I have a question as to how I can calculate how much overspeed I can get away with on an ac drive and motor. The motor is rated at 60Hz 12.7A @1775rpm. The drive is a 22A. Currently with the process running at 320FPM the motor pulls about 53% current or 6.73A continuously at 1450 shaft rpm. I want to find out how much I can overspeed above 60Hz while maintaining the torque required.
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Overspeeding AC drive/motor 1
- Thread starter jde777
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1
- #2
You are dealing with two limits. First, the motor,gearbox,machine has a maximum safe speed. That must not be exceeded.
Second, you must determine at what point the load torque becomes higher than the available motor torque. Obviously you can't operate above that point.
The second item above involves identifying the load machine's torque characteristic which can sometimes be difficult. It also involves the motor torque curve which can be calculated from the nameplate data. Take the hp or kw at motor base speed and calculate the torque. The motor will be able to produce that torque from zero to nameplate base speed. Above the base speed, the motor torque will fall off from the base speed value by the inverse ratio of overspeed. So, for example, if you go to 10% overspeed, that is 110%/100%. The inverse is 100/110 so the torque drops to the base speed value times 100/110. At 20%, it would be 100/120, and so on.
I would not expect to go any further than 90hz on any non-inverter-duty motor as the above derate rule starts to fall apart and the motor torque drops off even faster. I would limit my thinking to the 60-90hz range. If you are in a 50hz place, that would be 50-75hz range instead.
Second, you must determine at what point the load torque becomes higher than the available motor torque. Obviously you can't operate above that point.
The second item above involves identifying the load machine's torque characteristic which can sometimes be difficult. It also involves the motor torque curve which can be calculated from the nameplate data. Take the hp or kw at motor base speed and calculate the torque. The motor will be able to produce that torque from zero to nameplate base speed. Above the base speed, the motor torque will fall off from the base speed value by the inverse ratio of overspeed. So, for example, if you go to 10% overspeed, that is 110%/100%. The inverse is 100/110 so the torque drops to the base speed value times 100/110. At 20%, it would be 100/120, and so on.
I would not expect to go any further than 90hz on any non-inverter-duty motor as the above derate rule starts to fall apart and the motor torque drops off even faster. I would limit my thinking to the 60-90hz range. If you are in a 50hz place, that would be 50-75hz range instead.
- Thread starter
- #3
Thanks DickDV
I do not plan on going above 90Hz. The application is on a paper machine. This cylinder roll actually is a speed regulator while not lowered to the paper and becomes a torque regulator the moment it touches the paper and is used for watermarking. Just was curious as to how much more we can get out of the existing setup without changing gearin.
I do not plan on going above 90Hz. The application is on a paper machine. This cylinder roll actually is a speed regulator while not lowered to the paper and becomes a torque regulator the moment it touches the paper and is used for watermarking. Just was curious as to how much more we can get out of the existing setup without changing gearin.
Long discussion on this thread
motors, generators & controls engineering - Maximum speed a motor can be run at</a>
Good description at
Ed
motors, generators & controls engineering - Maximum speed a motor can be run at</a>
Good description at
Ed
- Thread starter
- #5
Thanks Fangas
DickDV in your example does that formula apply to AC motors?
According to the link by Fangas @90Hz I will only have 44.4% torque.
However if I use your example then I will have 66%.
Huge difference as the process runs at approx 40% torque.
DickDV in your example does that formula apply to AC motors?
According to the link by Fangas @90Hz I will only have 44.4% torque.
However if I use your example then I will have 66%.
Huge difference as the process runs at approx 40% torque.
- Thread starter
- #6
Skogsgurra
Electrical
The 'constant torque' wording is a misnomer. Available torque does not stay constant, but drops off with V/Hz squared.
On the other hand, cooling gets better when speed is increased. So, for a limited region above base speed, one can say that available torque is constant from a thermal point of view. But sooner or later (and dependent on NEMA Class) you will hit peak torque, which is also V/Hz squared dependent.
Say that you have a Class A motor, your peak torque will then be 2.7 times rated torque. So you can increase speed to sqrt(2.7) times base speed before hitting the peak torque. That corresponds to around 65 % above base speed.
But then, your slip will be in the order of 20 percent (huge) so your rotor heat will be immense. And the improved cooling due to higher speed will not help.
I am afraid you will have to understand all these subleties before deciding what speed increase is available. And, please note that NEMA B and C are a lot worse when it comes to peak torque. So you will not be able to run anywhere near 65 percent overspeed. A more realistic speed limit would be around 40 percent.
If, however, you can limit torque to half the rated torque (which seems to be a possibility when the egoutteur is in contact with the wire and stock) then you can probably go to 65 percent overspeed.
In my world, an egoutteur is driven lightly in order to avoid any force between its surface and the wire. So, it should be possible to run with minimal torque.
Drives are complicated sometimes and there is no YES/NO answer. You have to consult someone that does this for a living - or play it safe and specify a new drive and motor. Probably cheaper that way.
Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
On the other hand, cooling gets better when speed is increased. So, for a limited region above base speed, one can say that available torque is constant from a thermal point of view. But sooner or later (and dependent on NEMA Class) you will hit peak torque, which is also V/Hz squared dependent.
