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]

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

 

[DanDel]

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.

 

[electricpete]

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.

 

[DanDel]

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.

 

[electricpete]

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

 

[DanDel]

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.

 

[electricpete]

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

 

[DanDel]

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?

 

[electricpete]

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.
&quot;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&quot;.

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!

 

[jbartos]

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.

 

[DanDel]

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....

 

[electricpete]

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?

 

[electricpete]

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

 

[electricpete]

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).

 

[electricpete]

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?

 

[DanDel]

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.

 

[electricpete]

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?

 

[electricpete]

The &quot;assumed fraction...&quot; should be &quot;assumed fraction of TOTAL losses per phase....&quot;. 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.

 

[electricpete]

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