous plunger contact 50-1IOCB operate in this time limit?'.
We both know that the DC component(transient asymmetrical motor inrush current, as GE says) can actually hang around for significantly longer than 1/2 cycle, depending on the X/R of the system, which should be pretty high for a starting motor. The six-cycle delay should allow it to completely deteriorate(if the motor begins to accelerate), leaving just the LRC and not causing a trip. If the motor doesn't accelerate properly, the X/R stays high, the DC component lasts longer, and the current after the six cycle delay is still high enough to cause a trip to protect the motor.

I'm not sure, but I believe the reason that the high dropout Inst is provided in this relay is to wait for the DC component to deteriorate before a trip is actuated, not to trip because of it(within the six-cycle delay). The scenario as I see is this:

1. The high dropout Inst coil picks up within the first cycle of motor start because of the asymmetrical inrush.

2. The motor starts to accelerate, reducing the DC component.

3. Sometime before the six-cycle delay, the 'high dropout' rating allows the coil to dropout at 80-90% of pickup(as opposed to the 40-50% dropout for the standard Inst coil), which would be caused by the decreasing DC component, preventing an unnecessary trip.

After the delay, the only reason the current would be above LRC would be if the DC component was still present, indicating insufficient(or non-exsistent) acceleration, and then causing a trip.

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
Thanks Dan. Those are good comments. I agree with your discussion of the relay operation.  Also I never knew where the term high dropout came from but it makes a lot of sense when you explained it.

The one thing I am still a little unsure of is the exact significance of the 1/2 cycle peak. Certainly the relay/breaker will not interrupt in the first 1/2 cycle, but isn't it possible that first peak sets in motion events which will eventually lead to a trip even if the current were removed after the first peak?  It's a little academic and I'm sure there is no simple answer. One thing I think about is the fact that the force will in fact go to zero after the first peak and periodically thereafter.  It seems like tripping will depend upon the integral of force over the one-half cycle period between zeroe's of the force.

Since force is proportional to current squared, we would need to look at a quantity like the rms to evaluate the effect of a non-sinusoidal (dc-offset) waveform. By coincidence, the factor applied to LRC (let's say we call it sqrt(3)) is exactly the ratio between the rms of the fully-offset LRC waveform and a pure sinusoidal LRC waveform.

That can be found applying the definition of rms.

the sinusoidal LRC time waveform would be sqrt(2)*LRC*sin(w*t) where LRC is the rms current.

The offset waveform would be
Ifullyoffset = sqrt(2)*LRC*(1+sin(w*t))  (I'm a little sloppy with the phase angle but it doesn't affect the result).

The square would be
Ifullyoffset^2 = 2*LRC^2*(1^2+2sin(w*t)+sin^2(wt))

The mean square would be
<Ifullyoffset^2> = 2*LRC^2*(1 +  0+ 1/2)
                 = 2 * LRC^2 * 3/2
                 = 3 *LRC^2

The root mean square would be
sqrt(<Ifullyoffset^2>) = sqrt(3)*LRC

To me it seems a pleasing explanation.  I'm not sure if it's the right explanation for why we choose approx 1.7 (and of course it does depend on relays... I'm most interested in electromechanicals).  And even if it is the right explanation I'm not really sure it would prove anything about the time it takes for relay to respond.  But I'd be interested to hear your comments if you have any.

My original question has been answered.
Thansk





RE: GE electromechanical motor protection relay high-dropout flag

I think you may be over-analyzing the electrical portion of the system, where maybe we should look at the magnetic properties.
The Inst coil plunger actuates because of the magnetic field created by the current flow in the winding, however, I believe there is a shading coil which helps create enough magnetism to continue the attraction through the zero crossing, which continues the action of the plunger arm, so the force doesn't actually go to zero at any time after the first 1/2 cycle.
I'm sure we can dig up the equations for calculating the magnetic flux density which would include the current flow contribution, the shading pole circuit contribution, and apply that attractive force on the mass and moment of the plunger arm. I believe the actual answer to the response time may lie in this type of analysis. (My guess is that it is between the first and second cycle)
Wouldn't that be a wonderful question for the PE test?

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
That would be a stumper all right.

You are right I may be way off base on the sqrt(3) theory.
 
But it seems quite a coincidence that many references use 1.7, 1.7-1.75 and I have seen exactly 1.73*LRC (IEEE37-96-2000 as the recommended setting.  I am pretty sure that that number is not coincidence and that particular author came up with that number through similar computations as mine, although the assumptions and conclusions surrounding those calculations may be a little different. (it really doesn't prove anything other than that it suggests that  the critical  characteristic of a non-sinusoidal waveform - the dc offset inrush -  may be it's rms value, not it's peak).

