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I have an old lennox pulse furnace on the ground floor of a utility building, it is only used sporadically for air circulation not for heating. I replaced the blower motor about 4 years ago (original had bearing problems). This is a direct-drive configuration.

The motor is a 5-speed (EMERSON K55HXNHD-4656) 115V 10 amp. I have the motor wired using the slowest speed. Problem is, it runs very hot, frequently tripping it's internal heat overload when run for an extended period (4 to 6 hours). The current draw when its running is 12 amps. It draws about 1.5 more amps on medium speed and another 1.5 amps on the highest speed (and yes the speed does increase quite a bit for each speed). The run capacitor is 15 uf (and it is exactly that value when measured with a capacitance meter). I measure about 1 to 2 amps of current through the capacitor on the various speeds.

When I pull the fan assembly out of the furnace and run the fan on a bench (no ductwork involved) the current seems to go up a bit (to 14 amps) on slow speed.

When I pull the motor and run it (no fan, no load at all) the current is about 3.2 amps. The spindle turns freely, not much friction that I can tell.

The motor won't start without the capacitor, so I guess the cap is both a start and run cap? I don't know if the motor will keep turning if I disconnect the cap after it's started.

Bottom line: Why the heck does the current draw go so high when it's driving the fan?

This motor/fan is unusable in this configuration given that it acts more like a heater than a fan. I would like to be able to circulate air and hopefully do it by using more like 4 or 5 amps - is this not possible?

Or is this not a good motor to use in this situation? What sort of efficiency can I expect (what is the best I can do current-draw wise) if I obtain a different motor (with the same frame type) even if just a single-speed 1-phase 120v ?

To what extent could the capacitor be causing the motor to run hot? Is it possible the capacitor can test ok on a meter but in circuit it behaves differently? Or perhaps it might run more efficiently with a different value? What capacitor current should I read anyways? Should the cap current be X% of overall current draw, and if so what is X ?

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    These are great questions. Unfortunately there are too many of them all jumbled into one. While they are all related, it feels like there are far too many questions to be able to answer well and within the format of Home Improvement. Please edit to narrow the focus (your "Bottom Line" question is a good start), then feel free to ask new questions for the others. Refer back to this one so you don't have to repeat all the background (which is excellent BTW!) – FreeMan Jan 28 at 15:19
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You can measure current - Great!

You almost certainly need to (this will sound terribly counterintuitive, but you can use an ammeter and see it!) restrict the ductwork this is connected to. There might be a large master damper, or you might have only individual dampers, or perhaps you have to add dampers or put boards partly over registers.

You have already seen the current go up in "free air" - when you restrict the ductwork, you should see the current drop. When you have restricted it enough it will be drawing 10A on "high" - based on the information you have provided.

Bottom line - when it's driving the fan, it's doing work. When it moves more air, it does MORE work. When you restrict the airflow, it does LESS work and draws LESS current. People "intuitively" tend to think that restriction makes it "work harder" but that is not the case with a centrifugal fan. It's "counter-intuitive."

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  • Since I have the fan compartment door open when measuring current I am decreasing intake resistance but the output resistance is still there (ie all the installed ductwork). Still I can't believe that the fan is designed (with vanes and all) such that it can so easily load the motor like I'm seeing in a typical installation. Why not design the blower vanes (ie fewer of them) so it can't overwhelm the motor, especially on low speed? – Peggy Schafer Jan 28 at 3:30
  • Because what it's installed into varies. As the installer, part of your job is to check the current (preferably in a location where you don't have the door open) and adjust to suit how it performs in the system it's installed into. – Ecnerwal Jan 28 at 3:58
  • So - is it true that given the absurd 12 amp reading on "low speed" that if I want the current CFM being moved in the current configuration, that I have no choice but to accept the 12 amp current draw? Because blocking either intake or output will reduce current but presumably CFM will also go down. – Peggy Schafer Jan 28 at 4:02
  • I didn't install or spec this system. That aside, I wouldn't have thought that optimizing things like duct size, sealing joints, minimizing loss and resistance would actually work against you such that you'd have to, in the end, restrict airflow to prevent your motor from burning out. – Peggy Schafer Jan 28 at 4:06
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    "I replaced the blower motor" - you may not have installed/speced the original system, but your question states that you replaced the blower motor - not that someone else did. If that's not accurate, please edit the question. Adjusting the overall balance to limit the load on the motor is, in fact, normal. Homeowners "opening up all the dampers" is a common cause of burned out motors...I did warn that it's counter-intuitive. – Ecnerwal Jan 28 at 4:09

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