Specifying Motors - RPM??

Timeismoney08

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I was wondering how people go about picking the right motor and gearing combination.

If I need a 1/2hp motor and need to change the gearing to slow it down or speed it up, doesn't it change the torque value? How do you figure out what you guys need for your application?

Hope you can help! Just trying to learn.
 
I was wondering how people go about picking the right motor and gearing combination.

If I need a 1/2hp motor and need to change the gearing to slow it down or speed it up, doesn't it change the torque value? How do you figure out what you guys need for your application?

Hope you can help! Just trying to learn.

If you know what torque your load needs, you (or more likely a mechanical guy) can select a gearbox to match up to a standard motor speed of 3600, 1800, 1200, 900 rpm. Or 3000, 1500, 750 rpm for 50 Hz motors.

When we are installing equipment is it normally a tug-of-war between using a motor and gearbox combination that we already have spare parts for in our warehouse, but that is too large or for some reason has trouble fitting into the space that is available ... and purchasing the correct size, but then having to buy a spare gearbox, or gearbox and motor, for the warehouse ... finding somewhere in our crowded warehouse to store the new spare, assigning inventory numbers that make sense, getting it added to the corporate inventory ... etc etc.

Since we hate paperwork ... we have extreme cases where the motor and gearbox produce far too much torque for the load. But we add an extra coupler that is designed to break if we put too much torque on it ... sort of like a 'mechanical fuse' so we don't destroy our load. It's twisted logic, but it works.
 
In general, you don't pick the HP of the motor first, you determine the torque you need at the work shaft, then the speed you want to run the work piece at. From there you can figure out a combination of motor speeds based on available standards and then select a gear / belt / chain reduction ratio that allows you to use one of those standard speeds to get to where you need to be.


But yes, as you change speeds with a gear / belt / chain reduction, torque changes by the inverse of the speed change; i.e. you trade speed for torque or vice-versa.
 
First LAW of Thermodynamics: Energy is not created or destroyed.

Power is energy per unit time.

Therefore the horsepower throughout your system is always the same, minus losses to friction, of course. Those losses turn to heat.

T = hp x 5252 ÷ rpm, with T in ft∙lb

So as the speed goes down T goes up in the same ratio. Always.
 
First LAW of Thermodynamics: Energy is not created or destroyed.

Power is energy per unit time.

Therefore the horsepower throughout your system is always the same, minus losses to friction, of course. Those losses turn to heat.

T = hp x 5252 ÷ rpm, with T in ft∙lb

So as the speed goes down T goes up in the same ratio. Always.
... when using a mechanical method of changing speed. A VFD is different.
 
... when using a mechanical method of changing speed. A VFD is different.

No, it is not.

The only place that violates the First Law is a nuclear reaction, and that is governed by e=mc².

The equation for torque vs. hp and speed still applies. The power into the VFD and the power into the motor and the power into the load are all equal except for losses from resistance or friction. The total power is the same throughout the system. Always.
 
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Several years ago when I was working for a distributor a customer bought a VFD from me. The motor nameplate was 1 HP, 1750 RPM so that's the size VFD he bought. Upon installation he called to ream me out claiming I had lied to him about getting full torque over the entire speed range he needed to operate at. So, I paid a visit to the site and my first thought was that this was in no way a job for a 1 HP motor without a gearbox. I asked if he had removed a gearbox when he changed over to the VFD. He showed me what he had taken out. It was indeed a 1 HP motor, but it was an old Reeves Moto-Drive unit with a 5:1 fixed reduction after the variable pitch pulleys.
I understand the point jraef was trying to make in response to Tom's first post which is that with a VFD the available torque doesn't change with speed. But you can't ignore the law of conservation of energy. The OP's question reminds me of the way my old customer approached his task.
 
A quick question for you guys how do you calculate the torque on a motor at zero speed ?

Can you have a motor produce 150% torque with the motor stopped ?
 
A quick question for you guys how do you calculate the torque on a motor at zero speed ?

Can you have a motor produce 150% torque with the motor stopped ?


Yes, it can... but cooling may be an issue.
If you absolutely know you need 150% torque at zero speed, make sure your motor can handle the cooling. This may require a blower, or at the very least, a fan-over-fins end bell.
 
No, it is not.

The only place that violates the First Law is a nuclear reaction, and that is governed by e=mc².

The equation for torque vs. hp and speed still applies. The power into the VFD and the power into the motor and the power into the load are all equal except for losses from resistance or friction. The total power is the same throughout the system. Always.
Sorry, I didn’t mean the law of thermodynamics didn’t apply, only the last part of your statement where as speed decreases torque increases. That doesn’t happen in a VFD.
 
Sorry, I didn’t mean the law of thermodynamics didn’t apply, only the last part of your statement where as speed decreases torque increases. That doesn’t happen in a VFD.

Mutual misinterpretation, I guess - no problem.

I always start with the load by figuring torque and horsepower. Then I work backward to the motor demand. Then, whether there is a mechanical speed reduction involved or not you know the speed and torque demand at the motor and can stay out of trouble.
 

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