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DC offset in induction motor inrush current
2

DC offset in induction motor inrush current

DC offset in induction motor inrush current

(OP)

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?

RE: DC offset in induction motor inrush current

2

Quote:

Would worst cas current be when closing with voltage at zero or at peak
Worst case closing would be near zero of the phase-to-neutral voltage of the supply.  (even though there is no connection to the motor wye point).

Quote:

How do high effeciency motors impact the above equation resulting in typically higher inrush current?
One factor is that to achieve higher efficiency, the motors generally have lower rotor resistance. This results in longer L/R time constant (slower decay).  Which means less decay occurs by the time of the first peak.  There may be other losses such as stator losses and core losses that create the same effect to lesser extent.

Also in some cases the KVA code indicates higher LRC. That is taken into account for large motor relay calculations but sometimes not for breaker settings by NEC which work on multiple of FLA.

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

(OP)
Thanks Electricpete

Do I have the right equation above for modeling the transient inrush current with time over the first couple of cycles.  If not do you know where I can find the correct equation.

I want to use this equation to see how the current mannitude changes with resistance values in the motors and/or cables.  What would you recommend as the best approach to this?

RE: DC offset in induction motor inrush current

First I will say there are two different versions of the induction motor equivalent circuit.

1 - The version we are all familiar with is the "steady state" equivalent circuit.
2 - There is also a "transient" equivalent circuit, which is much better for predicting transients (although it still typically employs assumptions such as neglecting saturation and neglecting slotting effects). Attached is the induction motor transient equivalent circuit from Paul Krause's "Analysis of Electric Machinery".  

Use of the attached transient equivalent circuit requires choice of a radian speed "w" of the reference frame.  The typical choice is w = we (the frequency of the electrical supply), in which case our power supply voltage which is sinusoidal in the stationary reference frame becomes dc in the simulation reference frame.

The model on the slide has everything needed to do a state space simulation.  We can choose either the four currents or the four flux linkages as state variables (if we know one, we know the other per 4 equations in 4 unknowns at the bottom).

The magnetic effects are split into two types of circuit elements – the inductances and the dependent voltage sources..   In a synchronous reference frame under steady state conditions, the voltage across the inductances is 0.  So the entire steady state behavior is predicted by those dependent voltage sources, with the inductances in the circuit playing no role in steady state (other than as coefficients determine flux linkage as input to dependent voltage sources).   

So we know for a fact the dependent sources control the steady stage behavior with inductance elements playing no role.  I believe that for the first cycle of a DOL starting transient, the reverse is true... the inductance elements control the transient and the dependent voltage sources have little role.    Perhaps the fact that the voltage sources are initially zero helps us believe this assumption  (It is subject for validation and comment whether that is true and whether it can be intuitively deduced from the equivalent circuit).

If we accept the questionable premise that the dependent voltage sources play no role in the first cycle of DOL start, then we can cross the dependent voltage sources out of the circuit.  What is left are two d-q circuits with only the elements from the equivalent circuit including R2 which would be the value of the variable R2/s resistance when s=1.  If we reverse the dq transformation back into  a/b/c variables and assume balanced (*) conditions to support per-phase equivalent circuit, we recover the normal ("steady state") equivalent circuit.   (* if you are uneasy about applying the concept of "balanced" view to transient analysis, view it in the Laplace domain with your inductance elements as s*L rather than j*w*L, and your three voltage sources differeing by factor exp[-s*2*pi/3] and you can see it applies within the assumptions of linear system.).

So how valid is that approach especially given the questionable assumption?  It leads to analysis of the DOL starting transient using the steady state equivalent circuit which leads to the equation that you gave.  It is my perception that it is pretty close.  I know that NEMA MG-1 states the max theoretical transient inrush current is 2*sqrt(2) * I_lrc_rms, which is consistent with this approach.

What I would like to do is to do a few simulations to compare the two approaches.  That is the only way to answer the question that I know of.   

Quote:

Do I have the right equation above for modeling the transient inrush current with time over the first couple of cycles.
I am going to investigate how closely that matches the more exact transient approach.

Quote:

  If not do you know where I can find the correct equation.
The more exact transient analysis approach requires numerical simulation as described in Krause or other references.  I can provide spreadsheet or perhaps some on-line free references if you are interested in pursuing this approach.

