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]

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).

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(2B)+(2B)' ?

 

[LionelHutz]

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=V1...able for only one cold or one hot&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.

 

[electricpete]

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
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

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[electricpete]

A missing word is added below in bold:

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

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[electricpete]

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

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?


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[electricpete]

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.

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[electricpete]

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?

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[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.

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[LionelHutz]

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.

 

[electricpete]

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.

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[LionelHutz]

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