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Effect of reduced voltage start on thermal severity of a motor start
4

Effect of reduced voltage start on thermal severity of a motor start

Effect of reduced voltage start on thermal severity of a motor start

(OP)
To prevent hijacking another thread, I'll start a new one.

1 – What are reasonable procedures or thumbrules to evaluate whether a given motor driving a given load will be suitable  (from a thermal standpoint)  for starting at a voltage below what is specified during motor purchase ?

2 – Would you hazard any guesses of "typical" behavior
i.e. Reduced voltage start is more?/less? severe for small/large motor driving centrifugal pump load etc under various starting loading conditions (recirc is most typical for us... still creates a load)

I have tried to investigate this by looking at our plotted 80%-voltage and 100%-voltage plots of starting current vs time which are plotted on same scale as the motor thermal limit curves. The results are a little ambiguous to me:  if we use time to reach curve in seconds, than reduced voltage start seems less severe..... but I we use distance to reach curve measured against a log scale of time, then reduced voltage start seems more severe.  There seems some basis to use distance on log time scale because it represents a ratio of times and (under simplifying assumptions) a fraction of max allowable heat that is added.  Then when you really think about what those curves mean (different things in different regions), it gets even muddier.

I really don't know the answer (other than for completely unloaded starts where theory tells us soft start is beneficial) and would be interested in any thoughts or comments.
 

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RE: Effect of reduced voltage start on thermal severity of a motor start

2
(OP)
Also I notice that IEEE 620-1996 (IEEE Guide for the Presentation of Thermal Limit Curves for Squirrel Cage Induction Machines) gives thermal limit curves in a format that depends on motor voltage.  Specifically the acceleration/transition curve connects the locked rotor region to the running overload region.  The point of intersection of this acceleration/transition curve  with the locked rotor curve depends on voltage (since locked rotor current depends on voltage) and the point of intersection with running overload depends on votlage (since current at breakdown torque depends on voltage).

Do you apply soft start for large motor if the manufacturer has not provided a thermal limit curve for the reduced voltage?

Question/Whine: Why is it that IEEE and NEMA seem silent on the issue of evaluating reduced voltage start?
 

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RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Any thoughts?

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RE: Effect of reduced voltage start on thermal severity of a motor start

2
I calculate the I^2t of the reduced voltage start and then convert that back to an equivalent 100% voltage locked-rotor or stalled time. I will compare that to the published safe stall time of the motor.

In that previous thread, you had posted a link showing calculations where the reduced voltage start was calculated as taking longer to start than the allowed hot or cold safe stall time. As far as I'm concerned, that method is flawed because the motor heating at a reduced voltage is not the same as the motor heating at rated voltage. In other words, the safe stall time can be longer at a reduced voltage since the current is lower.

In my calculations, I have found that the calculated equivalent locked rotor time does not increase very much until the voltage is reduced enough that the motor comes close to stalling. However, the equivalent locked rotor time is always at least a little higher (like say 5% to 10% higher) with any type of reduced voltage start compared to the equivalent locked rotor time from a full-voltage start.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Thanks Lionel.

I noticed the same thing about using the same stall time at 2 different voltages after I posted that link.

Your approach makes sense. If I can say it a different way:
* Convert 100% Locked Rotor limit into i^2*t limit
* Compare computed accelerating I^2*t to the i^2*t limit.

I can see there are at least 3 conservatisms built in:
1 - locked rotor cooling is less than accelerating cooling
2 - if OEM takes any credit for cooling at all during locked rotor, the least cooling occurs during the shortest-duration locked rotor start 100%.  I presume that's why you picked 100%.
3 - The rotor resistance is highest at locked rotor.  This would affect integral of Energy = I^2*R*t.

So as long as you pass it's a good approach.  A little more pencil sharpening would be required if you don't.  I like that approach.
 

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RE: Effect of reduced voltage start on thermal severity of a motor start

Ya, that's basically correct. I wrote the starting analysis program I use, so I have it calculate the I^2t as it does the start calculation. Then, I just divide out the LRA^2 to get what I call the equivalent locked rotor time.

The main reason for picking this is because most motor manufactures will at least publish a safe cold and hot stalled time. It's hard to use a thermal limit curve when a start involves a ramping or a constantly changing motor current. I believe you could come up with an average starting current which, along with the starting time, could be used on the thermal limit curve. Of course, I can also just put different fixed current limits into the program and see what the starting times are.

