DC offset in induction motor inrush current 2

[rockman7892]

I am looking for an equation for explaining the DC offset that is associated with the inrush current when magnetizing an induction motor. The way I see it, the motor istelf would look like a series resistor and inductor circuit which would apply to the following equations:

Vsin(wt+alpha)= Ri + L di/dt where i is current.

Solving for I we get.

i = Vmax/Z [sin(wt+alpha-theta)-e^(-RT/L) sin(alpha-theta)

Is this a correct equation for showing the DC offset in the inrush current as a function of time t.

I know magnitude all depends on where on voltage waveform contacts are closed and residual flux remaining in the rotor. Would worst cas current be when closing with voltage at zero or at peak?

How do high effeciency motors impact the above equation resulting in typically higher inrush current?

 

[electricpete]

For an instantaneous magnetic breaker does the tripping magnitude depend simply on the peak of the transient current, or its RMS values? If peak do the peaks have to stay above certain values for more than one cycle? Current breaker settings are at 4000A. What kind of energy to these breakers need to see to trip?

That is the difficult I alluded to 12 Oct 10 15:02. Maybe someone can give you a good answer, but not me. I did quite a bit of research on that subject and came up fairly empty.

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[electricpete]

Some data points for consideration on your question of circuit breaker response (none conclusive).

Many references for larger motors (like the Buff Book I think) recommend a factor of 1.73*LRC for instantaneous setting. The basis is never described that I have seen, but it's fairly easy to backfit some logic. 1.73 = sqrt(3). If you take the worst case of a fully offset sinusoidal waveform, the rms is sqrt(3) times the rms of the same sinusoid with no offset. That suggest the rule was created by someone who thought the trip devices respond to rms.

The max instantaneous force in any plunger type element is proportional to max instantaneous current. There can be a spring and some inertia and in case of HMCP breaker a dashpot intended to alter the dynamics. I should back up and ask what kind of breaker you have... and what kind of trip unit it might have.

For molded case circuit breakers, I think the instantaneous trip setting tolerance is something like +/-25%. That throws a wrench in the works if you're trying to assure yourself the setting is high enough that you wont have a trip. Whether it represents lack of repeatability expected for a single circuit breaker I'm not sure.

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[electricpete]

If you take the worst case of a fully offset sinusoidal waveform, the rms is sqrt(3) times the rms of the same sinusoid with no offset.

Just to elaborate this point.

Let's say your non-offset waveform is
i(t) = sqrt(2)*LRC*sin(w*t). It has an rms value of LRC.

The fully offset waveform is
i1(t) = sqrt(2)*LRC*(sin(w*t-pi/2)+1)

The square of that is
i1^2 = 2*LRC^2*[sin(w*t-pi/2)+1]^2
i1^2= 2*LRC^2*[sin^2(w*t-pi/2)+2*sin(w*t-pi/2) + 1]

The mean of that is
<i1^2>= 2*LRC^2*[<sin^2(w*t-pi/2)>+<2*sin(w*t-pi/2)> + <1>]
<i1^2>= 2*LRC^2*[0.5 +0 + 1]
<i1^2>= 2*LRC^2*[1.5] = 2*LRC^2*3/2 = 3*LRC^2

The root of that is
sqrt(<i1^2>) = sqrt(3)*LRC

So of course it is sqrt(3) times the rms of the waveform without offset.


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[electricpete]

So, were the motor or breaker changed lately?

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[Skogsgurra]

Hello Rockman. Did you try 15 starts with that motor? What do thermal specs say about that?

I suppose that the 'high speed meter' is a scope or a recorder. What sampling speed does it have? And what bandwidth. Any filters in the signal path?

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[rockman7892]

I was finally able to pull the recorded transient files from the meter. I am attaching them below and in subsequent posts. The first (2) files are where the breaker tripped during starting as evident by the waveforms. The second (2) files are events where the motor started sucessfully.

Looking at these waveforms is difficult to see anthing that would explain why the breaker would trip in on case and not the others. Maybe the experts here can see something that I dont.

 

[rockman7892]

Second tripping event

 

[rockman7892]

Sucessfull motor start

 

[electricpete]

It looks like the successful start hit closer toward the worst-case closing angle had a higher peak than any of the others, and I would also judge a higher rms than any of the others if we are using an averaging interval of one cycle. That peak hit on phase C and was about 5200A peak.


Which probably means your peaks on A and B were a hair lower than in the other starts. I'm assuming this is molded case circuit breaker... as I mentioned the tolerances can be very wide. Maybe your A or B phase is tripping a little lower than the others.

You could pull out the breaker and check the tripping characteristics against specification and between phases, and possibly against another new breaker (might as well swap breakers while you're at it). Perhaps try several times to evaluate repeatability. As I remember there are different ways to evaluate instantaneous trip: slowly increase current and check for trip or apply pulses of successivly increasing magnitude. (both cases sinusoidal current). I know there are advantages and disadvantages to each approach but I don't recall specifically what those are (I think it was discussed on power forum). I'm pretty sure our guys use pulse method.

Let's talk about the setpoint:

If circuit breaker responds to peak, that is equivalent to a test sinusoid of 3676, which is below your 4000 nominal setpoint.

