Maximum speed a motor can be run at 4

[rockman7892]

I have a 1hp 480V 1800rpm motor being driven by a 1hp vfd. This motor is on a conveyor feeder application that is feeding material. It is not feeding enough material so production wants to increase the speed of the drive. The drive is already running with a max speed setpoint of 120hz.

I notice on the motor itself it has listed a safe maximum speed of 3600rpm. I am therefore hesitant to run this motor above this setting. When asked why we could not run above this 3600rpm rating the only thing I could think of was that if we go above this rating, the voltage to the motor will be limited and therefore the V/Hz will decrease and there may not be enough flux in the motor to produce the required torque. The motor may then not have enought torque to move the load and will lock up and thus burn the motor up. Is my explanation correct or am I way off?

So I am now looking into options for speeding this motor up. Option A will be to find a higher rpm motor, or a motor that can handle a higher drive frequency output.

Option B would be to purchase a larger motor that would have enough torque at a higher speeds in order to move the load.

Am I on track with the higher speed vs avaliable torque explanation being the limiting factor for not running this motor over 3600rpm or is there something else?

 

[edison123]

The centrifugal forces vary as square of the speed. So the motor will fail mechanically (shaft, bearing etc.).

 

[fangas]

You are correct. Above its base speed, the motor will be operating in the constant Hp range. In order to maintain that 1 Hp above 1800 RPM, the torque must go down.

*If* I did the math right, your 1 Hp motor developes about 3 Ft/Lb at 1800 RPM. At 3600, it will develope about 0.73 Ft/Lb.

You should really consider a 2-pole, 1 Hp replacement motor.

Ed

 

[rockman7892]

Thanks edison

Does it have anything to do with the torque as I mentioned above or strictly just mechanical limitations.

So it sounds like if it is a mechanical limitation, then then I would need to buy a higher rpm motor, or one that is capable of being operated faster?

This is going through a gearbox, so I imagine there would be limitations on the gearbox as well.

 

[rockman7892]

fangas

A two pole motor would have a synchronous speed of 3600rpm. Would be then be able to run this motor faster at lets say 120hz to get the higher speed required.

I come up with 3 lb*ft at 1800rpm and 1.45 lb*ft at 3600rpm by dividing the speed into the a HP of 1.

 

[edison123]

I was referring to only the mechanical limitations. Yes, all the connected equipments like gear box and the conveyor itself (!!!) would be highly stressed.

The torque is another issue as fangas says.

Your "management" needs some engineering education. This doesn't seem to me a good idea. It might be better to increase the conveyor size (with attendant motor + gearbox upgrade) to achieve more throughput.

 

[fangas]

Rockman,

Edison is correct. You start exceeding 3600 RPM with a motor, all manner of bad things are possible, mechanically.

The conversion to torque in the *approximate* constant Hp range is not linear. Being I'm a lazy sort, I'd like to direct you here;

http://www.powerqualityanddrives.com/torque_constant_horsepower/

There's a fine description of of what happens and why.

Ed

 

[edison123]

Nice link fangas.

 

[ScottyUK]

I think I uploaded this once before but it shows the behaviour of the motor above base speed quite well. The example is a 200kW 750rpm machine but the principle applies to any 3-phase induction machine.

http://files.engineering.com/getfil...5203c5f1b&file=VFD_ConstTorqueConstPower.xlsx

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If we learn from our mistakes I'm getting a great education!

 

[jraef]

Rockman7892,
Your torque calcs are a little off. The Tq./speed relationship is not linear above base speed. It varies by the square of the effective voltage / frequency ratio.

But to make it simpler to visualize, think of what happens when you apply lower voltage to a motor at base frequency. We already know that torque varies by the square of the applied voltage, so if for example you apply only 230V with a soft starter, the torque is reduced to .50² or 25% of rated torque. To now apply this to a motor running ANOVE base speed, we use the V/Hz ratio. A motor operating at 460V 60Hz will have a V/Hz ratio of 7.666, so at only 230V the V/Hz ratio is then only 3.833. If we calculate the V/Hz ratio when we operate that motor ABOVE base speed, the same effect applies to the torque. So at 120Hz, the V/Hz ratio becomes 460/120 = 3.833 again, so the effective voltage becomes the new ratio times the BASE frequency (3.833 x 60 = 230V), and the motor output torque is again reduced to 25% of rated. In your case, this appears to be enough because we can assume the motor is working (and not overloading), so obviously someone thought about this when sizing it in the first place.

Most motors are built to handle 25% more speed than what the 2 pole speed rating is. So that means you can run a 4 pole 1800RPM motor to about 4500RPM max. In that case though, the frequency applied to that 4 pole motor would need to be 150Hz. So your V/Hz ratio drops to 3.07, which is like having only 184V applied to the motor (3.07 x 60Hz), which is roughly 40%, and the output torque will drop to 16% of rated. So in your case, you will have increased the speed 25%, but cut the operating torque to 64% of what you have now!

If you change to a 2 pole motor, IT can still only be increased to 125% speed, but since it STARTS at 3600, that means the frequency only needs to increase by 25% to 75Hz. So apply the rules and the V/Hz ratio stays higher at 6.133, which corresponds to a torque equivalent of applying 368V, which is 80% of normal and thus the torque will end up at 64% of rated. So you will actually end up with MORE torque at the shaft than you have now. That of course is going to throw your gearbox ratios out the window however, so I doubt that is an option for you.

