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

[electricpete]

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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[edison123]

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

http://www.edison.co.in

 

[electricpete]

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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[LionelHutz]

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.

 

[edison123]

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

 

[LionelHutz]

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?

 

[edison123]

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

http://www.edison.co.in

 

[edison123]

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

Muthu

http://www.edison.co.in

 

[electricpete]

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)

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 .

What program do you use?

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)

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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[electricpete]

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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[LionelHutz]

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.

 

[electricpete]

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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[LionelHutz]

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.

 

[electricpete]

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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[electricpete]

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:
[tt]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
[/tt]
I will post transient model results - hopefully tomorrow.


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[edison123]

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

http://www.edison.co.in

 

[electricpete]

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.


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[edison123]

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

http://www.edison.co.in

 

[Skogsgurra]

Yes.

Gunnar Englund

http://www.gke.org

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

 

[edison123]

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

http://www.edison.co.in

 

[LionelHutz]

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.