ptonsparky
Tom
- Occupation
- EC - retired
What was the original setting of the breaker? If it was "A", the motor may not have reached peak long enough for the breaker to trip, until the load changed, be it bearings, windings or just to dang cold.
Circuit breaker tripping during motor start
- Thread starter mull982
- Start date
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I see what you are saying. I guess I was getting this confused with a transformer core still having residual magnetism when turing off a transformer at zero voltage due to current being at peak at this time and
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
A transformer core is the same as a motor, after several cycles the residual magnetism is gone. Unless the transformer is "immediately" re-energized it is not a factor.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090105-1559
Residual magnetism is a function of where in the hysteresis loop you turn off excitation. This residual remains until something causes demagnetization. Over short times (years) this does not change without some external excitation, such as: heat, mechanical, electrical, magnetic, etc. I have magnetic tapes from the 50s that still have good retention of the recorded material.
When new electrical excitation is applied to the magnetic circuit so that it forces the flux further up the hysteresis curve, then the magnetic circuit is driven more into saturation requiring more current. If the opposite polarity had been applied then the flux would move away from saturation, and current would diminish.
The rebalancing of the hysteresis curve occurs within a couple cycles. Both transformers and motors using ferromagnetic materials exhibit this characteristic. However, motors have a much bigger air gap and the effect is less significant.
In a transformer it is not unlikely to have a peak inrush current of maybe 10 times normal peak steady-state current resulting from this excessive saturation on some initial cycles. This can be largely eliminated by using an SCR switch for turn-off.
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Residual magnetism is a function of where in the hysteresis loop you turn off excitation. This residual remains until something causes demagnetization. Over short times (years) this does not change without some external excitation, such as: heat, mechanical, electrical, magnetic, etc. I have magnetic tapes from the 50s that still have good retention of the recorded material.
When new electrical excitation is applied to the magnetic circuit so that it forces the flux further up the hysteresis curve, then the magnetic circuit is driven more into saturation requiring more current. If the opposite polarity had been applied then the flux would move away from saturation, and current would diminish.
The rebalancing of the hysteresis curve occurs within a couple cycles. Both transformers and motors using ferromagnetic materials exhibit this characteristic. However, motors have a much bigger air gap and the effect is less significant.
In a transformer it is not unlikely to have a peak inrush current of maybe 10 times normal peak steady-state current resulting from this excessive saturation on some initial cycles. This can be largely eliminated by using an SCR switch for turn-off.
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I would think that the same applies for bogging down a conveyor with additional load when the conveyor is running. With the conveyor running under steady state even if a load is added that bogs down the coveyor or stalls the motor, it should not trip the breaker but rather the overloads. Even with the motor running loaded under steady state an infininte load can be added and the motor will never draw more than 352A thus not tripping the breaker but the overlaods instead.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090105-1721 EST
mull982:
On page 209 of "Alternating -Current Machinery", by Bailey and Gault, 1951, McGraw-Hill is a plot of a motor built with a high resistance and low resistance rotor. The only design difference is rotor resistance. The plot is of torque and line current on a graph of speed vs line current.
The current curves start at near 0 for no load and monotonically increase in current as speed decreases. This is true of both the high and low resistance rotors. The torque curve of the high resistance rotor motor is also monotonic. Not so of the low resistance rotor. The low resistance rotor torque folds back after reaching a peak at about 1200 RPM. Synchronous speed is 1800.
Monotonic means the slope of the curve never reverses. y = A + Kx (a straight line) is monotonic, dy/dx = K, a constant slope. x^2 + y^2 = R^2, a circle of radius R centered at x =0, y = 0, is non-monotonic. Any conic section (circle, parabola, ellipse, etc.) is non-monotonic. If you limit the range of a non-monotonic function then there maybe be regions that are monotonic.
Unless plugging, whatever you do with the load the maximum line current is at 0 RPM.
If you do not have quantitative data on the motor line currents, then you are just guessing.
Yes to your last post.
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mull982:
On page 209 of "Alternating -Current Machinery", by Bailey and Gault, 1951, McGraw-Hill is a plot of a motor built with a high resistance and low resistance rotor. The only design difference is rotor resistance. The plot is of torque and line current on a graph of speed vs line current.
The current curves start at near 0 for no load and monotonically increase in current as speed decreases. This is true of both the high and low resistance rotors. The torque curve of the high resistance rotor motor is also monotonic. Not so of the low resistance rotor. The low resistance rotor torque folds back after reaching a peak at about 1200 RPM. Synchronous speed is 1800.
