electric question follow-up II

[Julio Navarro]

Well, I got my new Leeson 3 something HP motor installed on the cyclone, I bought a tester and determined that the amp draw at full speed all gates but one closed was 17.2 amps continuously and the outlet read 240V, according to the tester
( http://www.sawmillcreek.org/showthread.php?t=41300 )

Julio,
The motor you have is probubly around 3 HP not know in the losses in the motor It's hard to tell exactly But here is how you can rate the comtinuous HP of any motor.

HP = Volts x Continuous amps/ 746 (Convert watts to HP) x effecncy.

Assuming 30% losses due to copper losses, windage, friction and depending on the air gap between the stator and the rotor you would ge the following.

230 x 15/746 x .7 =3.2 HP
if I plug these numbers in 240 x 17.2 /746 x .7 I get 7.9hp??

This cant be right?

[Rob Russell]

Adding some parens, it would look like this:

HP = ((Volts x Continuous amps)/ 746 [Convert watts to HP]) x efficiency

Using your numbers, that's ((240*17.2)/746)*.7 =~ 3.7 HP

[Julio Navarro]

Adding some parens, it would look like this:

HP = ((Volts x Continuous amps)/ 746 [Convert watts to HP]) x efficiency

Using your numbers, that's ((240*17.2)/746)*.7 =~ 3.7 HP

OK, that makes more sense, I just didnt copy Dons equation correctly.

Thanks Rob.

The motor is certainly quieter and smoother running than the HF motor but it still makes a heck of a powerfull suction. I cant imagine what the true 5hp would be like:eek:

[Don Baer]

The motor is certainly quieter and smoother running than the HF motor but it still makes a heck of a powerfull suction. I cant imagine what the true 5hp would be like:eek:

Julio,
The motor will draw whatever current it need to drive the blower so a 5 hp motor would be no differant then what you have now. If you feel you have more then enough suction you can close your blast gate a little and reduce the load on the motor.

db

[Rick Christopherson]

Adding some parens, it would look like this:

HP = ((Volts x Continuous amps)/ 746 [Convert watts to HP]) x efficiency

Using your numbers, that's ((240*17.2)/746)*.7 =~ 3.7 HP You either missed the powerfactor, or tried to combine it with the efficiency. They are not the same.

The proper equation (and numbers) is:
HP = (Volts x Amps x PF x Eff)/ 746

The typical powerfactor is 0.8 and the typical efficiency is also 0.8

Running these numbers, and you get 3.5 hp.