Why the 1.5hp limitation on 110v dust collectors?

[Dan Ness]

I sometimes see 2hp motors that run on 110V, but searching for cyclones that run on 110v everything is limited to 1.5hp. Is this an actual limitation of the voltage? Or just a limitation of the manufacturer? Or something else?

If I purchased a 2hp cyclone, and put a 1.5hp motor on it, would I get the same CFM, if both motors have the same RPM or is this not possible?

For example:
Grizzly 1.5hp Cyclone - 1000CFM - 12.5" Impeller - 3450RPM
Grizzly 2hp Cyclone - 1350CFM - 14.5" impeller - 3450RPM

So if I put the 1.5hp motor on the 2hp cyclone with the larger impeller will I get the higher CFM or will the motor not be able handle it?

[David Kumm]

Dust collector motors are among the hardest working in the shop and run at or near full load when gates are open. The impeller size and amp draw of the motor are matched so while it is possible to put a smaller motor on a larger impeller, you won't be able to open the gates to max the cfm without burning out the motor. ,That is an oversimplification but you get the idea. Because DC motors often start out under load as gates are usually open, they pull even more amps at start up than other machines which limits how much a 110 v circuit can handle. Dave

[Dan Hintz]

I sometimes see 2hp motors that run on 110V, but searching for cyclones that run on 110v everything is limited to 1.5hp. Is this an actual limitation of the voltage?

HP is a measure of mechanical power... electrical power is voltage x current. Since the typical 120V socket is not rated above 20A (it becomes more inefficient), you see a switchover to 240V systems as the current requirements go up. So, this puts an upper bound on the HP spec seen for 120V systems.

[Mark Burnette]

In addition to what David said DC's typically have large diameter heavy impellers and require lots of startup current (usually exceeding the breaker rating) until they reach operating speed. When I first installed my DC I wired it for 120 and it took about 2 full seconds to get up to speed, during which the lights dimmed. When changed to 240 it spins up almost instantly and the lights don't even flicker.

[Steve Milito]

1.5 HP is equivalent to a bit over 1100 watts at 100% efficiency.
Motors aren't 100% efficient, so a typical 1.5 hp motor will draw 15 or more amps at full load.

[Wade Lippman]

High efficiency 2hp motors that will draw 15a are available, but they are expensive.
Of course there is always the 2hp motor on the HF DC...

Rule of thumb is 1200w per HP for a "standard" motor.
But I have a 5200w 3hp motor on my grizzly cyclone and a 2900w 3hp motor on my SS; go figure.

[John Lanciani]

[QUOTE=But I have a 5200w 3hp motor on my grizzly cyclone and a 2900w 3hp motor on my SS; go figure.[/QUOTE]

Measured watts or just calculated using nameplate volts and amps? If measured, how did you measure? There is a huge difference...

[Wade Lippman]

Measured watts or just calculated using nameplate volts and amps? If measured, how did you measure? There is a huge difference...

The SS is just the nameplate; I had no reason to measure it. But it is unlikely they would claim 13a if it was actually more.
They cyclone is both nameplate and measured with an ampmeter, both 22a; they recommend a 40a circuit so I wanted to make sure a 30a was okay.

Actually now that I have all the ductwork in, I should go back and see how much current it draws when it is on a machine; probably rather less.

[John Lanciani]

The SS is just the nameplate; I had no reason to measure it. But it is unlikely they would claim 13a if it was actually more.
They cyclone is both nameplate and measured with an ampmeter, both 22a; they recommend a 40a circuit so I wanted to make sure a 30a was okay.

Actually now that I have all the ductwork in, I should go back and see how much current it draws when it is on a machine; probably rather less.

So you don't know what the wattage is, just the nameplate volt-amps. All you really know is that as marked, the Grizzly motor is much worse at converting electrical energy into work than the SawStop motor. The formula for wattage for a single phase motor is volts x amps x power factor x efficiency. Power factor is easy to measure with the correct equipment, efficiency number typically comes from the motor mfr. and includes things like stator and rotor design, bearing losses and windage; without these two you're just flapping in the wind trying to guess at wattage.

In all actuality the wattage is probably very close on both motors at full load, the higher amperage of the Grizzly motor x the lower efficiency & power factor will equal out.

[Wade Lippman]

So you don't know what the wattage is, just the nameplate volt-amps. All you really know is that as marked, the Grizzly motor is much worse at converting electrical energy into work than the SawStop motor. The formula for wattage for a single phase motor is volts x amps x power factor x efficiency. Power factor is easy to measure with the correct equipment, efficiency number typically comes from the motor mfr. and includes things like stator and rotor design, bearing losses and windage; without these two you're just flapping in the wind trying to guess at wattage.