Quote:

I want to use this equation to see how the current magnitude changes with resistance values in the motors and/or cables.  What would you recommend as the best approach to this?
You could certainly lump external resistances into R1 in the transient numerical simulation and (if it's valid to begin with) in the steady state equivalent circuit approach.

Note that if you had larger cables with significant inductance and wanted to model that external inductance, the same simple approach would NOT work.. you couldn't lump them in with L1 because L1 is involved in the transformations between flux and current.

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

My caveat about lumping external inductances into L1 applies to the broad application of the transient equivalent circuit accross a range of transients. I am not sure whether it would cause a problem in the specific case of DOL start or not. Probably not relevant to the question which concerns external resistances, but I just wanted to correct that statement.

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

OK, I have finished my review.  I conclude that the steady state equivalent circuit does an OUTSTANDING job of predicting the first few cycles of current, based on comparison to Krause's transient model.

The first part of the review was to look at the equivalent circuit and determine the analytical solution for A phase if we have a suddenly applied voltage of Van = cos(w*t-theta).  The solution is attached.  It is based on per-phase analysis of A phase using values of R and L which could very well be derived from the real and imaginary parts of the complex locked rotor impedance knowing locked rotor current and locked rotor power factor).

It is a complicated formula (perhaps it can be simplified using assumptions like R<<w*L), but it is at least an explicit formula predicting current as a function of time and parameters R, L, and closing angle theta.

 

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

The second part of the study was to compare the results of the above analytical formula (based on the steady state equivalent circuit) to the full Krause simulation.

For an example motor I used the same motor as described here:
thread237-280695: simulation to recreate measured wiggle in P.F. during DOL start
(if you look at attachments in that thread you can see current vs speed, torque vs speed, etc).

I already have the equivalent circuit parameters for that motor. So to find R and L, instead of starting with locked rotor impedance and angle, I simply used:
R = R1 + R2
L = L1 + L2
The specific values R2 and L2 are adjusted for deep bar effect based on the frequency seen at the rotor during start.  Note the magnetizing branch was completely ignored for simplicity (with negligible effect as we see later).

Attached are results of comparison for varying angles theta, where a phase line to neutral voltage is Valn = 1/sqrt(3)*VLL*cos(w*t-theta)).

The first slide shows theta = 90 degrees which is zero crossing of line-to-neutral voltage and gives max A phase current as expected. The other slides show 10 degree incremental variation in theta.

The Krauss solutions Ia, Ib, Ic are colored red, white, purple.  The analytical A-phase solution based on equivlanet circuit is colored blue.  

You can barely tell the difference between the red and blue curves until 10 or so cycles into the transient where the oscillations in the Krauss solution make them diverge.

I conclude that simplistic analytical solution of the equivalent circuit (as in my previous post) gives outstanding agreement with the Krause model.

And you use the analytical solution of previous post, add your external resistance along with R1, and your prediction should be very good.
 

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

A few more comments.

1 - The analytical solution of L/R circuit with suddenly-applied sinusoidal voltage posted 8 Oct 10 22:46 was correct, but the derivation using Laplace was too complicated and resulting solution also more complicated than needed.  I redid the solution using much simpler approach (attached).  The derivation  is simpler and resulting equation is simpler. I verified that it gives the same results (still matches Krauss).

2 - I may have been wrong to say that decreasing R2 was a big contributor to increased L/R time constant that makes high efficiency motors more likely to trip, since R2 varies by deep bar effect and may not be much lower during start for high efficiency motor than older motors (we need to keep R2 high during start to provide staritng torque).  It is more likely the other loss mechanisms more than R2.

3 - I have to admit I was shocked that the simple analytical analysis of R-L circuit matched Krause's simulation so well for the first few cycles.  I really don't have an explanation for that.  To my way of thinking it suggests that the inductances in the transient equivalent circuit (posted 8 Oct 10 12:31) dominate the transient while the dependent voltage sources don't affect the result at all... and yet the simulation shows the flux linkages and associated dependent voltages grow immediately with the current.  Does anyone have a good explanation why simple R/L analysis matches the transient model for the first few cycles even though there are some induced voltage effects (those dependent sources) that are not modeled by the simple R/L analysis?

=====================================
(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

(OP)
Electricpete

Thank you very much for your response.  I consider you very knowlegable on this subjet matter and appreciate your shared knowledge.

After digesting this all I have a few qeuestions:

1)  If I wanted to see the impace that cable resistance for long runs has on the inrush current can I use your revised il(t) equation by plugging in external values for R and plotting the resultant waveform?