 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
That makes good sense.  I am used to doing seeing the calculation it for assumed constant voltage (80% and 100%) when we also have thermal limit curve for those voltages.  Before your comments I had never stopped to think that using I^2*t from 100% locked rotor stall time would be conservative for all the other cases.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Thanks Muthu.  I will do a simulation of that test case for comparison.  If I get energetic there's lots more to do: add a load and track Integral of I^2*t. Could track integral I^2*R2(s)*t since R2 varies with s. Since there are thermal time constants there could also build thermal model of motor, but I don't think there is much standardization in terms of thermal modeling... will cross that bridge later.

I notice the current vs speed plot goes to zero at sync speed.  I think we had that kind of curve on some other motor posted on the forum, but doesn't seem realistic since there will remain magnetizing current.  Maybe the curve was generated from equivalent circuit with magnetizing branch neglected...doesn't make much difference in torque and doens't affect current until you get on the far right side of the current vs speed curve.

I also noticed 100%-voltage locked rotor time 75 seconds cold / 50 seconds initially hot.  That's a lot higher than any large motors in our plant. Maybe they make them beefier these days.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
You are showing a load which comes very close to the reduced voltage motor curve.   I'm guessing your acceleration curves did not include effects of this load, right? (if they did I would have expected acceleration extended by much larger amount)

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
SS is "soft start" which I assume is around 75% voltage based on ratio of LRC's.

What is CS?

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
I withdraw my comment about current going to zero.  Can't really tell from graph... it may go down to 30% or 50% FLA and just blends in with the vertical line.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Hmmm. I get time to start the motor unloaded at 100% voltage is between 6.5 and 7 seconds.  Looks like you got 3.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
I think I can prove 3 seconds is too low a starting time even without computer calculation.  Here is my hand calculation... if I'm wrong, then please let me know where I made an error

For your example motor:
Trated = 5908 N*m
J = 64 kg*m^2

The torque speed curve shows torque < 50% of rated for the entire interval 0 to 70% of sync speed.

Generously assume torque is 50% rated throughout this interval and compute the time to accelerate just to 70% of rated speed:

dw / dt = T / J = 5908*50% / 64 = 46.5 radians per second... per second

Convert above radians/sec into rotations per second:
(df/dt) = (dw/dt) / (2*pi) = 7.34 hz/second

The change in speed we're looking at is from 0 to 0.7*50hz = 35hz.

The time to change speed by 35hz / (7.34hz/second) = 4.76 seconds.

(and that's just the time to get to 70% speed, assuming we're at 50% Trated instead of  a little bit lower).

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Attached is my analysis of starting of Muthu's motor.

Slide 1: 100% voltage, no-load
Slide 2: 75% voltage, no-load
Slide 3: 100% voltage, loaded (*)
Slide 4: 75% voltage, loaded (*)
* Load torque is assumed to vary as speed squared, reaching a torque of 20% of motor rated steady-state torque at full speed. See slide 5 for load torque graph.

The 1st four slides are plotted on same scale for easy comparison.  The red curve is proportional to I^2*t.  You can see it is constant between the 1st and 2nd slide (both unloaded).  Then increases by about 10% when we add the load at 100% voltage (slide 3).  Then i^2*t increases yet another 10% when we reduce voltage to 75% while still loaded (slide 4).

So even with this very modest load (varying as speed squared... 20% of rated at full speed), there is 10% increase in I^2*t due to reducing voltage from 100% to 75%.  

This particular motor I thik has relatively low starting torque and relatively long start time compared to the ones I am used to. I think that is a tradeoff where the rotor resistance remains relatively low at starting which decreases the starting torque but allows longer start time.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

Thanks pete. I will study it and come back to you.

(Those nos. (starting current, time etc.) are as given by the motor oem and the SS oem.)

In this case, as given by the SS oem, the I^2t with ss is double that of DOL. Do you agree ? If yes, would it be better to not to use the SS at all ?

The problem got dropped on to my lap when the user was unable to start the motor even with SS since the utility side breaker kept tripping due to overload. The utility has set the O/L trip at 200 A, while this motor takes at least 500 A with SS. The utility is not willing to increase the time delay and the user is stuck. The user has taken only 2500 KVA as sanctioned demand. Is that enough demand for this motor ?

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Looking at the curves you posted, the reduced-voltage (75%) motor curve barely exceeds the load curve.   I can believe this would double the I^2*t. I will simulate that when I get a chance.

The KVA of motor at full voltage is 11,300.
To get it down to 2500, you'd need to reduce voltage to about 47%. That would clearly move the motor curve below the load curve which cannot give a successful start.

If they can't go above 2500kva or reduce load torque, I don't think it can be started with a constant level reduced voltage.   Maybe Lionel, Jeff or others know some rabbit to pull out of a hat but I can't see how it can be done.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)

Quote:

but I can't see how it can be done.
... with reduced voltage.  Of course you know vfd probably could work if there is $ for it.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Attached I did a simulation with the load shown on your graph.