Monotonic means the slope of the curve never reverses. y = A + Kx (a straight line) is monotonic, dy/dx = K, a constant slope. x^2 + y^2 = R^2, a circle of radius R centered at x =0, y = 0, is non-monotonic. Any conic section (circle, parabola, ellipse, etc.) is non-monotonic. If you limit the range of a non-monotonic function then there maybe be regions that are monotonic.
Unless plugging, whatever you do with the load the maximum line current is at 0 RPM.
If you do not have quantitative data on the motor line currents, then you are just guessing.
Yes to your last post.
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Last edited:
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Yes, some residual magnetism does exist in transformers.
According to Standard Handbook for Electrical Engineers 10th Edition, section 11-5:
If an idle transformer is energized at a time in the voltage cycle when the flux in the core would normally be other than the actual residual flux in the core, there will be an exponentially decreasing transient flux component in the core which has the effect of offsetting the initial flux wave. In extreme cases the peak flux may be more than doubled, exceeding saturation of the core, and causing peak magnetizing current several times rated load current. Magnetizing inrush is important principally because of the possibility of false operation of transformer protective relays.
I have not looked into it, but the decaying voltages of proportionately large motors connected to the transformer secondary should also be a factor in the amount of residual magnetism in the core after the primary is opened.
According to Standard Handbook for Electrical Engineers 10th Edition, section 11-5:
If an idle transformer is energized at a time in the voltage cycle when the flux in the core would normally be other than the actual residual flux in the core, there will be an exponentially decreasing transient flux component in the core which has the effect of offsetting the initial flux wave. In extreme cases the peak flux may be more than doubled, exceeding saturation of the core, and causing peak magnetizing current several times rated load current. Magnetizing inrush is important principally because of the possibility of false operation of transformer protective relays.
I have not looked into it, but the decaying voltages of proportionately large motors connected to the transformer secondary should also be a factor in the amount of residual magnetism in the core after the primary is opened.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090105-2148 EST
A good description of transformer magnetize current is presented at p 261-263 in "Electrical Circuits and Machinery", by Hehre and Harness, 1942, John Wiley.
A very well explained plot is shown in Fig 7-37 on page 262.
A discussion of inrush current is provided, but no plot.
On page 262 is a discussion on the meaning of power factor and a distorted current waveform.
"Examination of Fig. 7-37 will show that the term "power factor" or cosine of the angle between corresponding points of the impressed voltage and current waves is here meaningless. As represented, the zero points of the curves of voltage and current are 48 deg apart, while their maxima are 90 deg apart."
"--- If the power is divided by the product of the voltmeter and ammeter readings a value of power factor is obtained; but, for this to have any meaning, we must consider that the distorted wave of current has been replaced by an equivalent sine wave of current of equal effective value and proper phase."
Note: "effective value" means RMS.
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A good description of transformer magnetize current is presented at p 261-263 in "Electrical Circuits and Machinery", by Hehre and Harness, 1942, John Wiley.
A very well explained plot is shown in Fig 7-37 on page 262.
A discussion of inrush current is provided, but no plot.
On page 262 is a discussion on the meaning of power factor and a distorted current waveform.
"Examination of Fig. 7-37 will show that the term "power factor" or cosine of the angle between corresponding points of the impressed voltage and current waves is here meaningless. As represented, the zero points of the curves of voltage and current are 48 deg apart, while their maxima are 90 deg apart."
"--- If the power is divided by the product of the voltmeter and ammeter readings a value of power factor is obtained; but, for this to have any meaning, we must consider that the distorted wave of current has been replaced by an equivalent sine wave of current of equal effective value and proper phase."
Note: "effective value" means RMS.
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glene77is
Senior Member
- Location
- Memphis, TN
Jim,
Could Shallo be refering to the physical tripping mechanism, in which the lever slides across a small ledge?
In my curiosity, I have broken holes in the sides of CB, just to 'see'.
I have inspected CB where this ledge was not square edged, and the result was that a 20A CB would trip at 10A, constantly.
I have inspected CB where the points were badly pitted, again from the user making the CB into the motor control switch.
Those two possibilties (1) of wearing the square edge and (2) pitting the contacts, may be reasons for not using a OCPD as a switch. There are switch rated OCPD, which we use when the User requests the CB to be the controlling switch.
For one customer,
who insisted on using the CB as the controlling switch,
I noted both types of problems.
Comments are welcome.
Could Shallo be refering to the physical tripping mechanism, in which the lever slides across a small ledge?
In my curiosity, I have broken holes in the sides of CB, just to 'see'.
I have inspected CB where this ledge was not square edged, and the result was that a 20A CB would trip at 10A, constantly.
