Field Weakening 3

[d23]

All,

Is there a simple way to calculate motor size required when operating in an extended speed range for a squirrel cage type motor?

As an example assume the following:

I have a 380-volt, 50-hertz input supply to a ASD. For simplicity sake assume I can get 380 volts out of the drive.

At 50 hertz, 2975 RPM I have a 100 HP (177 lb.ft.) centrifugal load.

If I want to operate in an extended speed range at 63 hertz, 3748 RPM the load would be 200 HP (280 lb.ft).

If I could provide 479 volts at 63 hertz I would require a 159 HP motor.

(50 Hz / 63 Hz) * 200 HP = 159 HP

Because of field weakening I know I will require a larger motor than the 159 HP calculated.

How do I calculate the actual motor HP required?

Thanks
D23

 

[waross]

Your references to field weakening are confusing. Is this a conventional induction motor on a Variable Frequency Drive?
What HP do you need and what speed do you need it at?
What is the rated voltage of the motor and what is the maximum voltage output available from the VFD?
respectfully

 

[d23]

waross

It would be a conventional, inverter duty, 2-pole induction motor on a variable speed drive. I don’t have a specific application, but the question does come up from time to time in our company. I was looking for a formula to calculate HP required.

I was hoping the example information would offer enough data to use as an example.

D23

 

[waross]

I may be corrected by the drive experts but my understanding is:
For speed below rated speed the horsepower will be proportional to the speed. 50% speed = 50% HP. Same maximum current, same torque, 50% voltage and 50% speed.
If you need 50 HP at 50% speed divide 50HP by 50% for a 100 HP motor.

For speeds over rated speed with rated voltage the HP will stay at rated HP. (Rated voltage x rated maximum current)

However if there is extra voltage available to maintain the HZ/V ratio the horsepower may increase above rated HP.

For example if a 380 volt star motor is reconnected in delta, The new rated motor voltage will be 220 volts If the VFD is still fed from a 380 volt circuit the motor speed may be increased be a factor of 1.73 and the HP at 1.73PU speed will be 1.73 x rated HP.

In North America a motor may be connected for 230 volts and fed 460 volts at 120Hz. This will double the speed which will double the HP.

The key is the Volts per Hertz ratio of the motor. This is rated volts divided by rated frequency. 3.83 volts per hertz or 7.67 volts per hertz are common values in North America.
The motor wants to see this ratio within about 10%
If a motor is wired for 230 volts and the VFD fed from 460 volts, the motor will want to see 3.83 volts per hertz. With 460 volts available the VFD will be able to maintain the V/Hz ratio up to 480 volts. The 100 HP, 1800RPM motor is now running at 3600RPM and producing 200HP. 460 volts/120 HZ gives us the 3.83 V/Hz the motor wants.
respectfully

 

[Skogsgurra]

The V/Hz ratio at 63 Hz will be 380/63 = 6.03 V/Hz instead of 7.60 V/Hz.

The motor's torque at nominal slip is proportional to the V/Hz squared. So, if you want to run your motor at nominal slip*, you need a motor that can deliver (63/50)^2 times more torque than you need at 50 Hz. It will then deliver the needed torque at 63 Hz.

The load is "centrifugal", you say. That usually means that your torque goes up as RPM squared. So, your load torque goes up with RPM^2 and your available torque goes down with RPM^2. The effect is that your motor needs to have a 50 Hz rating that is (63/50)^4 times 50 Hz load. Or 100 HP times (63/50)^4 = 250 HP.

If, on the other hand, your torque does not increase with RPM squared, your 50 Hz rating will "only" need to be 60 percent higher, or 160 HP. But, make sure which load curve is true first!

There are some subtle differences between slip measured in percents or actual delta-RPM. There is also the fact that cooling improves when you are in field weakening. But, I would not try to profit from those secondary effects.


*You want to run the motor at nominal slip because slip is what determines rotor heating. And it is not a good idea to have more heat in the rotor than the machine was designed for. The heat will affect the bearings in the first place, and there is normally a quite small margin between NRL heat and the temperature where grease starts to coke.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[d23]

waross

“In North America a motor may be connected for 230 volts and fed 460 volts at 120Hz. This will double the speed which will double the HP.”

This is an interesting thought. I’ll keep this in mind. I may want to try this sometime however; with 2 pole motors I will have limits.

Skogsgurra

“You want to run the motor at nominal slip because slip is what determines rotor heating. And it is not a good idea to have more heat in the rotor than the machine was designed for. The heat will affect the bearings in the first place, and there is normally a quite small margin between NRL heat and the temperature where grease starts to coke.”

Slip is a valid point! Thanks

For field weakening or extended speed range we typically set the drive up where the maximum output voltage equals the input voltage (minus 10 to 20 volts) at the input frequency or maintain the constant V-Hz ratio you both describe. For the extended speed range we increase the frequency and maintain constant volts. I was looking in my Standard Handbook for Electrical Engineers and found a note about low voltage that resolved my question. “The maximum running torque of an induction motor will vary as the square of the voltage.” This statement turned on a light. I can calculate the motor voltage required base on constant V-Hz ratio. I can then calculate the percent voltage available, square that and multiply times the full power torque.

Thanks for your help!

D23

 

[DickDV]

My recommendation is to ignore hp and size the drive and motor entirely based on speed and torque for the motor and frequency and current for the drive.

The foundation for sizing any drive system is to know the load torque across the speed range. Having that, you choose a motor which has equal or greater torque capacity across the speed range remembering that induction motors are constant torque below base speed, constant hp from base speed up to some point (typically 90Hz but it varies by motor), and falling hp and rapidly falling torque above that point.

Once you have the motor, you translate motor torque into current and select your drive based on ampacity.

So, how about giving us load torque at several points in the speed range and we can evaluate the motor sizing for you.