New slide 8 shows the load torque-speed curve... goes from 0.1 at 0 speed to 0.5 at full speed sagging slightly below linear.

Slide 9 is 100% voltage start.  I^2*t is 5.5 on this scale (normalized by dividing by 800k... different than previous slides).

Slide 10 is 80% voltage start (it stalled at 75%... must not have had the curves just right).  The I^2*t is no 11 on this same scale as previous slide. Yes about double I^2*t in this particular case where the reduced motor torque just barely exceeds the load torque.

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

Thank you pete. Looking at those curves confirming the SS I^2t is almost twice the DOL I^2t.

The utility is not budging in increasing their O/L time delay. VFD pricing at that range will kill the client.

I am now verging on the idea of pony motor (11 KV, ???? KW) and using the presently useless SS for the pony motor. It means a new shaft and a new 11 KV, ???? KW motor, which will be lot cheaper than a VFD.

The question is what happens to the inrush current and duration when the main motor is energized at the rated speed ?

And, of course, what is the pony motor KW ?

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
I don't think there will be much transient when energizing the motor at full speed.   There is the decaying dc component which should be gone within a cycle or two.  Maybe a little bit of acceleration or deceleration if motor torque is not matching speed of load. I'm pretty sure it wouldn't come anywhere close to challenging the 2500 kva limit or the 200A overload.

The pony motor seems like it would be the challenge I think since as I understand you're starting the motor loaded.  I'm not sure how to go about sizing that to take account the load and the acceleration while taking credit for the fact that the poney motor runs a relatively short period of time.  Maybe others can comment on sizing the pony motor?

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

I ran some quick numbers. I was seeing 10+ seconds for acceleration time at 100% voltage. The time depends on the load inertia. The more inertia the more acceleration time.

Just some general observations.

That motor is a very poor candidate for starting that load and for use with a soft-starter. The motor starting torque is terrible being <50% of rated until about 70% speed.

There is no possible way that anything close to 2500kVA could be met with this motor and a soft-starter. Edison - I do not understand why you posted that the customer is meeting this requirement because they can't possibly be doing so. The 80% motor current is 417% so reducing the voltage to 80% means that the kVA draw begins at about 8580kVA and then reduces as the current reduces (which isn't much until >90% speed). Achieving 2500kVA requires that the motor current be reduced to about 150% of rated.

So, my conclusion is that the calculated starting data (design_sheet_1850) is just plain wrong. The motor current is shown as having a locked rotor current of about 690% at 100% voltage yet the motor data sheet clearly shows the locked rotor current is 560% at 100% voltage. Also, there is no possible way that motor and load would start in 3 seconds at 100% voltage or 8.5 seconds at 80% voltage. If this was part of the design work to qualify that equipment on the customer source then this is also a case of poor engineering.

The sizing of a pony motor is not too difficult. You look at the load torque curve and ensure the pony motor torque curve stays above this curve, while give some margin for maintaining accelerating torque and allowing for starting voltage drop. The big issue I see is that the load approaches 50% of the 1850kW motor rating meaning the pony motor needs to be either rated for 1/2 the size of the main motor or run in overload up to the point the main motor is energized. Once you get into a motor sized this large then the starting kVA of this motor will also be more than allowed.

electricpete - I'm not sure why but your curves show the motor current at 100% voltage and 0 speed is around 605% yet the motor data sheet shows 560% current at that point. Your slide with the 100% motor current curve showed 560% current.

Finally, I'm quite impressed you were able to do those simulation calculation in XL electricpete.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

Thank you very much, Lionel.

That the SS was poorly designed and engineered was my observation to the client. Also, the 2500 KVA sanctioned demand is too low. My suggestion was a minimum of 3500 KVA but the client is hesitating since he has to pay for the entire 3500 KVA, whether the consumes or not.

The whole package came from the compressor supplier, who seems to have no clue about the motor + SS issues.

I have attached the compressor curve. Could you possibly tell me what is the minimum HP pony motor that is needed to start this daymn thing ? Also, how much of an inrush current do you anticipate when the main compressor motor is energized at the rated speed ?

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

Those compressor curves are performance curves. You need a speed vs torque curve.

The motor speed has no bearing on the inrush levels, the speed just affects the duration of the inrush. So, the motor will draw locked-rotor current even if energized at full-speed. Actually, the full-voltage inrush is tyically a fair bit higher than the listed locked-rotor current for a few cycles until the magnetic fields are established.

You already have the soft-starter on site so just quickly ramp the voltage to the motor and you will experience a minimum inrush. Unless you plan or re-applying the soft-starter to the pony motor?