2) How does your revised final equation differ from that which I posed in my origonal post.  My initial thought was that you could come up with an eqivalent impedance consisting of a resistance R and reactance L for the motor model during locked rotor conditions (combining model impedances with s=1 to come up with overall equivelent impedance of motor) and then use these equivelent values to plug into the origonal equation that I posted.  Was this method not correct?  What was I negelecting by using this approach as opposed to the final formula that you have come up with?

3)  Worst case is when L-N voltage is at 0 crossing as you mentioned and is shown by the equations.  This ususally infers however that the arctan (wL/R) is 90degrees to put current at its maximum.  Does arctan (wL/R) always result in 90degrees during inrush since reactance is much greater than resistance during this time.  (In the equation that I posted this would be the alpha-theta component being equal to 90degrees)

4)  If the arctan (wL/R) (or alpha-theta) component equals to zero then it appears there will be no DC offset during the inrush.  Is this correct and is this the best case scenario.

5)  If comparing inrush as a function of resistance on two idential motors how much of a factor does residual flux in the rotor effect the results seeing that it is not accounted for in these equations.

Thanks again for the help.

 

RE: DC offset in induction motor inrush current

Quote:

1)  If I wanted to see the impact that cable resistance for long runs has on the inrush current can I use your revised il(t) equation by plugging in external values for R and plotting the resultant waveform?
Yes.

Quote:

2) How does your revised final equation differ from that which I posed in my origonal post.  My initial thought was that you could come up with an eqivalent impedance consisting of a resistance R and reactance L for the motor model during locked rotor conditions (combining model impedances with s=1 to come up with overall equivelent impedance of motor) and then use these equivelent values to plug into the origonal equation that I posted.  Was this method not correct?  What was I negelecting by using this approach as opposed to the final formula that you have come up with?
There are cosmetic differences because I selected my voltage source as cos(w*t-theta) and yours started with sin(w*t+alpha).  (your alpha is related to my theta)  I used cos originally because it was easier for me to coordinate with my model.   But to focus on your equation, I have redone the derivation starting with assumed voltage sin(w*t+alpha) attached.   My new equation has a sqrt(2) because I used an rms voltage where you used a peak -  (no problem).  Your theta corresponds to my arctan(w*L/R) – (no problem).  But there is one problem that your solution does not include an alpha-dependent multipler for the exponential decaying term.... so you will always have the same exponential decaying term regardless of closing angle... which is not correct.   The corrected version of  your equation would be at the bottom of what I attached to this post.

Quote:

3)  Worst case is when L-N voltage is at 0 crossing as you mentioned and is shown by the equations.  This ususally infers however that the arctan (wL/R) is 90degrees to put current at its maximum.  Does arctan (wL/R) always result in 90degrees during inrush since reactance is much greater than resistance during this time.  (In the equation that I posted this would be the alpha-theta component being equal to 90degrees)
Yes, it's a good simplifying approximation that arctan(wL/R) is close to 90 degrees.  Then if alpha = 0, the exponential decaying term is – sin(-90) = - - 1 = +1 which has the same polarity as the sinusoidal term during the first half cycle (the period when the decaying term has most influence) and gives the worst case highest peak.  This alpha = 0 is the same worst-case condition that we talked about closing near the zero of the phase to neutral voltage.

Quote:

4)  If the arctan (wL/R) (or alpha-theta) component equals to zero then it appears there will be no DC offset during the inrush.  Is this correct and is this the best case scenario.
Yes, if alpha = arctan(wL/R), then the sin(alpha- arctan(w*L/R)) =sin(0) = 0 and the decaying dc term is 0 and there is no dc offset in the inrush.   This roughly corresponds to closing at the peak under our assumption arctan(w*L/R) is approx 90.

Quote:

5)  If comparing inrush as a function of resistance on two identical motors how much of a factor does residual flux in the rotor effect the results seeing that it is not accounted for in these equations.
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.
 

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

(OP)
Great stuff ElectricPete.  Thanks!

Now that I have my head around the equations based on your explanations above I want to use them to try to model the inrush for a particular motor that I am looking at.  Motor is a 480V 200hp premium efficiency motor.  Hopefully tomorrow I will be able to find the datahsheet for this motor that I can post.