Does shading apply to plunger relays?

I will note that IEEE37-96-2000 describes the purpose of the high-dropout in similar terms to what I used above.
"When it is necesssary to set a direct triping 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. (they show figure with high-dropout at 1.1*LRC 0.1sec delay combined with another instantaneous at 1.7*LRC no intentional delay".

They have a discussion of locked rotor and stall protection and they don't say anything about use of high dropout or delay.

So who do I believe? I think I believe you. Here's why.

I am in the middle of troubleshooting some spruious trips on one of our motors. Looking at our history of old trips (I work at a plant which keeps very good records of maintenance), I saw one 7000hp motor that tripped on high-dropout during starting after a long maintenance period. Troubleshooting showed relays in cal, motor and cables tested good etc.  They tried to rotate by hand with strap wrench and required excessive force to break it free.  After that it started fine.  I have a written record of all of the above. I also have verbal reports from our operators that a similar scenario has repeated on this pump several time.  Now they turn it with a strap wrench before they even attempt starting after a long maintenance outage.  I don't know the root cause of the binding, but that's another story.

The main point is, that experience seems to support your discussion of high dropout as protection for locked rotor condition. And your explanation makes  good sense.  That is another valuable insight that I have gained from this discussion. Thanks!

RE: GE electromechanical motor protection relay high-dropout flag

Suggestions:
1. The DC offset is normally exponentially decaying. This is mathematically expressed over the exponential term in:
i(t)=(Vmax/|Z|)[sin(wt+alfa-theta)-exp(-Rt/L)sin(alfa-theta)]
which is Equation (11.2) on page 269 in:
William D. Stevenson, Jr., Elements of Power System Analysis, Third Edition, McGraw-Hill Book Co., 1975
2. Contact GE tech support for application notes and relay functioning details.

RE: GE electromechanical motor protection relay high-dropout flag

electricpete, thank you, this has been a very thought-provoking post for me also.
I do believe the relay plungers have shading coils, as I saw on p.8 of the GE manual(your link from the first post). Adjustment of the 'shading ring' is apparently used for pickup adjustment.
As I said before, I and the other engineers in my company have not seen this particular relay.
I basically used the information in that manual as background for an explanation of the how and why the High Dropout Inst works, (those 'insights' came to me themselves as I wrote my responses). I'm sure I never would have understood the relay action(if my explanation is indeed correct) without your questions.
Thanks again. Until next time....

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
One thing I did not see in the manual (or anywhere else) is discussion that the high dropout feature might have a benefit specifically for locked rotor protection.  That is a brand new idea and very interesting to me and appears to match some experience that we have.

But now that I am thinking about it some more, maybe the dc decay over 6 cycles is so much that there would be no significant residual dc, even at 0.2 power fact.

PF = 0.2 = R / sqrt(R^2+(w*L)^2)
where w = 2Pi*f
For R << wL (very lower power factor), sqrt(R^2+(w*L)^2)~w*L
PF = 0.2 ~ R/w*L.

dc component decays according to exp(-t*R/L).
In one cycle t=T=2Pi/w it will decay
exp(-(2PI/w)*R/L) = exp(-(R/w*L) * 2pi)
substituting in R/w*L ~ 0.2 we have
exp(-0.2*2Pi) = 0.28
That is the decay in one cycle.
For 6 cycles
exp(-6*0.2*2Pi) = 0.0005

From first look at the math it seems like there is no dc left after 5 cycles even at 6 cycles. Is anyone interested in checking my math?

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
Next to last sentence should read: "From first look at the math it seems like there is no dc left after 6 cycles even at pf=0.2"

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
The other way to find R/wL without the assumption R<<wL is

Pf = 0.2 = R / sqrt(R^2+(w*L)^2) = R/w*L / sqrt((R/w*L)^2+1)
Let x = R/w*L
0.2 = x / (sqrt(x^2+1)
  multiply by sqrt(x^2)+1 and square
0.04x^2 + 0.04 = x^2
0.04 = 0.96*x^2
x = sqrt(0.04/0.96) = 0.204      

(pretty close to previous approximation).

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
p.f. = 0.2 is used as typical average value for the duration of motor starting, but perhaps it is much lower at the initial locked-rotor condition?