In all actuality the wattage is probably very close on both motors at full load, the higher amperage of the Grizzly motor x the lower efficiency & power factor will equal out.

I'm just flapping? What on earth is wrong with you?!
The power factors and efficiencies are probably pretty close; and efficiency has nothing to do with wattage. Try to think before you post.

[Jim Neeley]

I sometimes see 2hp motors that run on 110V, but searching for cyclones that run on 110v everything is limited to 1.5hp. Is this an actual limitation of the voltage? Or just a limitation of the manufacturer? Or something else?

The *practical* reason, besides efficiency, is that large wire would need to be run and a less-than-common plug would need to be used. In other words, anyone buying their product would need to do a custom receptacle install, minimum. It's not unusual to see a 240v 15A or 20A receptacle installed (and they're available at the BBSs) but 120V 30 or 50 amp (no 40) receptacles are another issue.

If you want to see the different plug styles, google NEMA plug chart.

[John Lanciani]

I'm just flapping? What on earth is wrong with you?!
The power factors and efficiencies are probably pretty close; and efficiency has nothing to do with wattage. Try to think before you post.
I have 25 years of professional experience in electricity metering and power quality analysis, I'm pretty comfortable with my reply. Care to share your background?

Edited to add, Wade is correct that motor efficiency has nothing to do with watts input, it is used in the calculation for motor horsepower. Too many formulas, too little time...

[Dan Schrecongost]

The power factors and efficiencies are probably pretty close; and efficiency has nothing to do with wattage...

Actually it does. The lower the efficiency, the less useful output (work) you get from your motor for a given electrical input.

Getting back to the OP's question, What limits the HP motor you can serve with 120 volt power is current.
120 v x 20 amps = 2400 volt-amps x 0.80 power factor = 1920 watts electric input to motor.
1920 w x 83% efficiency = 1594 watts power output from the motor
1594 w / 746 watts per HP = 2.1 HP the largest motor you could power with 120 volt 20 amp circuit. 83% is probably the upper range for motors in this size range. Power factor could be as high as 0.90 but is usually between 0.75 to 0.85 on a fully-loaded motor.

Since motors briefly draw current 3x the nameplate and 15 amp circuits are common, 1.5 HP motors tend to be the largest you see wired for 120 volt single phase power.

[Dan Hintz]

And here I was thinking my simple answer was enough... ;)

[Ole Anderson]

Actually it does. The lower the efficiency, the less useful output (work) you get from your motor for a given electrical input.

Getting back to the OP's question, What limits the HP motor you can serve with 120 volt power is current.
120 v x 20 amps = 2400 volt-amps x 0.80 power factor = 1920 watts electric input to motor.
1920 w x 83% efficiency = 1594 watts power output from the motor
1594 w / 746 watts per HP = 2.1 HP the largest motor you could power with 120 volt 20 amp circuit. 83% is probably the upper range for motors in this size range. Power factor could be as high as 0.90 but is usually between 0.75 to 0.85 on a fully-loaded motor.

Since motors briefly draw current 3x the nameplate and 15 amp circuits are common, 1.5 HP motors tend to be the largest you see wired for 120 volt single phase power.

And you need to add in the issue of running a breaker at 100% load for long periods of time. I like the rule of 80% being OK. 80% x 2.1 hp = 1.68 hp on a 20 amp circuit using Dan's motor numbers.

[Cyrus Brewster 7]

And here I was thinking my simple answer was enough... ;)

Ha! Dan... your first answer pretty much covered it - although some of the banter is interesting.

[Michael W. Clark]

Hi Dan, I think the motor HP vs circuit size has been covered pretty thouroughly.

Regarding the 1.5 HP motor on the 14" impeller fan: Just because that fan has a 2HP motor on it from stock, does not mean that it has to have a 2HP motor. As long as the 1.5HP motor will start it (you may have to close off the gates to do so?) it could run the fan depending on your system. For example, if your system is 6 feet of straight 8" ductwork, you will have very low losses and the fan will draw high amps because it is moving a lot of air. However, if you system is 40 feet of 4" duct, you will have high losses, and the fan will require less break HP to turn at 3450 RPM. Your system is likely somewhere in between and you can find a spot where the 1.5HP motor will handle that fan. You may have to add some resistance to the system to do so. It isn't perfect, you will have to make some concessions. This is done on most industrial systems, just on a larger scale.

Mike