If the datasheet has the motor equivelent circuit paramaters can I use these to determine the combined R and X values for the motor under locked rotor conditions with S=1?

If the motor datasheet does not have circuit paramaters what is the best way to estimate R and L values to use for our equation?  Would measuring stator resistance with a micrometer help determine anything?

If I want to then figure how the added cable resistnace effects equation then can I simply add the cable resistance to the stator resistance when determing the overall motor eqivelent resistance?  Should I factor in cable reactance in series with equivelent motor reactance?
 

RE: DC offset in induction motor inrush current

Quote (electricpete):

compared to a device like trnasformer where all relevant flux loop paths are completely in iron without going thru air.
That is not 100% correct since leakage reactance paths for transformer can be in air/copper/insulation. However it doesn't matter because for unloaded transformer inrush the relevant reactance is the magnetizing reactance (not the leakage reactance) and the transformer magnetizing reactance flux path is completely in iron and therefore very sensitive to iron properties.

Quote:

If the datasheet has the motor equivelent circuit paramaters can I use these to determine the combined R and X values for the motor under locked rotor conditions with S=1?
Maybe.  But you have to look carefully to see if there is a correction provided for deep bar effect... X2 and R2 can change a lot of locked rotor to running conditions.
If you know locked rotor current and locked rotor power factor that would be all you need.... nothing else matters for this particular purpose.

If you have a broader interest in finding equivalent circuit parameters (not just locked rotor), I have a spreadsheet that I use , but it has evolved into a convoluted interface... fine for the developer of the spreadsheet (me), but probably frustrating for anyone else.  When I get a chance I will see if I can make it more user-friendly.  

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

(OP)
Attached is the datasheet I was able to dig up on this particular motor.  The datasheet does indeed show a locked rotor current of 1450A however it does not give a locked rotor power factor.

Is there any kind of deduction or assumption that can be made from this datasheet for determining a locked rotor power factor in order to determine the R and X values for this motor to be used in equations.

By the way this is a motor which contiuousily trips a 400A instantaneous only breaker everytime it is started.  I will have acess to a high speed meter tomorrow where I am going to verify inrush current drawn by the motor, but wanted to also understand from a theoretical standpoint what is happening and try to create a model to match the field measuremnts.

RE: DC offset in induction motor inrush current

200hp motor with nominal efficiency exceeding 94.5 is "energy efficiency" according to NEMA MG-1 Table 12-11.   The NEC gives you a little leeway to set the breaker instantaneous settings higher for energy efficient motors. Do you know the instantaneous setting?   And is it a molded case circuit breaker / combination starter/

The data you provided will allow a pretty good estimate of locked rotor power factor – I will do that tonight.   I have been down the road chasing instantaneous trips on an integral/medium size motor before.    Even with all my inclination to try to quantify things, I ended up concluding that we needed an empirical approach... increase the settings until you don't have trips (provided coordination and protection is maintained).   Some roadblocks: 1 – of course your measured peaks will vary depending on closing angle.  ***2 - Even though you can quantify the instantaneous peak, can you quantify how your breaker will respond to that non-sinusoidal (dc offset current)?  You use a sinusoidal current to calibrate the trip settings.   Will it trip you're your non-sinusoidal signal has the same peak as the calibration sinusoid? Or when it has the same RMS as the calibration sinusoid?   And what if you have an MCP with some kind of mechanical delay intended to ride through dc.... Can that thing be quantified.   Tough to tell. And those trip bands can be wide....
 

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

My "advice" from previous post is based on trips occuring immediately after motor or breaker replacement... not if trips start occuring out of the blue.

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

Attached is my spreadsheet which was used to fit your data.  The fit seems good based on matching the performance targets (*):

Quote (Summary of fitted model parameters):



