Connecting a motor in triangle with inverters, Torque untill 87Hz

I am sure Combo means delta connection.

I just happen to know that in some languages, what we call "wye" and "delta", is literally called "star" and "triangle". Combo is just not fluent in English techno slang, that's all.


That's indeed what I mean, delta.

I have 2 motors, both:

delta/star
220V/380V xA/xA 2.2kW

What I wanna do is, connect the motors in Delta; and the connect tot the frequency drives. I know that I shood connect in Star normally. Shat am I gonna do:

When u connect in delta, then u will haven 400 Volts on each winding (max); It Wil take more current, and the Torque wil be great untill 87Hz. Yes 87Hz, comes from sqrt 3 x 50Hz
OK, this is actually a valid concept for when you need to run a motor ABOVE the rated frequency and still need rated torque. Here is how this works.

First off, torque output from an AC induction motor is dependent upon maintaining a constant ratio of voltage and frequency, a V/Hz ratio. So a 400V 50Hz motor requires 8.0V/Hz (400/50), a 230V 50Hz motor requires 4.6V/Hz. So as long as you maintain that V/Hz ratio, the motor is capable of producing rated torque. The problem arises when you want to go OVER the base speed, in this case 50Hz, but your voltage is fixed, i.e. 400V. You cannot maintain the proper V/Hz relationship because you run out of V to increase with the Hz. So if for example you wanted to run the motor at 87Hz, 400/87 = 4.6V/Hz which is .575 of the design ratio, so that motor will now only be capable of delivering 57.5% of rated torque at that speed. But here is the workaround:


  1. You have a motor rated for 230V Delta, 400V Star.
  2. You have a 400V supply.
  3. You have a VFD rated for 400V.
  4. You strap (connect) the motor in DELTA (triangle) as if it is to get 230V.
  5. You then program your VFD to tell it that the motor is 230V 50Hz, which will create a 4.6 V/Hz ratio that the motor requires to get full rated torque.
  6. You then allow the maximum speed to be 87Hz instead of 50Hz, at which time the VFD will output 400V and we already know from the above that 400/87 = 4.6, so the motor is now capable of delivering full rated torque at a speed higher than the base speed.
This is done all the time by the way, totally legitimate. The only thing you MUST check is that the motor is physically capable of the higher speed. If it is a 2 pole motor for example, the base speed may be 2900 RPM, so operating it at 87Hz means it will be spinning at 5000 RPM and the bearings may not be rated for that additional speed. ONLY the motor manufacturer can answer that question, no two motors are built the same, so just because one motor is OK at that speed doesn't mean another one is.
 
OK, this is actually a valid concept for when you need to run a motor ABOVE the rated frequency and still need rated torque. Here is how this works.

First off, torque output from an AC induction motor is dependent upon maintaining a constant ratio of voltage and frequency, a V/Hz ratio. So a 400V 50Hz motor requires 8.0V/Hz (400/50), a 230V 50Hz motor requires 4.6V/Hz. So as long as you maintain that V/Hz ratio, the motor is capable of producing rated torque. The problem arises when you want to go OVER the base speed, in this case 50Hz, but your voltage is fixed, i.e. 400V. You cannot maintain the proper V/Hz relationship because you run out of V to increase with the Hz. So if for example you wanted to run the motor at 87Hz, 400/87 = 4.6V/Hz which is .575 of the design ratio, so that motor will now only be capable of delivering 57.5% of rated torque at that speed. But here is the workaround:


  1. You have a motor rated for 230V Delta, 400V Star.
  2. You have a 400V supply.
  3. You have a VFD rated for 400V.
  4. You strap (connect) the motor in DELTA (triangle) as if it is to get 230V.
  5. You then program your VFD to tell it that the motor is 230V 50Hz, which will create a 4.6 V/Hz ratio that the motor requires to get full rated torque.
  6. You then allow the maximum speed to be 87Hz instead of 50Hz, at which time the VFD will output 400V and we already know from the above that 400/87 = 4.6, so the motor is now capable of delivering full rated torque at a speed higher than the base speed.
This is done all the time by the way, totally legitimate. The only thing you MUST check is that the motor is physically capable of the higher speed. If it is a 2 pole motor for example, the base speed may be 2900 RPM, so operating it at 87Hz means it will be spinning at 5000 RPM and the bearings may not be rated for that additional speed. ONLY the motor manufacturer can answer that question, no two motors are built the same, so just because one motor is OK at that speed doesn't mean another one is.

This response was completely worth the zombie thread.
 
Hmm. I thought the calibration table contained in the powerstack data of the VFD decreases the torque above the rated data, regardless of frequency e.g if vfd set to 50Hz, decrease after 50Hz, 60 Hx decreases 10Hz higher. I best tell my employer this is not the case - and scrap my test results as they must be wrong. I didn't know it was really that simple and the results I've been getting show a roll off of torque at 60Hz for a vfd set to 50Hz. So what happens to all the voltage spikes above rated torque? does the V/Hz equation factor these in? or the extra flux as the winding fields start to overlap? zombie thread so... bye.
 

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