RE: GE electromechanical motor protection relay high-dropout flag

All the info I can find points to approximately 0.2 PF at startup, but I'm not sure if that is for sub-cycle time ranges. I suppose a time-domain analysis of a motor circuit model would tell what is actually happening. I'll try it tonight.
Meanwhile, my coord software draws a motor start curve which at the 0.01s time line is approximately 60% higher than the LRC, which it curves down to at 0.1s.  

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
It remains a problem of keen interest for me. I have another motor which I am troubleshooting recent trips. Details of these recent trips are sketchy (I don’t even know what flags were actuated) , but I do have some historical information recorded by plant computer.  I am trying to narrow down the scenario using a number of factors (more than I can explain here).  So I  am very interested in knowing whether trip due to binding on high-dropout is credible scenario or not.

Here is data on the other previous motor which is KNOWN to trip during startup and found high-dropout flag and later found the shaft difficult to break free.  

7000 HP 3600RPM motor FLA = 261 LR KVA Code = F  LRC = 1567 max

w *( L1+L2) = X1+X2  ~ V_L-G / I_LRC = 13,200/SQRT(3) / 1567 = 4.8 ohms

EFFICIENCY = 96.1
Total losses  ~ 4% * 7000hp = 280hp ~ 208,000 watts.

Total losses per phase = 1/3 times total losses = 70,000 watts

Assumed Fraction of Total I^2*R full-power losses per phase which are related to stator I^2*R – 30%
Total Stator losses full power per phase ~ 0.3*70,000 watts = 21,000 watts
Rstator ~ P/I^2 ~ 21,000 watts / 261A^2) ~ 0.3 ohms

Assumed Fraction of Total I^2*R full-power losses per phase which are related to rotor I^2*R – 20%
Total Stator losses full power per phase ~ 0.2*70,000 watts = 14,000 watts
Rrotor ~ P/I^2 ~ 14,000 watts / 261A^2) ~ 0.2 ohms

Note actual rotor resistance during starting will be somewhat higher due to skin effect.

Using the low-calculated value of Rrotor, we have

Rtotal = 0.2+0.3 ~ 0.2+0.3 = 0.5

Pf ~ Rtotal / w*L ~ 0.5/4.8 ~ 0.1

6 cycle decay = Exp(-2Pi*6 cycles * 0.1) ~ 0.03
It seems like that will still get me well below 110% of LRC and even lower if I consider skin effect on Rrotor.   By the way we don’t check the cal of the 0.1 sec delay so that may be in question as well.  Any commments?

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
The "assumed fraction..." should be "assumed fraction of TOTAL losses per phase....".  I don't have info on the loss breakdown for this motor.  I got 20% rotor losses and 30% stator losses out of a book as typical values for 2-pole large motor.

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
The other main question I am focusing on is mechanical binding, since the more recent motor tripped three times all following period of shutdown.  That is something that probably doesn't fit this forum. I have posted it here:

http://www.reliability-magazine.com/ubb2000/ubb/Forum2/HTML/002157.html

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

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?

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

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.

RE: GE electromechanical motor protection relay high-dropout flag

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.

RE: GE electromechanical motor protection relay high-dropout flag

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.

RE: GE electromechanical motor protection relay high-dropout flag

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

RE: GE electromechanical motor protection relay high-dropout flag

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!



 

RE: GE electromechanical motor protection relay high-dropout flag


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.

RE: GE electromechanical motor protection relay high-dropout flag

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.

RE: GE electromechanical motor protection relay high-dropout flag

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.\\\
 

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.


RE: GE electromechanical motor protection relay high-dropout flag

Electricpete,

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

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

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!

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

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!     

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

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!

RE: GE electromechanical motor protection relay high-dropout flag

'pete, one other point I though of while reading the follow-up comments; have you checked the calibration of the relay lately?

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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.

RE: GE electromechanical motor protection relay high-dropout flag

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.

RE: GE electromechanical motor protection relay high-dropout flag

Suggestion: Visit
http://www.geindustrial.com/multilin/notes/ref/indexapp.pdf
to see how GE categorizes and aligns relays for the motor protection, including IAC66M relay, namely:
49S/49/50 Starting, Overload and Fault, Single Phase . . . . . . . IFC66KD 2 or 3 2
or IAC66K 2 or 3 2
or IAC66M 2 or 3 2
or IAC66S 2 or 3 2
or IAC66T 2 or 3 2
or DIAC/DIFC/DSFC 2 or 3 *

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
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."

RE: GE electromechanical motor protection relay high-dropout flag

(OP)
last sentence should be "125%-200%" of full load current.  ("%" without the shift accidentally turned into "5").

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