If the circuit breaker responds to rms, we can calculate the rms of a fully offset sinusoid by remembering it has a peak twice as high as associated non-offset sinusoid, and an rms of sqrt(3) times as high as associated non-offset sinusoid. So the ratio of pk/rms must be ration of pk/rms of sinusoid times ration pk/pk of offset/non-offset divided by ratio rms/rms of offset/non-offset = sqrt(2) * 2 / sqrt(3) = 1.62. So your rms would be around 5200/1.62 = 3184.

In summary, not enough to trip on your nominal setpoint regardless of whether we think the breaker responds to peak or rms.

But there is that huge uncertainty band in there. Is the setpoint really adequaate? It could be that even though your start does not reach the nominal setpoint, the setpoint offset and repeatability errors are getting you.

Logical approach is to test the breaker. But if you don't find any problems you might still think about increasing the setpoint if electrical calculations allow.

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[electricpete]

Also for completeness I'll mention one other possibility that is unlikely, but always seems to come up during our "fault tree". It is possible that the breaker is not tripping, but instead responds to the vibration from closing of the contactor. We have actually thought about getting high-speed video to watch the closing sequence when we were on a similar witchhunt. We did drop several closed molded case circuit breakers from waste level (destructive test!) and some of them would trip open in the process, which proves in theory it may be possible, but that's a heckuva lot more jolt than I think the breaker would see when contactor closes.

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[electricpete]

In case my calculations weren't clear for comparing your waveform against the setpoint, here's what I did:
1 - I assumed 5200A peak was the highest.
2 -I assumed this waveform also represented the highest rms
3 - I assumed this was fully offset
4 - Since your trip setpoint is established by using a 4000A rms sinusoid, I converted your waveform to an equivalent sinusoid (whose rms will be compared to setpoint) in two different ways: 1- looking at the peak value of your actual waveform; 2 - looking at the rms value of your actual waveform.

1 - I converted 5200A peak to an equivalent sinusoid rms by dividing by sqrt(2). 5200/sqrt(2) = 3676rms. Since this is below your nominal setpoint 4000rms (established with a sinusoid), it tells us your breaker "shouldn't" trip if it responds to peak.

2 - I converted 5200A peak to it's own rms by dividing by 1.63. 5200/1.63 = 3184 (rms). Since this is below your nominal setpoint 4000rms (established with a sinusoid), it tells us your breaker "shouldn't" trip if it responds to rms.

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[electricpete]

I have heard discussion of transformer inrush being affected by residual magnetism, but I never heard anyone mention that for motors. I don't think it applies to motors. I can make some guesses (rationalizations?) for why that may be so. In order of relevance from my uninformed opinion:
A - * The transformer may be deenergized suddenly and all current and flux stops rapidly. In contrast, when we deenergize a motor there remains a current in the rotor which decays away. This will certainly tend to de-magnetize the stator core, since the decaying rotor current will cause the stator to see a decaying flux of alternating polarity.
B – The relevant reactance for the motor during inrush is the leakage reactance. For rotor slot leakage reactance, the relevant flux loop encircles a bar... and so a portion of it must go thru air. Since the loop path includes air and iron, the air tends to dominate the reluctance, and the effects of change in iron effective permeability (such as due to remnant flux) have reduced importantce compared to a device like trnasformer where all relevant flux loop paths are completely in iron without going thru air.
C – There is heavy saturation during energization of unloaded transformer for various reasons.... including the fact that the transformer operates further into saturation and the leakage reactances are smaller and magnetiing reactances higher. Also it is the magnetizing branch which saturates in a transformer since no secondary current, but magnetizing branch in motor will not saturate as much because rotor flux tends to cancel stator flux. Finally when we look at L/R time constant of a transformer during energization it is very high (due to high L associated with magnetizing branch)... in contrast for motor it is shorter since we have lower L (the leakage branch rather than magnetizing). So a small remnant flux can contribute a DC effect which exacerbates the saturation phenomenon in a transformer... but saturation is not much of a problem to begin with during motor energization. Maybe others can explain more especially on the transformer side.

I have another item to add to the above list (call it D) which is probably the biggest factor.
D - Remnant or Residual magnetism Br is dramatically lower for gapped core like a motor than for continous core like a transformer.
Below is excerpt from Magnetism Fundamentals:

http://books.google.com/books?id=Mg... of a slit on the magnetization curve&f=false

It shows a B vs H curve for iron piece on the left and for iron piece with series airgap on the right.

We could have guessed the shape of the graph on the right using our knowledge of series circuits... for electrical circuits the voltages add and for magnetic circuits the mmf's add. So we can arrive at the series curve by adding in the horizontal direciton the linear airgap curve to the iron B vs H curve.

What is the difference in the curves? saturation level is roughly the same, width of hysteresis loop is roughly the same (Hc). What is dramatically different is Br. Assuming the piece was driven into saturation and switched off without magnetization, the Br for the non-gapped core is very close to saturation, the Br for the gapped core is tiny.

This factor (D) is probably the biggest reason why residual magnetism plays so much more role for transformer energization peak current than for motors (followed by item A).

There was related discussion here thread238-120074
Reactorman made this exact comment at the end of the thread but it never really hit home for me until I saw the figure.


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