 

RE: Effect of reduced voltage start on thermal severity of a motor start

Thanks again, Lionel.

Yes, that was the compressor performance curve. I've asked for the speed/torque curve. I'll post it here when I get it.

Good point about the inrush current. The only question what is the time delay in the utility O/L time delay.

I was planning for a DOL start for th pony motor (provided it met the site restrictions) and doing away with the SS. But you raised an excellent point about using it for the main compressor motor after the pony motor does the job. The SS ramp is set for an initial 70% voltage (the SS OEM says it cannot be changed). In such an event, what would you advise the ramp time should be ?

(Wish I could tip you twice).

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

And sorry pete, for hijacking your thread. Hope you don't mind.

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)

Quote:

electricpete - I'm not sure why but your curves show the motor current at 100% voltage and 0 speed is around 605% yet the motor data sheet shows 560% current at that point. Your slide with the 100% motor current curve showed 560% current.
The slides are not well labeled, but the difference is that one is in %FLA and one is in amps.   560%  is equivalent to 605A (FLA = 108A)

Quote:

Finally, I'm quite impressed you were able to do those simulation calculation in XL electricpete.
Thanks.  I have in the past worked with old versions of Matlab and Maple but lately I am gravitating towards excel for everything because it just seems easier to access and manage the calculations once you get them all in excel.  This one was done with my general purpose Runge-Kutta ODE solver programmed in vba.   

That vba program may be a little overkill for this problem.....   I previously did the same thing (motor acceleration based on torque) in a much simpler way without vba in the spreadsheet attached to my post 11 Apr 09 21:46  in the thread linked here: thread237-238806: Induction motor modelling.  .  

What program do you use?

Quote:

The motor speed has no bearing on the inrush levels, the speed just affects the duration of the inrush. So, the motor will draw locked-rotor current even if energized at full-speed.
I'm not sure about that.  Use of the steady state equivalent circuit would lead to a different conclusion.
With s=1, the steady  state current is LRC.  The theoretical max peak inrush is 2*sqrt2*LRC  (where LRC is rms, sqrt2 corrects to peak, and factor of 2 represents worst case doubling due to decaying dc component).
With s~0.02 let's say, the steady state current is FLA or less. The theoretical max peak inrush is 2*sqrt2*FLA.  And the decay of the dc component will be faster if there is any load on the motor since L/R constant is smaller.
That is based on equivalent circuit, which I admit is not perfectly suited to transients.  But it's hard to see how locked rotor impedance has any relevance to the situation. I am pretty sure the max instantaneous peak would be closer to 2*sqrt2*FLA than to 2*sqrt2*LRC.   Don't you think?   
(fwiw I might try Krause's transient model on this scenario when I get a chance)

Quote:

And sorry pete, for hijacking your thread. Hope you don't mind.
Doesn't bother me.  It was good you gave an example motor to work with.  As it turns out your motor was very good example of how even the very modest load that I simulated the first go-round (slide 5.... torque~speed^2 maxing out at 25% of rated) resulted in noticeable 10% increase in total I^2*R heating  when we decreased voltage to 75%! (10% difference between slides 3 and 4).   Although I think your motor is atypical with that very low torque <50% rated through 70% speed.
 

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
There was a clumsy sentence fragment on my part. Let me clarify that "Don't you think" should have been "Don't you think so".  I was checking for possible agreement, not trying to throw an insult!

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

Mutha - seems odd that soft-starter is limited to a minimum of 70% voltage. Still, just a quick ramp of a second or 2 is all that is needed.

ElectricPete - I still believe the motor inrush current is the same if the motor is rotating. I probably worded it poorly. The first cycle inrush is mostly due to the stator impedance. It will drop quickly to running current if the motor is rotating at speed. It would drop to locked rotor current and then decay to running current as the motor accelerates if it is not running at speed.

And yes, this is a good example of how the motor heating will quickly increase as the motor voltage is reduced. With a reduction in voltage, there is not much margin between the motor torque and load torque and also these curves are close to each other for a large part of the acceleration. Both combined lead to a lot more motor heating. My numbers agree, the 80% start increased the motor heating by a little over 2x compared to full-voltage starting.

I use a custom program to simulate the motor start. It works quite well. I've seen and used some others like Etap but there are certain things that don't impress me about each one.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Thanks Lionel.   Good to see I am looking at the thermal model right and our numbers roughly agree.  I guess we still have a difference of opinion on one point:  I don't see how the transient of energizing a motor near rated speed has anything to do with locked rotor conditions (current or impedance).

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

electricpete, I posted that the same inrush current occurs regardless of the motor speed. The inrush current is also typically higher than the locked rotor current. So, you can assume the inrush will be at least the locked rotor current regardless of motor speed.

