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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[electricpete]

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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[electricpete]

Any thoughts?

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[LionelHutz]

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.

 

[electricpete]

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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[LionelHutz]

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.

 

[electricpete]

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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[edison123]

pete

Attached is the motor spec sheet and a soft-start design sheet for a 11 KV, 1850 KW motor. Tell me what you think of this.

Muthu
www.edison.co.in

 

[edison123]

Motor specs sheet

Muthu
www.edison.co.in

 

[electricpete]

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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[electricpete]

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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[electricpete]

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

What is CS?

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[electricpete]

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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[electricpete]

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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[electricpete]

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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[electricpete]

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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[edison123]

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

http://www.edison.co.in

 

[electricpete]

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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[electricpete]

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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[electricpete]

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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