Circuit breaker tripping during motor start

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mull982

Senior Member
I was having an conversation with a co-worker regarding the tripping of an instantaneous motor circuit breaekr during the start of a loaded conveyor. This conveyor has started fine in the past but recently is tripping the motor breaker during starting. He has told me that the reason the circuit breaker is tripping is because the conveyor is starting when loaded and is drawing too much current for the breaker.

I told him that even if the conveyor is loaded it should not trip the breaker but should trip the overloads. My reasoning is that a motor has a LRC when starting, and this LRC is the highest current a motor will ever draw no matter how much its loaded. Its basically the max current the motor will draw with the rotor locked down. The motor draws this LRC everytime the motor starts and is commonly referred to as starting current.

Because the motor breaker has to be set above this LRC in order to allow the motor to start, then the circuit breaker should never trip even with the belt overloaded because even if it was overloaded enough to lock the rotor the circuit breaker should not trip since the breaker should be able to handle the current with the rotor locked.

Since the breaker is an instantaneous setting, even if this LRC persisted for some additional amount of time it would not matter as long as it was below the setting on the breaker. The overloads should be what trips due to this additional current because the overload current is a function of time. Is my explanation correct?
 

jim dungar

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I told him that even if the conveyor is loaded it should not trip the breaker but should trip the overloads.
This would only be the case if the two devices were coordinated. There is no NEC requirement for this coordination.
 

don_resqcapt19

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Starting the conveyor loaded will increase the time that the motor draws the LRC. It could be that while the LRC is not in the instantaneous trip range of the beaker, if you draw that current for a longer time you may have moved out of the instantaneous trip range and into the short time trip range.
 
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The overloads will not respond as quickly as the circ breaker on excessive starting currents or locked rotor. Have the loads on the conveyor changed? Amount of product? Bearings? Binding?

Megger the motor and wires. Check the connections.

Then again breakers do fail, but that would be my last choice.
 
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glene77is

Senior Member
Location
Memphis, TN
Jim,

My perspective is:
(1) Fuses / Inst OCPD are for handling faults, Max current, fast acting.
(2) InvTime OCPD are for handling start currents, Locked Roter Current, thermally.
(3) OverLoads are for handling long term FLA, show response, set close to FLA.
Using this list, I check for Name-Plate Horse-power, LRC, and FLA.
Then I refer to 430 to check myself.

How do you see this little analysis?
 

mull982

Senior Member
Please list the motor size, voltage, FLA, LRA and breaker size.

I will be able to post these details tomorrow

Starting the conveyor loaded will increase the time that the motor draws the LRC. It could be that while the LRC is not in the instantaneous trip range of the beaker, if you draw that current for a longer time you may have moved out of the instantaneous trip range and into the short time trip range.

I understand what you are saying however the breaker is an instantaneous magnetic only breaker therfore it only has an instantaneous pickup and therefore no short time pickup. Therefore the way I see it, the LRC is either above or below the instantaneous setting and therefore either trips the breaker instantaneously or never trips it.

The overloads will not respond as quickly as the circ breaker on excessive starting currents or locked rotor. Have the loads on the conveyor changed? Amount of product? Bearings? Binding?

The way I see it, is the LRC is based on a constant impedance of the motor with the rotor locked. Therefore this is the maximum current the motor can draw. Even if the conveyor or motor is loaded more, this does not change the constant impedance of the motor when starting, and therefore does not increase the current, since this constant impedance is with the rotor locked.
 

ATSman

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All above comments are pertinent to the root cause of the tripping although what was not mentioned is the magnetic trip (instantaneous) setting of the circuit breaker. Is it set correctly? Over time this setting may have to be adjusted due to wear and tear of equipment. Westinghouse recommends the following procedure: Raise the inst setting to max then start the motor (loaded). Lower the setting one notch each time the motor starts until the breaker trips, then raise it one notch and verify it will not trip. This is the recommended trip setting.
If this checks out then if you have a Fluke 87 DMM and a clamp-on AM use the record mode and measure the inrush current of the motor. If this value is in the range of the breaker instantaneous range then the breaker can be defective. If so have the breaker tested with a high current test set to determine if the trip unit is tripping below its mag trip setting. If so, replace the trip unit or the breaker.
Other possibilities are bad motor bearings,worn drive belts.
 

shallo

Member
I do not think the breaker is supposed to be sized to the lrc. I think it should be sized to the motors flc times 250%. I'm not sure what kind of breaker it is but check the trip setting adjustment if it has it. Every time the breaker trips it gets weaker. It is possible it could just be weak. Having worked around conveyors some its been my experience that is usually the conveyor loaded beyond rating for start up or something mechanical. Make sure to check v belts to gear box. If they are loose and slipping it can cause motor to draw more current sometimes. I know how conveyors tend to stay in the industry awhile and get moved around and modified, For example I have seen a conveyor come new with a 7.5 hp motor and it needed to be up sized later and the breaker was never changed. Just a couple of ideas.
 

jim dungar

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PE (Retired) - Power Systems
Jim,

My perspective is:
(1) Fuses / Inst OCPD are for handling faults, Max current, fast acting.
(2) InvTime OCPD are for handling start currents, Locked Roter Current, thermally.
(3) OverLoads are for handling long term FLA, show response, set close to FLA.
Using this list, I check for Name-Plate Horse-power, LRC, and FLA.
Then I refer to 430 to check myself.

How do you see this little analysis?
I would have no problem with this, although I see little difference between 1 & 2 for most applications 10HP @480V and smaller.
 

jim dungar

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Every time the breaker trips it gets weaker. It is possible it could just be weak.