I have inspected CB where the points were badly pitted, again from the user making the CB into the motor control switch.
Those two possibilties (1) of wearing the square edge and (2) pitting the contacts, may be reasons for not using a OCPD as a switch. There are switch rated OCPD, which we use when the User requests the CB to be the controlling switch.
For one customer,
who insisted on using the CB as the controlling switch,
I noted both types of problems.
Comments are welcome.
Last edited:
Strahan
Senior Member
- Location
- Watsontown, PA
By increasing the setting on the magnetic trip unit you will only be delaying the real problem from showing itself I.M.O and from your posts I know you agree. We had a similiar situation at a much higher level and what it boiled down to was one of the lug plates in the magnetic trip unit itself had become loose. By increasing the setting which we did to keep production running caused the problem to appear. My suggestion would be to do some testing on the wiring to the motor and the motor itself (megger and so forth) if this all tests out fine attack the breaker FOP test and so forth connections ect...
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090107-0852 EST
glene77is:
I agree with your description.
The failure mode relative to the latch will be tripping at a lower threshold and this is not a safety problem.
The wear of the latch edges is a function of the material and hardness.
I use a number of QO breakers as switches that are routinely switched. Some of these probably at least once a day for the last 45 years. I have never needed to replace one of these. Is their trip point the same as original? I do not know, and it does not matter.
I outfit each of my benches with a main 2 pole breaker and following this another 2 pole for a 240 outlet, and 2 single pole units for two 120 circuits. Truly the primary purpose of these breakers is to provide a good switch that is convenient to install and use. Secondarily they are an overload device.
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glene77is:
I agree with your description.
The failure mode relative to the latch will be tripping at a lower threshold and this is not a safety problem.
The wear of the latch edges is a function of the material and hardness.
I use a number of QO breakers as switches that are routinely switched. Some of these probably at least once a day for the last 45 years. I have never needed to replace one of these. Is their trip point the same as original? I do not know, and it does not matter.
I outfit each of my benches with a main 2 pole breaker and following this another 2 pole for a 240 outlet, and 2 single pole units for two 120 circuits. Truly the primary purpose of these breakers is to provide a good switch that is convenient to install and use. Secondarily they are an overload device.
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- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
The NEMA design test is for thousands of mechanical operations, so I would not consider that all breakers inherently get weaker as they are used.
Yes, I know that a very few breakers do experience excessive wear on their latching mechanism, but they are part of the statistical group that says "99% accurate is close enough".
We continue to discuss this topic here and more and more scenarios come up.
One that got me thinking is what happens when a fan that is spinning backwards due to some draft caused elsewhere and then is given a start in the forward direction. Should this trip the breaker? The same question goes for any load spinning backwards.
My initial response it that it should not trip the breaker based on the same argument I've has all along in that the motor will never draw more than LRC so even if its spinning backwards it still will only draw LRC. The only thing I can see happening is that as the motor attemps to reverse the direction of the fan and turn it the right direction it will draw this LRC for a longer period of time. In this case it may either trip the overloads, or trip a time delay breaker but should not trip an instantaneous breaker due to what I have been saying about LRC.
Do you guys agree?
One that got me thinking is what happens when a fan that is spinning backwards due to some draft caused elsewhere and then is given a start in the forward direction. Should this trip the breaker? The same question goes for any load spinning backwards.
My initial response it that it should not trip the breaker based on the same argument I've has all along in that the motor will never draw more than LRC so even if its spinning backwards it still will only draw LRC. The only thing I can see happening is that as the motor attemps to reverse the direction of the fan and turn it the right direction it will draw this LRC for a longer period of time. In this case it may either trip the overloads, or trip a time delay breaker but should not trip an instantaneous breaker due to what I have been saying about LRC.
Do you guys agree?
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
I agree, only duration should change, not magnitude.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
LRC is not the only cause of tripping of instantaneous trips on CBs. That is why the rules in 430.52 have exceptions and allowances for when it is proven to NOT work when a breaker is set per the chart. LRC can be defined as the maximum STARTING current, but not the maximum INRUSH current. True inrush is indeed essentially the same as a transformer and that can cause breakers to trip magnetically. Newer energy efficient motors are becoming notorious for that in fact.
As to spinning backwards, my experience is that it depends. Some motor designs have higher residual magnetism than others and will in fact be generating a small amount of voltage when spinning on their own, such as a fan windmilling. I have measured it when troubleshooting repeated VFD failures and by implementing a braking cycle just prior to starting, we were able to eliminate the damage to the VFD transistors. So in theory, that out-of-phase power could indeed exacerbate any natural inrush current issues and cause higher than normal instantaneous currents when the motor is energized. I have seen it proven out by an engineer (who is much more willing than I to crunch numbers) as potentially being as much as 2000% of FLA if the contactor is closed in the right point of the phase angle.
