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MCCB Instantaneous setting calibration 2

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EddyWirbelstrom

Electrical
Feb 17, 2002
214
Does the magnetic element of a thermal-magnetic MCCB, or the Instantaneous element of MCCB electronic trip-unit respond to the rms asymmetrical short-circuit current or the peak short-circuit current ?
ie. Is the MCCB trip setting calibrated for the rms asymmetrical short-circuit current or the peak short-circuit current ?
 
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In my understanding, TCC are given for RMS values. In a short circuit, RMS current will be the result for both the symmetrical AC current and the DC current.
 
My understanding is that the instantaneous trip responds to peak current but the setting is calibrated in RMS current. Generally a ratio of full load current. 500% to 1000% of RMS

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Thank you YIGO and waross.
Attached excel sheet calculations determine required MCCB instantaneous setting to avoid trip on motor DOL ( Direct-On-Line ) start given motor LRC as multiple of FLC and motor starting power factor ( at instant of motor DOL start ). The spreadsheet protection does not have a password.
Although for a motor with LRA / FLA of 6 and a starting power factor of 0.2, the attached spreadsheet calculates an MCCB ( or Motor Circuit Protector ) instantaneous setting of 9.4 x FLA.
(Assuming the instantaneous element has a -20% pickup current tolerance)
In practice I have found that instantaneous settings of 12.5 x FLA and below will result in an instantaneous trip on motor DOL start. To prevent tripping I am using an instantaneous setting of 15 x FLA. Why is this so ?
 
 http://files.engineering.com/getfile.aspx?folder=2d83679b-8c35-49a0-aa07-2edef47ab8b1&file=MCCB_Instantaneous_setting_calibration_01.xlsx
My thought is that this is an unknown.

Standards require calibration of instantaneous circuit breakers with a sinusoidal current. (Obviouisly we characterize that sinusoid by an rms.). But they don't specify how the breaker should respond to a non-sinusoidal current to my knowledge. (other than vague definition "no intentional delay").

If it is an electromechanical trip device, there are dynamics to consider. We might assume a certain instantaneous em force required to overcome spring and initiate trip, in which case we would focus on peak instantaneous current. But I recall some of the breakers (cutler hammer hmcp) also incorporate a mechanical damping element whose purpose is to help it ride through the first quarter cycle of a normal motor start. While that might be interpreted as conflicting with "no intentional delay" it is in fact used and it introduces a dynamic behavior that cannot be characterized simply as responding to an instantaneous value imo (it also depends on time history).

So I tend to think most approaches for setting instantaneous trip have evolved as experience/empirical approaches rather than theoretical. NEC guidelines are multiples of FLC seems to ignore change in kva code. After energy efficient motors came out, they added different requirements for those in recognition of observed behavior even though the difference between energy efficient and older motors is not captured in the kva code (it has more to do with the dc decay constant to my understanding). NEC has also included requirements that allow increasing the setting (within certain limits) if trips occur which again reflects an approach which is to a certain extent empirical or experimental.



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(2B)+(2B)' ?
 
From the Canadian Electrical Code.
Does the NEC have any similar provisions?
"Instantaneous-trip circuit breakers (see Appendix B)
When used for branch circuit protection, instantaneous-trip circuit breakers shall be
(a) part of a combination motor starter or controller that also provides overload protection; and either
Instantaneous-trip circuit breakers (see Appendix B)
(b) rated or adjusted, for an ac motor, to trip at not more than 1300% of the motor full load current or at not
more than 215% of the motor locked rotor current
, where given, except that ratings or settings for trip
currents need not be less than 15 A; or"

From Appendix B
"The intent of this Subrule is to allow an increase in the trip setting above 1300% for motors with high locked
rotor currents that will trip on the asymmetrical inrush at the 1300% rating.[/b] For example, a motor with 800%
locked rotor current would result in a trip setting of up to 1720% of full load current. Higher locked rotor currents
are common in energy-efficient motors.The intent of this Subrule is to allow an increase in the trip setting above 1300% for motors with high locked
rotor currents that will trip on the asymmetrical inrush at the 1300% rating. For example, a motor with 800%
locked rotor current would result in a trip setting of up to 1720% of full load current. Higher locked rotor currents
are common in energy-efficient motors."

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
That would be for an instantaneous (magnetic) only breaker. Generally different rules for breakers that also include the thermal element.
 
davidbeach said:
That would be for an instantaneous (magnetic) only breaker. Generally different rules for breakers that also include the thermal element.
The magnetic (Instantaneous Trip or IT) elements are the same either way though if the breaker frame is the same, but the IT settings on any given breaker are never more than 10x the frame size. What using the different rules for a T-M breaker does for you is to allow a larger frame size, which will then have a higher setting available to use.

