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Motor Instantaneous Trip setting ~ 1.73*LRC  why?

I have seen many reference which advocate setting the motor instantaneous trip setting in the range of 170%175% of motor locked rotor current.
In trying to understand the basis for that number, I use the following logic.
Threephase ungroundedwye or delta motor should act similar to onephase LR circuit upon sudden application of a voltage... worst case phase closing would give a current that goes from 0 to 2*I_LRC (ignoring any decay of the DC component). If I compute the rms value of this waveform, I see it is composed of DC component of rms magnitude I_LRC and ac component of rms magnitude I_LRC/sqrt(2). Total rms is the sqrt of sum of squares of two components or I_tot_rms = sqrt(I_LRC^2+(I_LRC/sqrt2)^2) = sqrt((1+0.5)*I_LRC^2) = sqrt(3/2)*I_LRC. If I compare this to the symmetrical (ac) portion of LRC which has rms of sqrt(1/2)*I_LRC, I see the ratio is sqrt(3).
Is that the basis for the sqrt(3). If so, why would we use the ratio of the rms currents instead of the ratio of the peak currents (which is 2.0). Doesn't an instantaneous element respond to the peak value of current?
It occurs to me the other possible explanation is that they are assuming some decay of the DC component prior to the first peak. But without knowing the L/R of the circuit, that seems nonconservative (might lead to trip during starting). 

I'm not aware of any relay that responds directly to peak values. The fastest relays I know of operate in 1/2 cycle. Takes at least that long a period of integration to get an rms value. If your reasoning is correct, doesn't it put the setting close to the edge? A little CT error, relay error and you're tripping every time you hit it on the zero crossing. 

Found a reference. ANSI C37.96 says 165 to 187%. 150% accounts for asymetrical current (rms). 1025% of that amount is added as a safety margin. Or they suggest setting relay so it will trip on starting. Slowly raise until three to five starts are made without tripping, then add another 10% margin. The 150% is fairly close to the 155% from C37.41 Annex C, which allows for decay in the first half cycle. Mr. Elmore says 2X LRC for 1500 HP and above with no other explaination. 

jbartos (Electrical) 
30 Jul 01 18:04 
Suggestion: If you happen to know the motor thermal curve, the setting can be 1.8 x LRA, for example, for a threephase induction motor; however, the motor protective device characteristics shall be carefully scrutinized to be under the motor thermal characteristics with some design margin left. After all, it is the "Science and Art of the Protective Relaying" or viceversa. Obviously, the lower multiple of LRA is used, without any nuisance motor tripping, the better for the motor. 

electricpete, I am curious about the references that you mention. Personally, I can't think of any good reason to size the instantaneous trip of a motor contol circuit for a value higher than the known LRA of a motor except to provide a margin for variation in other factors (ie. voltage, load, etc.). I would think that a good margin would be no more than 2540%, the less the better. This is, of course, assuming that you know the actual values for motor LRA and CB inst. trip.
I say this keeping in mind, for example, that some magnetic trip elements in molded case circuit breakers have fixed settings whose accuracy for calibration testing is only rated at perhaps +15%/20% of the setting. Specifically, if a setting of 1000A resulted in a trip anywhere between 800A and 1150A, it would be considered within specification.
Anyway, before I prove my ignorance, please let me know more what you are thinking. 

Thanks for all the good answer.
rhatcher  there is a decaying dc offset which is over and above the sinusoidal locked rotor current. This depends on the phase angle at which the bkr is closed.


Stevenal  don't electromechanical plungertype relays respond to the instantaneous value? They are not concerned with true rms, only with whether peak magnetic force exceeds the spring?
I guess the response may be slowed/filtered by the mechanical elements... but there is no intentional damping in an instantaneous element (by definition). 

A quick easy exercize to see the decaying dc offset.
Model the starting motor at an inductance Llrc
Voltage v(t) Vm*sin(2*Pi*f*tau)
Close the breaker at time t0 with initial current i0=0.
i(t)=1/Llrc*int(v(tau)dtau) i(t)=(1/Llrc)*int(Vm*sin(2*Pi*f*tau),tau=tclose..t) = [1/2/Llrc*cos(2*Pi*f*tau)/Pi/f*Vm](tau = tclose..t) = 1/2/Llrc*cos(2*t*Pi*f)/Pi/f*Vm +1/2/Llrc*cos(2*tclose*Pi*f)/Pi/f*Vm
The first term is sinusoidal with expected max magnitude of Vm/(2*Pi*f*Llrdc).
The second term is "dc offset" which takes on a max value of Vm/(2*Pi*f*Llrdc) if we choose tclose=0.
In realworld scenario there is a resistance which will cause the dc offset to decay away as exp(t*R/L).
Although it's not immediately obvious, it is not too hard to show that the neutral voltage of ungrounded threephase motor remains zero during starting if we assume all elements are linear (no saturation). Therefore singlephase analysis applies to starting of ungrounded wye motor. (also can be shown for delta).
I have an actual plot of a motor start captured with a digital storage oscilloscope. It behaves very much as predicted (sinusoidal plus decaying dc), except that there is a kind of "glitch" in the current approx 90 degrees after each peak. It seems this might be related to saturation... but if so, then why doesn't it occur at the peak of the current (especially if the motor is acting like an inductive load => load current should be in phase with excitation current which should be in phase with flux). If I get a chance I'll try to figure out how to post it so you can see what I'm talking about.