Say that you have a Class A motor, your peak torque will then be 2.7 times rated torque. So you can increase speed to sqrt(2.7) times base speed before hitting the peak torque. That corresponds to around 65 % above base speed.
But then, your slip will be in the order of 20 percent (huge) so your rotor heat will be immense. And the improved cooling due to higher speed will not help.
I am afraid you will have to understand all these subleties before deciding what speed increase is available. And, please note that NEMA B and C are a lot worse when it comes to peak torque. So you will not be able to run anywhere near 65 percent overspeed. A more realistic speed limit would be around 40 percent.
If, however, you can limit torque to half the rated torque (which seems to be a possibility when the egoutteur is in contact with the wire and stock) then you can probably go to 65 percent overspeed.
In my world, an egoutteur is driven lightly in order to avoid any force between its surface and the wire. So, it should be possible to run with minimal torque.
Drives are complicated sometimes and there is no YES/NO answer. You have to consult someone that does this for a living - or play it safe and specify a new drive and motor. Probably cheaper that way.
Gunnar Englund
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
I think Gunnar's post is a bit confusing in that I believe his term "available torque" means peak torque. Peak torque does indeed come down as the square of the overspeed.
Continuous rated torque, however, does come down as the inverse of the overspeed so you should have 66% continuous torque available at 1.5 times base speed. As Gunnar points out, there won't be any overload torque available at that maximum speed, however. That shouldn't be any problem with your application.
Continuous rated torque, however, does come down as the inverse of the overspeed so you should have 66% continuous torque available at 1.5 times base speed. As Gunnar points out, there won't be any overload torque available at that maximum speed, however. That shouldn't be any problem with your application.
- Moderator
- #9
There is a techniques that often doesn't work. Increasing the voltage will allow you to maintain the V/Hz ratio and gain HP as you go over speed.
The higher voltage combined with the PWM waveform and the associated harmonics and ringing will possibly destroy the motor in a short time.
One technique is to reconnect the motor for a lower voltage than the applied voltage.
The windings of an inverter duty 460V motor are all suitable for 480 volts. If a 230/460V motor is connected for 230 Volts it will often withstand 480V at 120 Hz. I understand that some manufacturers of compressor skids are using this technique to double the horsepower of their motors.
Another technique is to interpose a transformer between the drive and the motor. The transformer will filter out almost all of the PWM hash and you will then be able to deliver relatively clean over frequency/over voltage to the motor and maintain the V/Hz ratio above base speed.
I have seen these techniques but I have not been party to the design. There may be something that I don't know about them.
It is a field for you to research.
Gunnar is one of our drive Gurus. Be guided by his suggestions.
Bill
--------------------
"Why not the best?"
Jimmy Carter
The higher voltage combined with the PWM waveform and the associated harmonics and ringing will possibly destroy the motor in a short time.
One technique is to reconnect the motor for a lower voltage than the applied voltage.
The windings of an inverter duty 460V motor are all suitable for 480 volts. If a 230/460V motor is connected for 230 Volts it will often withstand 480V at 120 Hz. I understand that some manufacturers of compressor skids are using this technique to double the horsepower of their motors.
Another technique is to interpose a transformer between the drive and the motor. The transformer will filter out almost all of the PWM hash and you will then be able to deliver relatively clean over frequency/over voltage to the motor and maintain the V/Hz ratio above base speed.
I have seen these techniques but I have not been party to the design. There may be something that I don't know about them.
It is a field for you to research.
Gunnar is one of our drive Gurus. Be guided by his suggestions.
Bill
--------------------
"Why not the best?"
Jimmy Carter
rockman7892
Electrical
DickDV
When you are talking about max torque, or avaliable torque are you referring to th rated torque of the motor at FLA or some other torque above this value such as the breakdown torque?
If you are referring to the max avaliable torque as the breakdown torque then will the rated or continuous torque which is the rated torque at FLA decrease as an inverse of the speed as you mentioned?
If you look at the motor currently running at base speed and it is operating near full load current is it safe to assume that it is operating near full rated torque and you cannot increase speed any further because torque will decrease, current will increase and the drive will trip. So to put it simply if already operating near full load then there is no room to increase speed?
When you are talking about max torque, or avaliable torque are you referring to th rated torque of the motor at FLA or some other torque above this value such as the breakdown torque?
If you are referring to the max avaliable torque as the breakdown torque then will the rated or continuous torque which is the rated torque at FLA decrease as an inverse of the speed as you mentioned?
If you look at the motor currently running at base speed and it is operating near full load current is it safe to assume that it is operating near full rated torque and you cannot increase speed any further because torque will decrease, current will increase and the drive will trip. So to put it simply if already operating near full load then there is no room to increase speed?
rockman7892
Electrical
On the other hand lets say the drive and motor are running at base speed and at base speed the motor current is 80% of full load current.
Since the motor has head room for another 20% current increase can we say that this 20% current increase will corrospond to a 20% torque decrease on the rated load torque and thus we will have enough room to increase the the speed by 20% to still operate at the motor rated torque?
Since the motor has head room for another 20% current increase can we say that this 20% current increase will corrospond to a 20% torque decrease on the rated load torque and thus we will have enough room to increase the the speed by 20% to still operate at the motor rated torque?
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