Motor sizing

[JeanMicheling]

I'd like to know what is the best way to size a motor in function of a tool inertia. My tool inertia is 72lbf-ft^2 and referring on motor's chart, a 75 hp would be required. I find that really big for a tool that just need to reach full speed and their will be no external applied to the tool (because it's a bench test). Does someone hava an idea?

 

[jraef]

How the heck did you arrive at 75HP? Even if it were a 3600RPM motor I see that it would provide 110 ft. lbs. of FLT, and upwards of 250 ft. lbs. of Breakdown Torque! Are you trying to use a 5500RPM motor?

First determine the torque and speed you need to RUN the tool, irrespective of acceleration. That will give you the minimum HP required from the motor. Then calculate the Average Accelerating Torque from that motor using the formula:

AAT = [(FLT + BDT)/2] + BDT + LRT
------------------------------
3

Then see if it starts fast enough for you by using the fallowing formula:

Seconds to Full Speed = WK²(ft.lbs.) x Speed Change (RPM)
--------------------------------------------------
308 x AAT

If it doesn't accelerate fast enough for you, go to the next size motor and start over until it does.

And don't forget to factor in any gear reductions or increases!

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[jraef]

LOL, I guess "fallowing" is a legit word according to the Google spell checker! Of course, I mean "following".

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[itsmoked]

Leaving the dirt to lay around fallow I would guess.

 

[waross]

How about a VFD?

 

[JeanMicheling]

Jraef,

I found 75 hp following the salesman recommandation. he said that the wk^2 value on the motor's chart is the maximum inertia a motor can carry. I don't think he's right (I'd say it's the rotor inertia?). You said I should determine the torque I need to run my tool but as I said, at full speed (3600RPM) there is no torque except motor's restriction. So, once the tool has reach its full speed, no more external torque will be applied to the motor. I tried to use the formula you gave me and I have some interrogations. First, the wk^2 calculation, it says that it's motor rotor inertia+load inertia*load rpm^2/motor rpm^2. How the load speed could be different than the motor speed? Secondly, the Seconds to full speed doesn't tell me which motor size I should choose. Theorically, even if you have a tool with a lot of inertia, the smallest torque will be able to accelerate it to full speed. In practice, you will agree that if your motor is too small, it will overload before reaching its full speed. So, how can I deal with that? Is there a minimum time to full speed that can use as a thumb rule? Can I have this information with the motor specs?

Thanks,

 

[CJCPE]

It appears that your salesman's recommendation may be based on information in NEMA MG1. Briefly summarized:

Three phase induction motors described in MG1 shall be capable of accelerating load WK^2 referred to the motor shaft as listed in table 12-7. Under conditions: motor not to be damaged by heating during starting, rated voltage and frequency, load torque = rated at rated speed and proportional to speed^2 below rated speed, start once if motor at rated temperature, twice if motor at room temperature.

The listed WK^2 figure for a 75 Hp, 3600 RPM motor is 71 lb-ft^2.

You might assume that for no-load starting, a smaller motor could safely start the same inertia as long as the starting time does not exceed the time required to start the inertia listed in the table under the load conditions given in MG1.

With a variable speed drive, you could limit the starting torque and current to the motor's rated torque and current and safely take as long as you want to accelerate the load. You could as small a motor as you want, but at some point the load windage and friction would become significant.

 

[Marke]

Hello JeanMicheling

Smoe motor data sheets do list the maximum starting inertia of the motor, others do not.
When you start an induction motor, there are very high slip losses in the rotor of the motor. The total of the slip losses is dependant on the effective load inertia as seen by the motor shaft. (you need to adjust for any speed change) There are major differences between rotors when it comes to their ability to absorb this slip energy. Some motors are rated at less than 10 seconds Locked Rotor Time and others can be higher than 40 seconds for motors of the same rating.
The figures are typically quoted as maximum load inertia, or as maximum Locked Rotor Time, or as maximum full voltage start time.
The maximum load inertia is very different from the rotor inertia often quoted in data sheets.
If the motor is doing next to no work once it is up to speed, then your selection needs to be on the rotor thermal capacity, not KW rating.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[aolalde]

I have the same criteria exposed by Cjcpe. Motors are manufactured to comply with the applicable standards (NEMA, IEC, JIS etc) .The standard defines the maximum load inertia that a motor is capable to accelerate successfully at full voltage and frequency.
The temperature rise for the stator and rotor windings is a function of the starting current handled and time, the larger the inertia the longer the required accelerating time, as per Jraef formulas.
A Variable Frequency Drive could help to reduce the motor size as far as the speed-torque curve of the load , including windage will not exceed the motor and driver capacity (increasing frequency synchronous start with reduced current).

 

[jraef]

Yes, but if he does not need variable speed at all, a VFD is a waste of money compared to just selecting the correct motor to begin with.

JeanMicheling,
You mention not needing ANY torque once at full speed? There must be something you don't understand, or I am not understanding something about your "tool". Torque at speed = work. Speed alone does nothing, unless your only goal was to get to a speed and then coast back down for some reason, but I would hardly call that a "tool", maybe just a test bench accelerator. If that is all it is, then lets go from there. The term "tool" implies it will be doing something. So at 3600RPM, your tool must be doing something that requires torque at the tool surface, a grinder, cutter, polisher, mixer etc. etc. All of those things require torque! So what is that "something" and how much torque does it require at 3600RPM? If you cannot answer that basic question, you need to get someone involved that can instead of a motor salesman.