electricpete,
I understand that there will be a DC offset (asymmetrical current) of the same type encountered during a short circuit. Had I noted Stevenal's post of 7/30 before posting on the same day, I would have seen the answer I was seeking in terms of a reference and a reasoning behind it.
I am still wondering whether this guideline applies to all classes of short circuit protection. For example, a current limiting fuse may act within the first cycle and I will accept that some vary fast acting relays may respond as quickly. But, I would think that most devices would take 35 cycles to respond and that most DC offset would have decayed by that time.
As well, I am trying to decide in my mind what classes of devices would see the DC offset. For example, a fuse or a peak sensing solid state trip (as found on LVCB's) would certainly see the current based on magnitude alone. But I would think that a true RMS sensing solid state device (newer LVCB trips) would not sense the DC.
Anyway, can you shed some light on these questions? 

Pete,
It takes energy to move that plunger. We can talk about instantaneous v, i, and p; but instantaneous energy is always zero. Make the plunger and spring light enough, and little energy is needed. I still have not heard of a relay faster than a half cycle. Anyone got some specs that say otherwise? Seems silly to try to shave fractions of a cycle off the relay end, then hook the thing up to a three cycle or more breaker.
A quarter cycle or faster electromechanical might tend to wear quickly, moving back and forth for each half cycle of ordinary steady state operation. 

Pete, Now I'm curious, so I answered my own question. See http://www.abbus.com/abbrelays/pdf/ILsIBs/Cs41/417661l.pdf for a plunger style instantaneous relay. Timing curve is on page 5. Looks like 1/4 cycle for the SC at the high current end, 2.8 cycles at the low end. Maybe they've avoided the wear problem by lengthening the time at the low end. You could crowd the LRA with this one, because you'd have 2.8 cycles of DC decay. Bye. 

jbartos (Electrical) 
2 Aug 01 18:17 
Suggestion: Consider the total current It where there is a dc offset, Idc, and Iac=Irms symmetrical. It = sqrt( Idc**2 + Iac**2) This is closely related the the switchgear Close & Latch current. Further the protective device is from the motor (or generator) the smaller the DC offset is. In some cases, it is neglected. This is covered in the ANSI/IEEE C37 Switchgear series. 

I see the following words from IEEE37.96 section 7.2.10.3:
[regarding instantaneous overcurrent relay for motor protection]
"The inrush curent value is multipled by a factor to account for assymetrical current value that may be obtained. .... The factor varies up to a maximum of 1.73"
In my mind this confirms that the 1.73 is exactly the sqrt(3) and was derived in the manner shown in my first message. (not an empirical number or sum of margins for various factors). But we're still left with all the questions about exactly what the various types of instantaneous protective devices might respond to.
I do know in the case of an one electronic 50/51 relay, the vendor told me it would respond to the peak value of the waveform including ac plus dc. I'm not so sure about other types of relays.


jbartos (Electrical) 
20 Aug 01 7:53 
Suggestion to the previous posting: See IEEE Std C37.0131997 "IEEE Standard for AC HighVoltage Generator Circuit Breakers Rated on a Symmetrical Current Basis" paragraph 7.3.5.3.6 "Required Closing, Latching, and Carrying Capability where: Ipeak/Isym=sqrt2 x [e**(t/133) + 1] = 2.74 with t approximately 1/2 cycle in milliseconds for 50 Hz or 60Hz respectively. It is better to present more mathematical derivations, rather than your mind. 