For normal overload operation (below full rated fault currents) breakers do not get weaker each time they trip. The NEMA standard for mechanical and electrical full load operations of breakers is in the thousands of operations for breakers smaller than 225A.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090105-0923 EST

mull982:

Your post #9 seems quite correct. And tonytonon's suggestion of measuring current should be your first step. Check the current on each line. You might even need to use some sort of recorder and simultaneously monitor all three lines.

Correlating the starting current measurements with what is expected should route you to the motor or the breaker as the problem. Maybe shorting conductors to the motor are a possibility.

It is clearly not a problem from bearings, belts, or load on the motor because you are correct that maximum starting current could exist forever and never trip a properly sized and correctly working instantaneous breaker.

.
 

mull982

Senior Member
Please list the motor size, voltage, FLA, LRA and breaker size.

Motor is a 50hp 480V motor. The FLA for the motor is 57.7 and the corrosponding LRC for the motor is 352A.

The breaker used is an Siemens type HEM instantaneous only type breaker with a current rating of 100A. This 100A rating corrospongs to adjustable settings of 300A - 1100A. The adjustable settings are as follows:

A - 300A
B - 500A
C - 700A
D - 900A
E - 1000A
F - 1100A

The breaker has since been cranked all the way up and has not seemed to cause any problems.

090105-0923 EST
It is clearly not a problem from bearings, belts, or load on the motor because you are correct that maximum starting current could exist forever and never trip a properly sized and correctly working instantaneous breaker.

Agreed. My initial point that I was trying to show my co-worker was that since the motor had a LRC of 352A, this motor would never draw above 352A no matter how much it was loaded (unless of course there was a fault). Even if this motor was loaded with infinite additional load, this motor would never draw more than 352A.

With that being said, theoritically a circuit breaker setting of "B" corrosponds to 500A and is above the 352A LRC. Therefore with a setting of 500A this breaker should never trip due to any starting current or any associated load current. Technically with this 500A setting the conveyor could be loaded any given amount, and the breaker should not trip since its above 352A. Since this is an instantaneous only pickup this 352 could last any given amount of time or unitl the motor burned up.

The electricians have since turned the breaker setting all the way up which I have instructed them is not the ideal solution, and am trying to explain to them why a setting of "B" should work. The problem has seemed to has ceased at least for now but I'm still intereste in providing them with the correct answer.

I'm not sure other then a breaker going bad what would have caused this change. Is it possible that during previous starts the motor never reached LRC and now with additional load it does? Does a motor always draw this max LRC no matter how much it is loaded? It was my assumption that it does.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
090105-1215 EST

mull982:

If there was a shorted winding in the motor it might, probably would, cause a higher starting current and running current.

Assume that this is not the case and there are not other shunt currents, then:

If the breaker is truely an instantaneous breaker, on a half cycle basis ( for extremely short times there is no breaker that is instantaneous -- I would conjecture that a 1 microsec 1000 A pulse would not trip your breaker in the 300 A range.), then for at least 1/2 cycle (8.3 MS), for your size motor, there will be no difference in the starting current for 0 to full load on the conveyor.

The magnetic saturation state of the motor at turn-off will have some effect on initial current at turn-on. This is a random function from one cycle to the next. Probably not a significant factor and has nothing to do with failure now compared to no tripping previously.

You really need to instrument the motor to make sure the motor's currents are what you expect. If the motor temperature and currents are normal, then you are probably looking toward the breaker for an answer.

It would make sense to move down to the 500 A setting of the breaker as a trial, but do not avoid doing the instrumentation.

.
 

lx2077

Member
does the belt work if it is not loaded...completly empty...also megg the windings the motor fields may be open and in this case it ma be cheaper to replace the motor than have it rewound and fixed,either way keep us informed
 

mull982

Senior Member
If there was a shorted winding in the motor it might, probably would, cause a higher starting current and running current.

.

This makes sense. I guess a shorted winiding would cause a higher starting current.


If the breaker is truely an instantaneous breaker, on a half cycle basis ( for extremely short times there is no breaker that is instantaneous -- I would conjecture that a 1 microsec 1000 A pulse would not trip your breaker in the 300 A range.), then for at least 1/2 cycle (8.3 MS), for your size motor, there will be no difference in the starting current for 0 to full load on the conveyor.
.

So what you are stating here is confirming what I suspected in regards to the motor drawing the same starting current independent of the load and with normal conditions it will never be greater than 352A. Is it always at or around this LRC value or can it be under this value.

The magnetic saturation state of the motor at turn-off will have some effect on initial current at turn-on. This is a random function from one cycle to the next. Probably not a significant factor and has nothing to do with failure now compared to no tripping previously.
.

I understand that there can be some residual magnetism left in the stator depending on where on the waveform the motor is turned off (zero crossing voltage being wort case). This residual magnetisim would cause the inrush transient portion of the starting current (current lasing for about .1sec) to be greater and like you mentioned be independent on a case by case basis.

You really need to instrument the motor to make sure the motor's currents are what you expect. If the motor temperature and currents are normal, then you are probably looking toward the breaker for an answer.

I'm assuming here you are referring to steady state currents and temp. If these are normal during steady state running then you would suspect starting currents should be normal as well. If not normal during steady state then most likely there will be an issue with starting as well.
 

jim dungar

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Location
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PE (Retired) - Power Systems
I understand that there can be some residual magnetism left in the stator depending on where on the waveform the motor is turned off (zero crossing voltage being wort case). This residual magnetisim would cause the inrush transient portion of the starting current (current lasing for about .1sec) to be greater and like you mentioned be independent on a case by case basis.
The time that residual magnetism exists is measured in cycles. It is a concern only when plugging and jogging a motor. Once an AC induction motor comes to rest residual magnetism is absolutely not a possibility.
 
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