The steady state equivalent circuit will not show the inrush current correctly.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Thanks. I don't see why we would expect the same inrush regardless of speed. My thought is that we would expect higher inrush when energizing from rotor-stationary condition than when energizing from near rated speed.   

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
The first step to do transient analysis is to build a model.  These are basically the same equivalent circuit parameters as used in the steady state equivalent circuit (but they will be used in a different way).   There are also two small modifications to the equivalent circuit I used:
1 – resistor R_NL is placed in parallel with the input voltage to simulate portion of the no-load losses (the portion that doesn't vary with load).
2 – R2 is the low-frequency value near rated speed.  For other speeds including locked rotor condition, a correction is applied to R2 which increases it account for deep bar effect based on an equivalent rectangular bar depth which is also a fitted parameter of the model.  A similar correction is available to apply to L2, but I did not use it because the correction is smaller and the effort to incorporate it into the transient model would be too much (affects the transformation between flux linkage and current).  I will post a link to the formula I used for that correction if anyone wants it.

The results are shown below:
Results            
===============Model Parameters Solution================            
Name    Value    Units    Description
R_NL    3,635    ohms    Resistor simulate portion of No-Load losses - connected direct in parallel with the source
R_1    0.2332    ohms    Stator Resistance
X_1    5.6801    ohms    Stator Reactance
R_2    0.2301    ohms    Rotor Resistance refd to stat
X_2    5.8517    ohms    Rotor reactance refd to stator
X_M    232.8    ohms    magnetizing reactance
FullLoadSlip    0.0038    none    Full Load Slip
BarDepth    0.0426    meter    Equivalent Depth of rectangular rotor bar


============Selected Inputs ====================================            
VLL    11000    volts    Line To Line Voltage
SyncSpeedRPM    3000    RPM    Synch Speed in RPM (like 1200, 1800, 3600 etc)                

=========== Model Performance Against Targets==============                            
PerfVariable    CalcValue    Units    TargetValue    FractError    WtFactor    WtdSqdFractError    Comment
FullLoadAmps    108.05    Amps    108    0.0004    1    1.76471E-07    
FullLoadEff    0.9743    none    0.974    0.0003    1    8.31141E-08    
FullLoadPF    0.9107    none    0.91    0.0007    1    5.5495E-07    
FullLoadPower    1,826,453    watts    1,850,000    -0.0127    0.01    1.6201E-06    Redundant with previous 3
FullLoadTorque    5,836    N*m    5,906    -0.0119    0    0    Redundant with 1st 3
HLEfficiency    0.96173514    none    0.966    -0.0044    0.1    1.94919E-06    
HLPowerFactor    0.854856754    none    0.87    -0.0174    0.1    3.0297E-05    
LRC    554    Amps    604.8    -0.0834    0.1    0.00069529    
LRT    2,951    N-m    2,953    -0.0008    0.1    6.06664E-08    
NoLoadCurrent    26.69243514    Amps    21.6    0.2358    0    0    20% FLA for 2-pole (not used)
BD_Tq    15,791    N-m    14,766    0.0694    0.1    0.000482038    
X2overX1    1.03022015    none    1    0.0302    0.0001    9.13257E-08    Thumbrule - split is not critical to model
R2overR1    0.9868    none    1    -0.0132    0.01    1.73276E-06    Thumbrule
X1overXm    0.0251    none    0.05    -0.4972    0    0    Thumbrule- NOT USED
Full Load Slip    0.0038    none    0.003    0.2722    0.0001    7.41071E-06    Constrained to 5 rpm from nameplate
Bar Depth    0.0426    m    0.02    1.1295    0    0    Very lightly weighted guess

                    SWSFE    0.001222177    

I will post transient model results - hopefully tomorrow.
 

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(2B)+(2B)'  ?

RE: Effect of reduced voltage start on thermal severity of a motor start

pete - Thank you for all the work. You're a math genius.

Lionel - The SS man says that the 70% ramp start voltage and the ramp time cannot be adjusted. Is that correct ? I wanted the ramp voltage to start at 40% and the ramp time to 2 minutes just for the motor decoupled run. He is saying that would require a new SS. I was under the impression that these parameters were settable.

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Well the results of my simulation (attached) suggest that Lionel was 100% right - we see roughly the same transient spike whether DOL start or applying voltage to a motor already at speed.  I should have known better than to doubt him.

Slides 1-9 are normal DOL start of Muthu's motor with the specified load applied (0.1 to 0.5 of full load torque varying linealry over the speed range 0 to full speed)
1 – Speed vs time
2 – torque vs time
3 – torque vs speed
4 – Ia vs t
5 – Ia vs t (zoom-in)
6 – Ib vs t
7 – Ib vs t zoom-in
8 – Ic vs t
9 - Ic vs t zoom-in.

