imcjoek
Mechanical
- Sep 7, 2007
- 241
Hello,
I have a problem which keeps coming back to me in my line of work involving how to size an AC vector motor.
We build many winding machines. Usually these are "center wind" in that they drive the shaft and the material wraps around the spindle.
We try to hold a constant tension on the material, and also a constant linear speed of the material as it winds. This leads to the following problem:
As the size of the winding roll gets bigger, the torque the motor must overcome is greater. (Constant tension, with a bigger moment arm).
As the size of the winding roll gets bigger, the speed of the roll also must decrease. (To hold constant linear material speed).
This results in the drive needing to supply a lot of torque at low speeds, however the power required is very low. The speed must also be achieved when the roll is small (and so this limits my pulley reduction based on motor max speed).
I believe this is referred to as a constant-power load?
My question then,
Let us say that I have a job which requires 40ftlb torque at 180RPM and 3ftlb at 1750RPM. These are both approximately 1HP.
Even a 2.0HP motor however will not even stand a chance, since it cannot make 40FT-LB at 180RPM. So clearly I cannot size on power required alone.
If I go by the motor figures I can use a 5HP since its lock rotor is about 50ft-lb. So it can physically "do it".
However I will be greatly exceeding its "full load" torque (15ft-lb) for a long duration of time. Will this be an issue considering a 5HP motor is only being asked to output about 1HP? I have seen some websites that suggest sizing base on highestTorque x highestRPM. This seems ridiculous as I end up needing a 15HP motor! However that DOES put the torque below its "full load".
I realize "full load torque" is at rated speed, but what I dont understand is how you size a motor that does not operate at that rated speed 90% of the time.
The motors we typically use are Marathon "Black Max" vector duty AC, run with Parker/SSD drives.
I would like suggestions. Thank you.
I have a problem which keeps coming back to me in my line of work involving how to size an AC vector motor.
We build many winding machines. Usually these are "center wind" in that they drive the shaft and the material wraps around the spindle.
We try to hold a constant tension on the material, and also a constant linear speed of the material as it winds. This leads to the following problem:
As the size of the winding roll gets bigger, the torque the motor must overcome is greater. (Constant tension, with a bigger moment arm).
As the size of the winding roll gets bigger, the speed of the roll also must decrease. (To hold constant linear material speed).
This results in the drive needing to supply a lot of torque at low speeds, however the power required is very low. The speed must also be achieved when the roll is small (and so this limits my pulley reduction based on motor max speed).
I believe this is referred to as a constant-power load?
My question then,
Let us say that I have a job which requires 40ftlb torque at 180RPM and 3ftlb at 1750RPM. These are both approximately 1HP.
Even a 2.0HP motor however will not even stand a chance, since it cannot make 40FT-LB at 180RPM. So clearly I cannot size on power required alone.
If I go by the motor figures I can use a 5HP since its lock rotor is about 50ft-lb. So it can physically "do it".
However I will be greatly exceeding its "full load" torque (15ft-lb) for a long duration of time. Will this be an issue considering a 5HP motor is only being asked to output about 1HP? I have seen some websites that suggest sizing base on highestTorque x highestRPM. This seems ridiculous as I end up needing a 15HP motor! However that DOES put the torque below its "full load".
I realize "full load torque" is at rated speed, but what I dont understand is how you size a motor that does not operate at that rated speed 90% of the time.
The motors we typically use are Marathon "Black Max" vector duty AC, run with Parker/SSD drives.
I would like suggestions. Thank you.