Eng-Tips: Help for your job, not for your homework Read faq731-376

 

[JeanMicheling]

Jraef,

Sorry about the word ''tool'' but it's what it is. We put the tool on a bench test to run a spindle in. Thus, as I said, no load will be applied to the ''tool'' or if you prefer we can call it a wheel of inertia. Also, there will be a VFD for sure because we need to control the speed. I'll use a belt to drive the spindle. But since the spindle need to run at 4200 rpm in some cases, the ratio can't be so big. Can I still use the wk^2 formula even though my motor runs faster then 3600 RPM? It seems that there is no way to exactly know which motor I need in this case especially if I use a VFD.

 

[Marke]

Hello JeanMicheling

If there is a VSD, then everything changes. The inertia is only relevant in terms of the maximum rate of acceleration and deceleration that you require.
The maximum load inertia of the rotor is only relevant where the motor is under high slip conditions and this does not occur with a VSD.
The issues are : Residual torque required at full speed and the acceleration/deceleration times, otherwise you can use a very small VSD and motor with a long ramp up time. The smaller the motor, the longer the ramp up time. The limit is really the residual torque requirement at maximum speed.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[JeanMicheling]

Hello Mark,

When you say residual torque, you mean torque other than motor's restriction? We have a 2 hp 145T design motor at my shop with a vfd. They tried once to run the spindle with the tool but couldn't. I was not working there at that time but the mechanic told me that the spindle didn't want to accelerate. Is it possible the drive wasn't set correctly?

 

[Marke]

Hello JeanMicheling

There will be a torque output required to keep the spindle turning and if you torque output is less than that, you will not get anywhere. If it is free to turn bu hand, then you should be able to use the small motor and drive with a very long ramp time and see how far you get. I would set the acceleration time very long initially, you can always shorten it later. The first thing to do is to get the machine up to speed, or as high as it will go and this will give you an indication of the requirements.
If the VSD was set up with a short acceleration time, it would give up the gohst pretty quickly.
My suggestion is to set the VSD up for a long accel ramp and for a free wheel stop. i.e. the drive shuts down rather than trying to pull the speed down. Once you have got things going, you then have some numbers to work with.
If the drive has a current limit or a torque limit, set these up to 100% - 120% and that should prevent any trips.
I think that it is a case of trial and error at this stage.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[CJCPE]

If you have a 2 Hp, 3600 RPM motor, your VFD should provide 3 lbs-ft or torque without exceeding the full load motor rating. If you use 3 lbs-ft as the average accelerating torque, the formula provided by jraef will tell you that it will take 280 seconds to accelerate to 3600 RPM.

You must have bearings external to the motor. They represent a small friction load. At 3600 RPM, the spinning tool must also represent some windage load. The windage plus friction may not be negligible compared to 3 lbs-ft.

If you have a belt and sheaves, they introduce additional windage and friction that needs to be considered. More importantly, the inertia of the tool must be reflected to the motor shaft using reflected inertia = load inertia*load rpm^2/motor rpm^2 as you mentioned previously. The inertia of the motor, sheaves, belts etc. should be added to the total, but that probably doesn't add very much compared to the 72 lb-ft^2 tool.

Considering windage, friction and a 3600:4200 RPM sheave ratio the 2 Hp VFD acceleration ramp may need to be set for 7 minutes or longer. It may or may not have sufficient adjustment range for that. If the VFD can get the tool up to speed, as Mark pointed out and I will emphasize FREE WHEEL STOP IS ESSENTIAL. The tool must be allowed to coast down. If you try to ramp it down with the drive it will trip. If it is not well protected, it could fail catastrophically.

 

[mc5w]

At 1 place where I worked we has two 300 HP motors on soft starts and they would take about a minute to get up to full speed. 1 soft start went senile because of invisible lightning along the lines of St. Elmo's Fire. 480 volt ungrounded systems tend to blow up motors because of destructive static electricity buildup similar to how 97% of lightning damage in a telephone cable plant comes from silent lightning and invisible lightning.

One of these motors was a belt driven crusher and the other a direct drive blower for the baghouse. In the case of the crusher we were using a 300 HP motor as a 125 HP motor because of the torque peaks that crushers have.

 

[JeanMicheling]

Could you tell me what are the bearings in that crusher spindle?

 

[aolalde]

In reality the accelerating torque should be calculated for small speed increments (dw) and the load (at least windage) torque substracted from the motor torque.
AATn = MTn-LTn

Then the time calculation integrated for the full accelerating speed range.
The load torque normally increases with speed.

 

[JeanMicheling]

Aolalde,

The windage load, is it the load produce by the tool air restriction?

 

[waross]

In this instance I think you should consider it to be the total air restriction, or drag of all your rotating parts. Think of the parts acting as little fans, moving air.

I try to look at the whole picture, and some times some of the factors become imaterial. I think that the two inportant factors here are windage, and acceleration time.
While windage is a factor, it is not a serious factor at 3600RPM or 4200 RPM.

Suggestion; Decide on an acceptable maximum acceleration time and post it. I'm sure someone will be willing to work backwards from that to give you a motor size. I strongly suspect that the nature of your load is such that the limiting factor will be acceleration rather than the windage.
Also, with a VFD control, you may be able to direct connect the load to the motor with out the need for belts, and still attain 4200 RPM. Belts can be more of a load than windage with small HP aplications in this speed range.
yours