Slides 10-18 represent the same motor with same load applied, which is initially at 49hz at time 0 when it is energized.
10 – Speed vs time
11 – torque vs time
12 – torque vs speed
13 – Ia vs t
14 – Ia vs t (zoom-in)
15 – Ib vs t
16 – Ib vs t zoom-in
17 – Ic vs t
18 - Ic vs t (zoom-in)

The timing is such that phase A does not see a large transient. However phase B and C see a transient of approx 1200 – 1300A true peak.  It shows up on slides 7/9 for the normal DOL start as well as on slides 16/18 for the energization from 49hz.

I would have never have thunk it.  I did know from experience that an open-transition wye-delta start sees a current spike upon energizing the delta, but I always thought that had to do with residual current.
I can't quite reconcile the reason for this large jump given that the rotor circuit is close to an open circuit.  I'm going to ponder it a little more.  If anyone wants to explain some more, please do.
 

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

RE: Effect of reduced voltage start on thermal severity of a motor start

pete - I think the back-emf is zero at the time of energizing the motor, irrelevant of the speed. Would that explain the inrush ?

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

Yes.

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

RE: Effect of reduced voltage start on thermal severity of a motor start

pete/Lionel

I have the motor moment of inertia (64 kgm^2) and the compressor moment of inertia (48.7 kgm^2). Is that enough for calculating the pony motor capacity ?

 

Muthu
www.edison.co.in

RE: Effect of reduced voltage start on thermal severity of a motor start

Mutha - It does seem very odd that the soft-starter can't be adjusted. I have no idea what make or model you have though.

The moment of inertia will basically modify the acceleration time. What I mean is that on the same system (motor and load characteristics), more inertia will take more time to accelerate. You also need the load speed vs torque curve during starting to properly pick the pony motor. Then, a motor needs to be picked and a simulation of the start performed. The main motor will also have to be added as a load, probably simulating it as around a 60kW fan load would suffice.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
An interesting thing to note is that the steady state induction motor equivalent circuit predicts the DOL starting current very well, including the magnitude of the decaying dc component.

Based on the success of applying the steady state induction motor equivalent circuit to DOL starting, I had an (unrealistic) expectation that I could do the same for starting from near full speed.

So it suggests that I should go back and review why we are apparently successful in applying the steady state equivalent circuit to starting current in one case but not in the other...

For DOL start, the circuit looks like series combination of elements R1,  L1, L2, R2/1  (magnetizing inductance has very little role for staritng current)

For the start from full speed, the circuit was [unrealistically] expected to be  R1,  L1, L2, R2/s where s is small and creates an open circuit.   But R2/s = R2 + R2*(1-s)/s where the latter term represents an induced voltage.  As Muthu and Gunnar pointed out, that induced voltage would initially be zero, so for predicting the early behavior we should discard that element R2*(1-s)/s, in which case the circuit now looks very much like the circuit we used for DOL start.

Strictly speaking the steady state equivalent circuit has no business being applied for transient analysis of either case (start from locked rotor or full speed).  Krause has a transient equivalent circuit for induction motors which shows all the equations of the transient model as attached.   The q circuit contains voltage sources dependent upon the flux linkage of the d circuit... and the d circuit contains voltage sources dependent upon the flux linkage of the q circuit.   During start, all of these voltage sources are initially zero because the opposite-circuit currents are zero.   If we eliminate these voltage sources (due to zero voltage), we are left with a q circuit that looks identical to the steady state equivalent circuit... this explains the mystery why we were able to succsessfuly use the steady state equivalent circuit to predict transient effects of a DOL start. And also why we have the same initial response starting from rest or full speed (because in both cases all voltage sources are zero).

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

RE: Effect of reduced voltage start on thermal severity of a motor start

I never got back to this but I wanted to add another note about that textbook method linked in the other thread.

http://books.google.com/books?id=V1SXEnhCN88C&amp;pg=SA2-PA47&dq=This+motor+is+therefore+suitable+for+only+one+cold+or+one+hot&hl=en&ei=9_RFTNruLsG78gbnsYDWBA&sa=X&oi=book_result&amp;ct=result&resnum=1&;amp;ved=0CCgQ6AEwAA#v=onepage&amp;q=This%20motor%20is%20therefore%20suitable%20for%20only%20one%20cold%20or%20one%20hot&amp;f=false

The way this method adds the torques at the different speeds and then divides by the number of torque values to calcuate an average accelerating torque means an assumption was made that the motor is accelerating at a constant rate.

