Turning cyclone on and off

[wwhitney]

Hello,

As I understand it from reading other forum threads, the start windings of the Leeson motor will burn out if the motor is power cycled too frequently. This is because the thermal protector only protects the run windings, and because the startup current is so large, which means there is a lot of heat generated in the motor on startup.

So how long does the system have to be idle for it to be worth shutting off? I mean, tens of minutes, or hours?

Also, is the correct thing to do when the cyclone is idle but running to close all the blast gates and starve it of air? It seems like this would minimize noise, and based on what I've read, minimizes the current draw by the motor. Minimizing current draw saves energy, but perhaps it increases the heat generated in the motor? [E.g. 80% efficient at 20 amps draw = 4 amps * 240V of heating; 0% efficient at 6 amps draw = 6 amps * 240V of heating; but I don't know if these calculations are accurate.]

Thanks, Wayne

 

[Matt]

This is because the thermal protector only protects the run windings, and because the startup current is so large, which means there is a lot of heat generated in the motor on startup.

Yes that is correct.

So how long does the system have to be idle for it to be worth shutting off? I mean, tens of minutes, or hours?

I would say if you aren't going to do anything for 15 mins then shut it off...if you plan to connect to automatic blast gates or something of that nature then i would set a delay timer for 10 mins (6 times an hr.)

Also, is the correct thing to do when the cyclone is idle but running to close all the blast gates and starve it of air? It seems like this would minimize noise, and based on what I've read, minimizes the current draw by the motor. Minimizing current draw saves energy, but perhaps it increases the heat generated in the motor? [E.g. 80% efficient at 20 amps draw = 4 amps * 240V of heating; 0% efficient at 6 amps draw = 6 amps * 240V of heating; but I don't know if these calculations are accurate.]

That would be right at 0 percent effecienty. I Don't know if that is right though as it takes some energy to spin the motor itself so it can't be 0. That not to say that it's 82 percent effecient either. At any rate the motor will idle for a long time before getting warm and also there is a cooling fan on the motor itself so it is correct to close all the gates when not in use to conserve energy and also to help cool the motor unless it turns out that the motor is 0 percent effecient at no load.

You have broght up an interesting question that we will look into and add more informatin when we get it.

Hope this helps,
Matt

After Cruising the net I have found that efficiency ratings are wild (above me)....but it appears that it is rated from the no-load state after a warm-up period. I will contact leeson to see which method (or standard) of testing they employ and get back to this thread in few days.

 

[wwhitney]

That would be right at 0 percent effecienty. I Don't know if that is right though as it takes some energy to spin the motor itself so it can't be 0. That not to say that it's 82 percent effecient either.

I assume that when the stated efficiency at full load is 82%, then that means that 18% of the electrical energy is being converted into heat in the motor.

Now when the motor is spinning an impeller that is just churning air in a closed container, then any electrical energy that is converted into motion of the impeller is quickly converted into heat in the air in the container. So in that sense I would say the efficiency is 0.

However, I imagine the heat dissipated in the air in the closed container is not so likely to damage the motor, compared to the heat generated inside the motor windings itself. So even if the total heating at no load is greater than the total heating at full load, the motor heating seems like it would be less.

Anyway, I'm just feeling my way in the dark here, so it would be interesting to hear what Leeson says.

Cheers, Wayne