The issue is essentially the same as what happens when you transition a Y-Delta open transition starter from Y to Delta; the re-application of power while the motor is still in regen can cause massive spikes that can damage equipment or pop fuses and trip CBs. It won't happen every time, but under the right circumstances it can be really really bad. Somewhere I have a paper written by Richard Nailen about this which shows the mathematical probability of it, if I can find it or a link to it, I will post it here later.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
090107-1407 EST
mull982:
If you reverse drive a motor then the current will be greater than locked rotor. This would be an extrapolation of the curve I mentioned in an earlier post:
On page 209 of "Alternating -Current Machinery", by Bailey and Gault, 1951, McGraw-Hill.
Our local engineering library, U of M, is one of about five that Google is scanning in its entirety. I believe 7 to 8 million volumes. Bailey and Gault has been scanned. In this case apparently from The University of Wisconsin, rather than Michigan. Other than for a few random parts of the book you might not find it useful. However, on the used market it is readily available. My original new cost was $6. One used book store on the Internet has it for $15. Should be in many engineering libraries.
However, using Google Books is not a useful experience for me. The snippets are too short, curves are not displayed when specifically selected. I can legally copy more with a Xerox machine than Google provides in their snippets. On the other hand very rare old books out of copyright are supposed to be fully available.
You can run your own line current test to some extent with maybe a 1 HP motor. Locked rotor is fairly easy. Next you could rotate forward, then immeadiately change to reverse. You might want a current recording instrument for the tests.
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mull982:
If you reverse drive a motor then the current will be greater than locked rotor. This would be an extrapolation of the curve I mentioned in an earlier post:
On page 209 of "Alternating -Current Machinery", by Bailey and Gault, 1951, McGraw-Hill.
Our local engineering library, U of M, is one of about five that Google is scanning in its entirety. I believe 7 to 8 million volumes. Bailey and Gault has been scanned. In this case apparently from The University of Wisconsin, rather than Michigan. Other than for a few random parts of the book you might not find it useful. However, on the used market it is readily available. My original new cost was $6. One used book store on the Internet has it for $15. Should be in many engineering libraries.
However, using Google Books is not a useful experience for me. The snippets are too short, curves are not displayed when specifically selected. I can legally copy more with a Xerox machine than Google provides in their snippets. On the other hand very rare old books out of copyright are supposed to be fully available.
You can run your own line current test to some extent with maybe a 1 HP motor. Locked rotor is fairly easy. Next you could rotate forward, then immeadiately change to reverse. You might want a current recording instrument for the tests.
.
davesparky
New member
Why would you want to run a conveyor off of a circuit breaker anyway? You need a motor starter that can take the large initial current surge when starting a conveyor.
Yes I understand that there is an inrush associated with motor starting, which is mostly reactive current used for establishing magnetic fields. However because this current is a reactive current it is independent of any loading which corrosponds to real current. Therefore I would expect this current to be a function of the magneict properties of the motor at the time of energizing as well as voltage waveform and not have anything to do with the load connected to the motor.
This inrush typically lasts for about .1s so if the breaker has a delay on the instaneous setting it might not ever pickup on it.
So basically you are saying that the only the greater inrush associated with a fan spinning backwards is mostly due to out of phase voltages as a result of re-gen and not really a load issue. In other words this large current transient is due to out of phase voltages, and not real current as a result of a torque trying to spin the fan in an opposite direction?
How much residual magnetisim exists in a motor? I always thought that because an induction motor needed an excitation field that when the source was removed after some given amount of time the field would no longer exist and therfore the motor could not produce any re-gen? You are saying that even after the motor being off for an hour there may be some residual magnetism in the motor that if the fan starts spinning it can start to re-gen a voltage?
Right I understand this effect, and I usually refer to it as plugging a motor. Basically the way I understand it is that when in Re-Gen the magnetic fields and re-gen voltage phasors are at particular phase angles, and if you supply a sourch voltage which is at a different phase angle then the resulting current magnigute will be very large as the different phase angles try to lock together. This current will be almost like a short circuit current and will trip most breaker settings?
The way I understand it though is that this is only the case when there is a strong resisual magnetic field still existing in the motor in order to produce re-gen voltage. How long does a magnetic field in a motor take to decay after de-energizing?
I would love to read that paper if you ever come across it.
Please see attached page from Siemens web re HEM circuit breaker settings. I guess your electricians will have to trust the manufacturer setting recommendations.
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