In real life, the motor would accelerate slower during the low torque periods and it would accelerate quicker during the high torque periods so I believe that more "weight" should be given to the low torque numbers and less "weight" given to the high torque numbers to calculate the real average or equivalent accelerating torque value. It's also possible to have data where the motor would really stall yet this method would show the motor would accelerate.

I guess I'm just saying that a more involved method besides calculating an average accelerating torque is required and be careful you don't get caught if you do an average accelerating torque calculation like that example shows.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
I agree.   Using an average torque is a rough and and easy approach, but a very inaccurate approach.   There two better ways to attack the problem of determining starting time:

1 – As described in  thread237-238806: Induction motor modelling.
Rotational equivalent of a = F / M:
dw/dt = (Telec-Tmech)/J     
dt = [J/(Telec-Tmech)] dw
time =  [J/(Telec-Tmech)] dw, w=0..wfinal    
Integrate the area under the inverse of accelerating torque vs frequency
Very specific instructions and spreadsheet given in the linked thread

2 – Simulation using any ODE solver to solve the initial value problem which amounts to integrating with respect to time:
dw/dt = (Telec-Tmech)/J     
w(0) = 0

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

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
A missing word is added below in bold:

Quote (electricpete):

time =  Integral{[J/(Telec-Tmech)]} dw, w=0..wfinal    

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

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
I didn't mean to jump past your comment– it was a good point that using "average accelerating torque" (averaged over the speed range) is not the corrrect way.    You gave one explanation.  I'll give an alternative one that leads to the same conclusion:

comparing then google book approach to the 1st of the 2 approaches I described above (2 Sep 10 10:59 ):

     In both cases we are looking at some kind of averaging with respect to speed.:

         In the EASA method,  the time is proportional to the inverse of the average torque.

        In my 1st method above, the time is proportional to the average of the inverse of the torque.

     The "inverse of the average" is definitely not the same as the "average of the inverse.", and that's why the google book approach is wrong.    The average of the inverse will weight the lower torques higher than the average of the torques as required.   If instead of the "speed-average torque", we were to use a "time-average torque" that would be the correct type of average accelerating torque to use for estimating speed.  In a time-averaged torque the lower torque values would again receive a highger weighting because we spend more time there, and the result would match the average (with respect to speed) of the inverse method, which is correct.     That corresponds to the same observation you mentioned that we spend more time at the lower torques so they should be weighted higher than a simple speed-averaged torque.

And now that I think about it, this "error" might be even a little more widespread that I thougth.  Look at the following  EASA on page 27.30 of the link below related to unloaded motor: they use the term "average accelerating torque" http://www.joliet-equipment.com/pdf/easahandbook.pdf

Quote (EASA Handbook):


TIME FOR MOTOR TO REACH OPERATING SPEED (IN SECONDS)
Seconds = Wk^2 * SpeedChange / (308 * Average Accelerating Torque)
Average accelerating torque = {   [(FLT + BDT)/2] + BDT + LRT   } / 3 }
Where: BDT = Breakdown torque
FLT = Full-load torque
LRT = Locked-rotor torque
I'm sure everyone here has seen that formula before.   I have always known it was approximate, but I also assumed it was at least based on sound reasoning. Now I'm not so sure.

If the term "average accelerating torque" is intended to describe a simple an average with respect to speed, then it's wrong.  

So what exactly is this formula  ({   [(FLT + BDT)/2] + BDT + LRT   } / 3) supposed to represent, and where did it come from (what assumptions/approximations) ?

If someone approximated the torque speed curve as going linearly from LRT at s=1 to BDT at s=0.333, and then linearly from BDT at s=0.333 to FLT at s~0.... then that torque speed profile would result in the expression above for speed-averaged torque.  But then again s=0.333 sounds way to low for BDT,..... and not sure how FLT even comes into the mix since the formula involving motor torque (without load) cannot even be applied to loaded start.

If we wanted one summary torque statistic to properly represent the whole curve in an unloaded accelerating time calculation,  it would be the inverse of the speed-averaged-inverse-torque.   If you did a stepwise approximation to the torque-speed curve with N intervals, then the inverse of the speed-averaged inverse torque would be a product of N torques in the numerator over a sum of of N torque products (each product involving N-1 torques) in the denominator.   If you did a linear approximation to the torque speed curve, then the summary statistic would include natural logarithms similar to the spreadsheet linked above.  The EASA expression doesn't look like either of those.  At this point, without knowing the exact origin of that EASA expression I am inclined to believe they were just trying to convey a simple average torque with respect to speed, which is downright wrong and tends to steer people in the direction of the google book (averaging torque over speed) which is again wrong.

