I am trying to use a DuraPulse GS21 VFD to run a 2HP motor for the purpose of screwing in a fitting and stopping or stalling a 100 Ft-LBS. The mtr nameplate doesn't show max torque and I have calculated this thinking I could get it but when i run the motor at low RPM around 2 RPM i can stop the shaft but the torque is like 14 ft-lbs.
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Can I get 100 FT-LBS of torque from a 2HP Ironhorse 3
- Thread starter LSnyder
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Do a little research on the law of the lever and then apply that to circular levers, often called gears or pulleys.
Apart from that you have answered your own question.
With the most common type of motor, locked rotor torque is about 150% of full load torque. This increases to about 2005 of full load torque at around 75% of rated speed, or 25% slip. The 25% slip is the important factor.
If you set your speed/frequency at 25% of full load speed you will have 25% slip and about 200% of full load torque.
In practice the VFD protection circuits may not allow this to be realized.
You may be able to adapt one of these and have all the torque you may ever need, now and into the future.
$84.99 at Amazon
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Ohm's law
Not just a good idea;
It's the LAW!
Apart from that you have answered your own question.
With the most common type of motor, locked rotor torque is about 150% of full load torque. This increases to about 2005 of full load torque at around 75% of rated speed, or 25% slip. The 25% slip is the important factor.
If you set your speed/frequency at 25% of full load speed you will have 25% slip and about 200% of full load torque.
In practice the VFD protection circuits may not allow this to be realized.
You may be able to adapt one of these and have all the torque you may ever need, now and into the future.
$84.99 at Amazon
Link
--------------------
Ohm's law
Not just a good idea;
It's the LAW!
-
2
- #3
If you know the maximum HP (and you do), then Torque = HP x 5250/RPM
So assuming a common 4 pole motor spinning at 1750RPM, the Full Load Torque of that motor is likely
2 x 5250 / 1750 = 6 lb. ft. of torque.
AC induction motors generally have the ability to provide a PEAK torque of 200-220% of the FLT for a few seconds, so that is likely why you can see 14 lb.-ft. of torque. The motor is however basically at the edge of stalling there, so operating at that level for very long can damage it.
When you use a VFD to change the speed, the BEST you can hope for is that the VFD allows the motor to continue providing its rated torque at the lower speeds, so you can STILL get 6 lb.-ft. at 2RPM and 14 for a few seconds. You will never get to 100 lb.-ft. of torque with this motor unless you use "mechanical advantage", as in a gear box.
Gears will "trade" speed for torque. So if you need 100 lb.-ft. and your motor supplies 14 at its peak, then you need at LEAST 100/14 = 7.4 as the minimum gear reduction ratio. I would go for at least 10:1 because there are some losses in the gears, but if you need that 100 lb.-ft. for more than a few seconds, I suggest using a 17:1 ratio. That then means if the motor is at full speed, the final shaft RPM will become 1750/17 = 102 RPM, so then that's where your VFD will help to get that down to the slower speed without losing that torque.
" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
So assuming a common 4 pole motor spinning at 1750RPM, the Full Load Torque of that motor is likely
2 x 5250 / 1750 = 6 lb. ft. of torque.
AC induction motors generally have the ability to provide a PEAK torque of 200-220% of the FLT for a few seconds, so that is likely why you can see 14 lb.-ft. of torque. The motor is however basically at the edge of stalling there, so operating at that level for very long can damage it.
When you use a VFD to change the speed, the BEST you can hope for is that the VFD allows the motor to continue providing its rated torque at the lower speeds, so you can STILL get 6 lb.-ft. at 2RPM and 14 for a few seconds. You will never get to 100 lb.-ft. of torque with this motor unless you use "mechanical advantage", as in a gear box.
Gears will "trade" speed for torque. So if you need 100 lb.-ft. and your motor supplies 14 at its peak, then you need at LEAST 100/14 = 7.4 as the minimum gear reduction ratio. I would go for at least 10:1 because there are some losses in the gears, but if you need that 100 lb.-ft. for more than a few seconds, I suggest using a 17:1 ratio. That then means if the motor is at full speed, the final shaft RPM will become 1750/17 = 102 RPM, so then that's where your VFD will help to get that down to the slower speed without losing that torque.
" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
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