To ensure then that you get at least the torque you have now, without having to change the gearbox, you would need to increase the motor HP size by 56%. If by chance you still had a little overhead left in the amount of torque you are using, you might get away with a 1-1/2HP motor, but if it were me, I'd go to 2HP.

But before you ASS-u-me the motor can run at 125& of 2 pole speed, you must ask the mfr. I would also do the same with the gearbox mfr as well.


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[electricpete]

But before you ASS-u-me the motor can run at 125& of 2 pole speed, you must ask the mfr

I think we already have the manufacturer guideance on this subject:

I notice on the motor itself it has listed a safe maximum speed of 3600rpm

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[DickDV]

jraef, in the constant hp range, the motor's torque goes down as a function of the inverse of the overspeed, not the square of the V/Hz ratio.

The breakdown shortterm torque comes down as the inverse of the overspeed squared.

Thus, the torque at 3/2 base speed is 2/3, at 2/1 base speed is 1/2, etc.

The breakdown torque at 3/2 base speed is 4/9, at 2/1 base speed is 1/4.

If you plot these curves, they come together at double speed so effectively you have no overload torque available anymore.

 

[waross]

Reconnect the motor for 230 volts and set the V/H ratio for 230 volts. Continue tofeed the VFD with 480 volts. At 120 Hz, your V/Hz ratio will allow 200% voltage or 480 volts. With the extra torque you may be able to change the reduction ratio of the mechanical drive to the conveyor to increase the conveyor speed.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

The others have already posted the effects on over speeding a motor so I just will strongly second what Bill just posted. This will effectively do the same thing as purchasing a 2hp, 3600rpm, 2-pole motor. This should also provide enough torque to be able to over speed the motor further if you decide to attempt that instead of varying the gearbox and/or belt ratio.

 

[rockman7892]

O.K. after a couple times of reading through everything that was posted I think I finally understand the concept and did some calculations of my own.

Using the existing 1hp, 4-pole, 1800rpm motor I current have I came up with the following toruqe values for the different overspeeds:

@60 HZ and 1800rpm Torque = 2.91 lb*ft
@120 HZ and 3600 rpm Torque = .727 lb*ft
@150 HZ and 4500 rpm Torque = .4656 lb*ft

If I upgrade to a 2hp, 2-pole, 3600rpm motor as others suggested I came up with the following values:

@60 HZ and 3600 rpm Torque = 2.91 lb*ft
@75 HZ and 4500 rpm Torque = 1.86 lb*ft

I notice that the 4500 rpm torque for this motor is much larger for that of the 1hp motor at this speed.

I was trying to run some calculations for the suggestion that Warros made but I'm not sure that I approached it right. Heres what I did.

If the motor is re-wired for 230V then new V/Hz ratio is 3.83 at 60hz. So if we set the drive up for a V/Hz ratio of 3.83 then at 120Hz the V/Hz ratio would be 4. I used the following calculations.

Torque at 60hz = (2.91)(3.83/3.83)^2 = 2.91 lb*ft
Torque at 120hz = (2.91)(4/3.83)^2 = 3.17 lb *ft

The torque value for 120hz seems much larger than that of the previous two cases, but I'm not sure if I'm doing this right by using the torque of 2.91 which was calculated for a 460V 4-pole motor. If I'm off base can you please explain the theory behind wiring the motor for 230V and supplying VFD with 480V?

The issue I have with the 230V operation is that the motor is only a three lead motor wired for 460V. I guess I would have to send it out to be re-wound at 230V, which in that case it may be worth just buying a 2-pole 2hp motor.

 

[rockman7892]

The one thing I thought of after posting was determining how much torque is being used in the current operation at 120hz.

If I were to measure the current being drawn by the motor and calculate the % of full load current being used (factoring in reactive current)by the motor and equate this to a % torque could I find aproximately how much torque if being produced by the motor? Could I then compare this torque to the calcualted value I have for 4500rpm and see if at 4500rpm we will still have enough torque to move the load? If I see that we do then could we run the existing motor up to 4500rpm.

Is the 125% or 2-pole speed the rule of thum for all motors? This is the first motor I have ever seen a max rpm listed before. Since all 2-pole motors are 3600rpm then does this mean that the maximum any motor can be run at on a 60hz system is 4500rpm? What if a faster speed was required, would you have to find a higher speed motor at nominal frequency?

 

[fangas]

for the typical off the shelf, commercial motor, the bearings, balance and other mechanical things are the same as use in a 2-pole. Hence, the 2-pole reference.

Ed

 

[rockman7892]

In regards to my questions about monitoring the torque I went and looked at the VFD manual and see that you can meter torque current, estimated torque, and commanded torque. This should give me an idea of the torque being used to move the load at the current speed of 1200rpm. Maybe I could then compare this toruqe value to the calculated torque value for higher speed as described above?

 

[rockman7892]

I was able to look at the drive an find an paramater that showed the torque being used. At the particular time I looked at the drive the motor was being run at 65HZ and the drive showed that is was using 11% of the avaliable torque. If I watch this torque paramater and see what it is showing when the drive is run at 120HZ should this be a good estimate of the torque required for the motor?

How does a motor have an rpm of faster than 7200 rpm (1 pole)?

 

[edison123]

Speed = 120xFrequency/Poles. Minimum poles = 2. Increase the frequency beyond 120 Hz to go over 7200 RPM. Of course, the ratio V/Hz has to be held constant to provide the required torque. Look at waross's suggestion on how to over 60 Hz.