What do you guys think about this mysterious formula {   [(FLT + BDT)/2] + BDT + LRT   } / 3 } ?
What does it represent?
Where did it come from?
 

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

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
I guess I'm bending over backwards to give EASA the benefit of the doubt, but it obscures my meaning.


What seems to be the obvious apparent answer to me seems to be is that the expression  Average Acceleraing Torque = ({   [(FLT + BDT)/2] + BDT + LRT   } / 3)   was intended to represent a speed-averaged torque.... and therefore whoever came up with the equation and applied it to acceleration of an unloaded motor didn't really think about what they were doing.  And seems to have been repeated again and again by people that never realized it was a fundamentally flawed approach.  I'm certainly one of those people....  I never stopped to think carefully about that particular formula until Lionel's comments highlighted the silliness of averaging torque with respect to speed for an acceleration calculation.

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

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
In fairness to developers and users of the formula, I should acknowledge that we all expect/accept a tradeoff between accuracy and simplicity in our formulas.  (inaccurate formula is someomtes preferred over more accurate formula if the inaccurate formula is simpler).

But to follow up a little further on the folly of the equation  A.A.T = ({   [(FLT + BDT)/2] + BDT + LRT   } / 3) .....

Note that BDT plays a big role in that number... it accounts for half (1.5/3) of the average.    But if we were doing it the right way, BDT accounts for only a very small contribution of the equivalent uniform accelerating torque, because
1 – the higher torques should be weighted less than the smaller torques
2 – bdt is an extreme max value on the curve... not representative of any interval on the curve
3 – the duration of the torque peak associated with bdt is usually only 10% or 20% of the speed interval.
Put all that together and I'd say you could almost leave bdt out of any simple equation without much loss of accuracy.

But it doesn't end there.  Put that together with an assumed linear increase from LRT to BDT. (rather than lingering near LRT like a lot of curves), and a final value of FLT rather than No-load torque... and we have a recipe for massive overestimation of equivalent uniform accerating torque.... and massive underestimatino of accelrating time.   

If I am going to use an inaccurate equation for the sake of simplcity, I'd like it not to err in one direction and particularly not in the non-conservative direction!  And underestimating accelerating time is the non-conservative direction in any practical scenario I can think of.

All that said, I'm not sure what I would come up with for a "better" equivalent uniform torque that can be used to estimate unloaded acceleration AND and that would have comparable simplicity to the EASA one.  I'd have to think about it.  Any suggestions on that one?

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

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)

Quote (electricpete):

and a final value of FLT rather than No-load torque.
I'll withdraw that particular criticism.  Full load torque is a fair endpoint for an acceleration calculation since at that point current has dropped to fla.

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

RE: Effect of reduced voltage start on thermal severity of a motor start

Don't forget that the ESA formula assumes no required load torque which is another non-conservative assumption.

I've never really though about a simple formula. The main problem I see is that it's really impossible to calculate the equivalent average accelerating torque unless you actually know the acceleration time, which first requires more math to calculate.

I guess if you wanted a simpler method then I would suggest breaking the torque curve into parts where the curve is fairly linear and calculate the acceleration in a piecewise manner over those parts of the curve. Then, add the acceleration times together. You could likely break most torque curves into about 5 pieces and do a reasonable job. More or less the same idea as integration but on a more coarse level.

I bet most people never though that calculating the details associated with the acceleration of an electric motor would be so involved.
 

RE: Effect of reduced voltage start on thermal severity of a motor start

(OP)
Agreed, definitely only applies to unloaded calc which I don't think EASA mentioned, but I expect people would figure out.

Attached is an example calc for 200hp 3600 rpm motor whose curve is shown here:
http://www.reliance.com/pdf/pdf/aced/W06884-A-A001.pdf

Using my attached spreadsheet:
1 - the start time is 1.14 seconds as calculated using many segments as shown on tab "main".  I believe that is pretty close to exact.
2 - the start time is 1.12 seconds if we break the curve into 4 segments and as shown on tab "4segments".  Maybe that is lucky, but I'd expect to be within 10% in any case.
3 - the start time is 0.76 seconds using the EASA formula as shown on tab main on top right hand side (only 67% of the more exact solution from 1).
(in all cases assumed unloaded, full-voltage start).
A pretty big error for #3 imo, and in the wrong direction.  
I think in 90% of cases you would get closer than the EASA formula if you just used AAT = LRT.  A simpler formula than EASA which would get you closer than EASA.  I'm not saying LRT is a great estimate, but better than the EASA formula which again is just... silly.  

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

RE: Effect of reduced voltage start on thermal severity of a motor start

Good example. That is exactly what I had in mind when I posted to try breaking the curve into 5 part curve.

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