===============Model Parameters Solution================            
Name    Value    Units    Description
R_NL    296.7664742    ohms    Resistor simulate portion of No-Load losses - connected direct in parallel with the source
R_1    0.022912846    ohms    Stator Resistance
X_1    0.115435668    ohms    Stator Reactance
R_2    0.014675133    ohms    Rotor Resistance refd to stat
X_2    0.115438237    ohms    Rotor reactance refd to stator
X_M    3.074880359    ohms    magnetizing reactance
FullLoadSlip    0.012499082    none    Full Load Slip
BarDepthR    0.039887569    meter    Equivalent Depth of rectangular rotor bar - Used for deep bar correction of R2
BarDepthL    0.040885101    meter    Equivalent Depth of rectangular rotor bar - Used for deep bar correction of X2
============Selected Inputs ====================================            
VLL    460    volts    Line To Line Voltage
SyncSpeedRPM    1800    RPM    Synch Speed in RPM (like 1200, 1800, 3600 etc)
BarMat    Aluminum        Select Aluminum or Copper for use in deep bar correction                
factor_SPT    1.155819684        Slip at Peak Torque calculated from R2 and L2 but cannot apply deep correction since S_PT not known before calculated => Apply iterative correction based on slip that produces peak torque in Model Output                
factor_SHL    0.949378029        Slip at Half Load initially assumed half of full-load slip.  But slip is actually somewhat non-linear with power -> correct iteratively based on slip that produces half power in model output.                 
                            

=========== Model Performance Against Targets==============                            
Perf Variable    Calculated Value    Units    Target Value    FractionalError    Weight Factor    Weighted Squared Fractional Error    Comment
FullLoadAmps    233.3548675    Amps    233    0.001523037    1    2.31964E-06    INPUT
FullLoadEff    0.960204839    none    0.961    -0.000827431    1    6.84642E-07    INPUT
FullLoadPF    0.862277005    none    0.845    0.02044616    1    0.000418045    INPUT
FullLoadPower    153938.0632    watts    150749.0128    0.021154702    0    0    Redundant - not used
FullLoadTorque    827.0035752    N*m    798.5767431    0.035596869    0    0    Redundant - not used
HLEfficiency    0.969661061    none    0.957    0.013229949    1    0.000175032    INPUT
HLPowerFactor    0.739480191    none    0.746    -0.008739691    1    7.63822E-05    INPUT
LRC    1493.580016    Amps    1493.580044    -1.90975E-08    1    3.64716E-16    INPUT
LRT    1916.583864    N-m    1916.584183    -1.66895E-07    1    2.78538E-14    INPUT
NoLoadCurrent    83.25511366    Amps    69.9    0.191060281    0    0    INPUT - low weight due to outlier - may not model no-load losses well due to zero-slip assumption but friction/windage cause slip
BD_Tq    2072.87712    N-m    2076.299532    -0.001648323    1    2.71697E-06    Low weight - model estimate of BDT not good
X2overX1    1.000022249    none    1    2.22492E-05    0.01    4.95028E-12    Thumbrule - split is not critical to these performance parameters
R2overR1    0.640476214    none    1    -0.359523786    0    0    No target imposed
X1overXm    0.037542351    none    0.05    -0.249152987    0    0    No target imposed
Full Load Slip    0.012499082    none    0.011111111    0.12491741    0    0    No target imposed, but constrain to within 2.5 rpm from nameplate
BarDepthR    0.039887569    m    0.015    1.659171295    0    0    No target imposed

                    SWSFE    0.00067518    

Starting Power Factor    0.432976227        
The predicted starting power factor of this model is 0.43.  Maybe it seems a little high... 0.2 is often thrown around as a standard number, but there is certainly variability among motors, and in general motors with unusually high locked rotor torque will tend to have unusually high starting power factor., which leads me to another point.

* I ran into an oddity I have not encounted before.  If you go to the ModelPlot tab and look at torque vs speed, the flat shape between locked rotor torque and breakdown torque looks unrealistic.    Locked rotor torque is 240% and breakdown torque is 260% by your data sheet.... that's a little bit odd.  But I suspect your real curve must dip in between locked rotor torque and breakdown torque.  I haven't ever seen my model produce those dips... maybe it is a result of using rectangular bar assumption for deep bar correction whereas actual bars may have many other characteristics.... also no input data is fitted for any slip between locked rotor and breakdown, so it's not surprising the model doesn't do good in that area.

So, knowing the model is not good between locked rotor and breakdown, should we believe predicted locked rotor power factor?  I believe we can make a fair case that any model that correctly predicts locked rotor current and locked rotor torque will come somewhat close to predicting locked rotor power factor, using some rough assumptions.  We assume the locked rotor current is primarily inductive and determined to a first  approximation soley by the leakage reactances (neglecting magnetizing reactance and resistance).  So if we have current right, we have the leakage reactances right. Since magnetizing branch is small we also predict the rotor current I2 reasonably well.  And we know torque is 2*I2^2*R2/(2*pi*Felectrical).  If we have I2 right and we have torque right, then we  must have predictred R2 well.   Based on deep bar effect, we can say losses in R2 at locked rotor condition are larger than all other losses and dominate the loss picture.  So we have correctly predicted the losses.  If we correctly predicted the total current and correctly predicted the losses, then we have correclty predicted the power factor.   There were a lot of assumptions and approximations in that logic but I would bet you a beer it's at least within 0.05 and bet you a sixpack that it's at least within 0.1 from the true starting power factor.
 