jbartos – I welcome comments that challenge my statements or conclusions. That’s a good way for me to learn. But I hope we can keep the tone professional.
I honestly don’t understand your most recent post. I have a series of statements/ questions aimed at exploring why we’re looking at this differently.
I don’t understand why a discussion of circuit breaker ratings is relevant to a motor instantaneous setting thumbrule expressed as a percentage of locked rotor amps. (Of course the setting must be within circuit breaker ratings, but that consideration is evaluated separately, and that calculation would not involve any multiple of locked rotor current.)
I do see that the equation you have provided attempts to describe the value of an ac component plus decaying dc component at a point in time. That is the same form that I have described above. I agree that ½ cycle after circuit breaker closing is the relevant time where the higest instantneous value of current magnitude will occur. If I plug 8.33 in for "t", I get 2.74 (as you say). What does that have to do with the basis for the “1.73” in IEEE37.96 section 7.2.10.3 as quoted in my 8/18/2001 message?
I believe that the basis for 1.73 in IEEE37.96 section 7.2.10.3 is rms value of the offset waveform as described above and NOT the result of a decaying dc calculation for the following reasons: #1 – It is relatively unlikely that any calculation of decaying dc offset would come out to be almost EXACTLLY sqrt(3) (to three signficant figures as given in IEEE37.96) #2 – The quoted section above states that “the factor varies up to a MAXIMUM of 1.73”. Since the R/L is unknown, the maximum value is arrived by assuming no decay.
I would like to repeat my proposed explanation of where the 1.73 is coming from. Hopefully I can make this explanation a little more complete and understandble.
PROBLEM: Calculate maximum total rms current due to starting a motor of known locked rotor current with rms value I_lrc (under nominal voltage conditions). Source voltage linetoneutral is at nominal v(t) = Vm*sin(2*Pi*f*t). For simplicity ignore source impedance.
SOLUTION: To find MAX current, assume the load is 100% inductive (resulting in a dc component which does not decay). Model this with an equivalent wyeconnected inductance L_lrc = Vm / [(sqrt2*I_lrc)*2piF)]. For this balanced system, only one phase needs to be analysed.
i(t)=1/Llrc*int(v(tau)dtau) i(t)=(1/Llrc)*int(Vm*sin(2piF*tau),tau=tclose..t) = [Vm/(2piF*Llrc) * cos(2piF*tau)](tau = tclose..t) = Vm/(2piF*Llrc) * cos(2piF*t) +Vm/(2piF*Llrc) * cos(2piF*tclose)
The above is a general for for closing at any time t_close with current = 0 at time t_close. Highest current occurs when tclose=0 and is calculated as
i(t)= Vm/(2piF*Llrc)*cos(2*t*Pi*f) +Vm/(2piF*Llrc) = sqrt(2)*Ilrc*cos(2*t*Pi*f) +sqrt(2)*Ilrc
Inspection of this indicates it is simply the normal sinusoidal locked rotor current of PLUS an dc offset equal to peak value of the normal sinusoidal locked rotor current.… so it goes from zero to twice the normal locked rotor current peak.
Find the rms content of the above waveform. Since the dc and ac components are orthagonal, the total rms content is the square root of sum of squares of the rms values of these individual orthagonal components:
I_total_rms=sqrt([Idc_rms]^2 + [Iac_rms]^2) = sqrt( [sqrt(2)*Ilrc]^2 + [Ilrc]^2 ) = sqrt (3* Ilrc^2) = sqrt(3)*Ilrc (wher Ilrc is an rms value).
If the total rms calculation at the end is a source of confusion, I can provide alternate proof of that part using the definition of rms = sqrt ( 1/T*integral (I(t)^2)
If there is a flaw in my logic or part that doesn’t make sense, I’d like to hear it. If there is any other proposed explnation for using 1.73, I’d like to hear that too.


It's worth mentioning that I ignored any saturation effects in that motor starting analysis. I think that saturation does occur due to the transient, but it’s a necessary assumption.
I think the discussion of singlephase analysis as applied to motor starting transient is worth discussing…. . .
We take it for granted that we can apply singlephase analysis on a balanced 3phase balanced system in STEADYSTATE. But what about during transients?
The answer is that we can apply singlephase analysis to motor starting transient IF voltages are balanced and IF we again ignore saturation. This is based on the fact that the neutral voltage of an ungrounedwye will remain zero.
Here’s a quick poof by Laplace transform method: Assume motor connected ungrounded wye. Each leg has impedance R+jwL Apply voltage v1(t), v2(t), v3(t) at t=0. (Laplace transforms of these voltages are V1, V2, V3)All currents and voltages are zero prior to t=0.
The sum of the currents at the wye neutral point must be zero
(V1VN)/(R+sL) + (V2VN)/(R+sL) + (V3VN)/(R+sL)= 0
multiply by (R+sL)….
V1 +V2 + V3 + 3*Vn=0
Note that V12+V2+V3 is sum of a balanced set of time functions v1,v2,v3 whose sum must be zero.
Vn=0


jbartos (Electrical) 
20 Aug 01 18:44 
Suggestion to the previous posting marked by ///\\\: I do know in the case of an one electronic 50/51 relay, the vendor told me it would respond to the peak value of the waveform including ac plus dc. I'm not so sure about other types of relays. ///50/51 relays are traditionally associated with switchgear. I addressed this item in my previous posting. May I suggest that you brush up on switchgear before you go to motors since a lot of damage could be done to them.\\\ 

jbartos  Once again you come back to switchgear. I say there are a number of aspects which go into the selection of motor instantaneous trip. The aspect that gives rise to the 1.73*LRA thumbrule is that the motor should not trip during a normal start. This particular aspect is not dependent on circuit breaker ratings in any way since the current drawn during a normal motor start is independent of the switchgear.


jbartos (Electrical) 
21 Aug 01 13:34 
Suggestion: It appears that the induction motor acceleration curves have not been addressed yet. To avoid any nuisance tripping, the motor protective device characteristics should not cross this curve, at least not much. 