One reason it is difficult to pre-calculate the IT settings is that there are numerous factors that go into the instantaneous current. One of those is the magnetizing current at the instant the motor is connected. Studies have shown that can be as high as 2200% of the motor nameplate FLC. This is because in the very first instant the windings are connected, there is no inductance yet, only the resistance of the magnet wire in the windings. So the current will rise at the available fault current level, minus the conductor resistance. Depending on when in the sine wave that connection is made, that can last upward of 1/2 cycle, long enough to make a difference in the total inrush value seen by the magnetic elements. That's why I generally just use whatever works without nuisance tripping, up to (but not over) the legal limits. I don't sweat the details as to why it has to be that high, it's just too variable.

When people started experiencing this first hand is when motor manufacturers were forced to start increasing efficiency. They did so of course by reducing losses and one major way to accomplish that was to reduce the I2R losses inside of the stator by reducing the winding resistance. So when "Energy Efficient" motors first started hitting the market in the mid 1980s, those used in replacing older motors began blowing fuses and tripping breakers on instantaneous even though the previous older motors had never given them any problems. Here in the US, the NEC had to add another exception to the IT settings for motor circuits, allowing for up to 1700% trip settings instead of the previous 1300% limit if it can be shown that the lower settings cause nuisance tripping. On EE motors, it almost invariably does.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington
 
One reason it is difficult to pre-calculate the IT settings is that there are numerous factors that go into the instantaneous current. One of those is the magnetizing current at the instant the motor is connected. Studies have shown that can be as high as 2200% of the motor nameplate FLC. This is because in the very first instant the windings are connected, there is no inductance yet, only the resistance of the magnet wire in the windings. So the current will rise at the available fault current level, minus the conductor resistance.
I don't quite view it that way. The motor stator and rotor windings do resemble large inductor during start with suddenly applied voltage. This model predicts the dc offset fairly well. The supply impedance plays a role only to the extent the motor terminal voltage voltage droops during start. If it droops from 100% to 80% then the effective supply impedance is on the order of 25% of motor starting impedance (assumes both are roughly the same phase angle which is close to inductive).


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(2B)+(2B)' ?
 
From the Canadian Electrical Code.
Does the NEC have any similar provisions?
Thanks for your post. Canadian code you posted makes more sense to me. From my memory the NEC guidance for instantaneous setting of mcb's feeding motors is based entirely on full load amps (and efficiency classification), with no regard for starting current or kva code.

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(2B)+(2B)' ?
 
Thank you for all the very informative posts.
Before the advent of modern high-efficiency motors I set motor MCCB instantaneous pickup settings at 10 x FLC / 0.80 = 12.5 x FLC without any tripping on motor DOL start.
This corresponds to the Canadian code for inst setting of 1300% FLC.
For high-efficiency motors I have had to increase the MCCB instantaneous pickup setting to 15 x FLC.
This corresponds to the Canadian code for inst setting of 2.15 * LRC where LRC = 6.97 x FLC.
The instantaneous elements of the MCCBs I refer to have no intentional time delay and are to the IEC Standards IEC60947.2
This matter is discussed in the following link to the Schneider publication ‘Complementary technical information’ :

In future I will use the empirical value of 2.15 x LRC for motor MCCB instantaneous pickup settings.
 
It always ends up being an interesting discussion on this topic both at our plant and on the forum.

I nitpicked on jraef's theoretical explanation of inrush but that was silly because we both agree it's more empircal than theoretical. His discussion does bring up a point I glossed over that there are not only uncertaintities in predicting response of the instantaneous unit but also in predicting exact nature of the motor current. The critical time period is that first half cycle as he correctly said (not first quarter cycle that I referred to). I have seen a lot of traces and often there are slow wiggles in the dc offset not predicted from the simplest L/R model, perhaps better predicted by a dynamic model. I have not seen the peak inrush get above 2*sqrt2 times the locked rotor current when locked rotor current was adjusted for voltage or estimated based on looking at peak-to-peak distance in current trace a few cycles after start.