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(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

Quote (electricpete):

We assume the locked rotor current is primarily inductive and....
Actually that assumption is not required.  If we have the stator current right, we only have to assume the magnetizing current is small (compared to LRC) to conclude that we have rotor current I2 approximately right.   Then again with I2 and torque, we know we have R2 right (from the equation for torque in terms of I2 and R2).  The weakest part of the argument would be the assumption that starting rotor resistive losses dominate the losses.  I think they are well over half, but still a pretty gross approximation.

=====================================
(2B)+(2B)'  ?

RE: DC offset in induction motor inrush current

(OP)
Thanks for the spreadsheet and power factor.  I am looking through it all trying to understand it.

I was able to get ahold of a high speed meter today and measure the transient inrush current.  Unfortunately when I downloaded the transient plots to my computer the files got corrupted so I'll have to redownload them tomorrow to post.  Heres what I saw.

On the particular motor that is tripping I saw peak currents of between 3500A to 5600A with the worst case scenario being where voltage being near zero as discussed above.  During the testing of probably about 15 starts or so we were only able to see the breaker trip twice.  The strange this was that when the breaker tripped we did not see much difference in peak currents from when the breaker didn't trip.  

For instance one time when the breaker didn't trip and motor started the first cycle peak current was about 5600A while the second cycle peak current was 3200A as the DC offset decayed towards steady state locked rotor current.

For another instance where the breaker did trip the first cycle peak current was 4800A with the second cycle peak current being 3300A as the transient decayed.  I find it strange that the higher peak current for one start didn't trip the breaker while the lower one did.  We saw similar numbers on identical motors right next to the problem one that did not trip the same breaker.

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?  

Also in a couple of instances we noticed that the motors which are on pumps were starting with the pump outlet valve closed.  I dont think however a closed outlet valve will have any bearing on the inrush transient since we are already dealing with locked rotor conditions.  Do you agree?

I'll try to post waveforms tomorrow.

RE: DC offset in induction motor inrush current

Quote:

Also in a couple of instances we noticed that the motors which are on pumps were starting with the pump outlet valve closed.  I dont think however a closed outlet valve will have any bearing on the inrush transient since we are already dealing with locked rotor conditions.  Do you agree?
I agree in general valve position makes no difference for instantaneous trip.  The one exception that was suggested to me on the forum when we were having our trips was that a pump which happens to be spinning backwards can draw higher instantaneous peak.

Also, bouncing contact might cause unusual peak, but should be evident in your traces.

 

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RE: DC offset in induction motor inrush current

NEC limits you to 17*LRC = 17*233= 3961 or next higher setting
for energy efficient motors. Looks like you are out of luck in NEC space, but that doesn't mean the setpoint isn't the problem.

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RE: DC offset in induction motor inrush current

Quote:

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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RE: DC offset in induction motor inrush current

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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RE: DC offset in induction motor inrush current

Quote (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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RE: DC offset in induction motor inrush current

So, were the motor or breaker changed lately?

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RE: DC offset in induction motor inrush current

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

RE: DC offset in induction motor inrush current

(OP)
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.

 

RE: DC offset in induction motor inrush current

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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RE: DC offset in induction motor inrush current

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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RE: DC offset in induction motor inrush current

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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RE: DC offset in induction motor inrush current

Quote (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=MgCExarQD08C&;pg=PA144&amp;lpg=PA144&dq=the+influence+of+a+slit+on+the+magnetization+curve&source=bl&ots=oD369G9-MU&sig=shkM_4s0MLJ1ViEMHsgHME6qEqA&hl=en&ei=oc4xTdzbI8nUgAeiz_jeCw&sa=X&amp;oi=book_result&amp;ct=result&resnum=1&ved=0CBMQ6AEwAA#v=onepage&q=the%20influence%20of%20a%20slit%20on%20the%20magnetization%20curve&amp;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: Power transformer inrush current
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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(2B)+(2B)'  ?

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