If you assumed the breaker instantaneous element responded to instantaneous then you would establish a (sinusoidal rms tested) setting with some margin (*) above 2*LRC.
Ical = 2*sqrt2*LRC*sin(2*pi*f*t+theta) (without the margin)
Istarting = sqrt2*LRC*(1+sin(theta).
Max(Ical)=Max(Istarting) = 2*sqrt2*LRC
where theta = 2*pi*t+phi, LRC is rms value of steady state locked rotor current

If you assumed the breaker instantaneous element responded to rms value of the waveform, then you would set it up for some margin above sqrt(3) times LRC ~ 1.7*LRC
Ical = sqrt(3)*sqrt2*LRC*sin(theta) (without the margin)
Istarting = sqrt2*LRC*(1+sin(theta).
rms(Ical) = sqrt3 * LRC
rms(Istarting) = rms{sqrt2*LRC*(1+sin(theta)} = sqrt2*LRC*sqrt{Mean{1+2*sin(theta)+2*sin^2(theta)}} =
…. = sqrt2*LRC*sqrt{1+0 + 1/2} = sqrt2*LRC*sqrt{3/2} = sqrt3*LRC
rms(Ical) = rms(Istarting)=sqrt3 * LRC

There are a lot of documents for larger motors mentioning factor about 1.7 ~sqrt3 for instantaneous trip so it does bring RMS to mind. One could make an argument that if the dynamics of the system make the average force (rather than instantaneous force) relevant, than rms would be relevant (because instantaneous force depends on i(t)^2 so average force depends on Irms in the same way that average power depends on rms).

* Let's talk about that margin. We neglected any decay during that first half cycle which gives some margin to avoiding trip although difficult to quantify. We neglected breaker setting tolerance which is not conservative with respect to avoiding trip. We should add on another 20% or so to account for variability in breaker behavior unless we specifically test/screen the breaker to be installed in the application and reject breakers which trip on the low end.

A design engineer at our plant recommended (as a corrective action for some instantaneous trips upon start after installing new motors) that we should gather traces of starting current waveforms for many of our small critical mcc-fed motors to verify we have margin to avoid instantaneous trip during start. I would have been the one assigned to coordinate the effort. They viewed it as a simple excercize but I knew better and didn't like the idea. I said I would only agree if we agreed ahead of time on what would be the criterion. That effectively killed the effort very quickly because we could not agree on an approach. The design guys were looking to keep the setpoints low (I think they had code in mind… I don't think we have any problems maintaining selective trippign with upstream devices) while I as an engineer directly supporting maintenance/operations was looking to keep the setpoints high. My approach was something like three starts (in attempt to get close to worst case closing angle) plus the instantaneous-responding assumption plus allowance for breaker variability.


=====================================
(2B)+(2B)' ?
 
How does this sound, Pete?
The worst case is full DC offset. At that point, the current is basically limited by the winding effective resistance.
How about measuring the winding effective resistance and using that to calculate the worst case peak current?
First half cycle versus first quarter cycle. I understood that the current peak was reached at the end of the first quarter cycle.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Worst case instaneous peak (i.e. with full dc offset) is 2*sqrt2* LRCrms.
It depends primarily on inductance: LRCrms ~ Vterminal/(X1+X2).
stator winding resistance (along with rotor resistance and Xm parallel to rotor branch) introduce a small deviation from this approximation LRCrms ~ Vterminal/(X1+X2)

The worst case peak occurs approximately half cycle after closing. Consider closing a voltage V = Vmax*sin(2*pi*f*t) onto a pure inductor at t=0, V=0 (and I=0 initial ocndition). Inductor current is the integral of voltage and continues to increase from initial value for as long as voltage remains positive. The sin wave remains positive for a half cycle, so the worst case current peak is reached approx half cycle later.

My two paragraphs above used a mental model (that motor can be represented as an impedance for purposes of predicting worst case instantaneous peak) to prove the behavior predicted by the model... which might be viwed as slightly circular logic. I can say the results are consistent with accepted behavior (for example NEMA MG1 mentions worst case instantaneous peak current 2*sqrt2*LRCrms, I'm pretty sure references about motor starting transient will mention half cycle. The model is basically the steady state induction motor equivalent circuit with suddenly applied voltage. I have done transient simulation of worst case suddenly applied voltage using Krause induction motor transient model and the steady state equivalent circuit model (which is a misapplication of the steady state equivalent circuit model). Surprisingly (for me anyway), results of this misapplication of the steady state model match the transient model pretty well over the first few cycles. I have viewed many measured starting waveforms and they tend to agree with results predicted when accounting for closing angle... particularly highest magnitude peak (one of the three usually comes close but not over that prediction) and time to the higest peak (the highest of three current peaks is usually latest in time and gets close to half cycle after close).

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(2B)+(2B)' ?
 
Most of the sample wave forms I have seen were current based rather than voltage based.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Most of the sample wave forms I have seen were current based rather than voltage based.
Bill - Please explain what you are talking about. I have no idea what point you are trying to make and how it relates to my post.

I did mention voltage because I modeled the motor as an inductor. Worst case (highest) current of an inductor closing onto single phase sinusoidal voltage occurs if you close in at zero crossing of the applied voltage. Worst case (highest) current of a given phase of an induction motor occurs if you close in at the instant that the associated line to neutral voltage would be zero.

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(2B)+(2B)' ?
 
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