10 amps vs 15 amps

[Dave Lehnert]

This is a question based on "Just curios " than need.

Looking at Grizzly bandsaws. The model G0513 and the model G0513X2. Each has the exact same motor, 2hp. But the G0513 recommends a 10 amp Circuit and the G0513x2 a 15 amp circuit.

Now the G0513 has aluminum wheels. The G0513x2 has cast iron wheels.
Would the cast iron wheels need more amps to get going?

Link to the bandsaw page.

http://www.grizzly.com/images/pdf/Grizzly_BandsawComparisonChart.pdf

[glenn bradley]

I notice in the individual manuals they both call for a 15 amp breaker at 220v.

[Leo Graywacz]

A 2HP motor is only going to draw 8 amps @ 220vac. I have never seen a 10 amp circuit in a house or shop. Just 10 amp cords. Put both on a 15 amp circuit, you'll have no problems.

[Rick Moyer]

If you're still really curious, call Grizzly.

[Bruce Wrenn]

It's a misprint. 2008 catalog, pages 96 -97 shows exact same current draw (20/10) for both machines.

[Chip Lindley]

A 2HP motor is only going to draw 8 amps @ 220vac. I have never seen a 10 amp circuit in a house or shop. Just 10 amp cords. Put both on a 15 amp circuit, you'll have no problems.

Only 8A/230V? These 2hp motors on machines in my shop....

Rockwell RC33 planer 12.0 FLA.
Rockwell HD shaper (Doerr motor) 12.0 FLA.
ShopFox moulder 12.0 FLA
Delta 1-Stage DC 10.2 FLA

Perhaps you are thinking of a 1.5hp motor
Delta DJ20 jointer 8.4 FLA
Rockwell 12" RAS 7.0 FLA

A *dedicated* 15A circuit with 12ga wiring is sound. But that circuit is Full! Nothing else can be on that circuit but one 2hp motor at a any given time!

Always look at the FLA draw of a motor instead of seller claims of *2 HP, etc* Deceptive advertising may try to sell a 1.5hp motor as a *2 HP* motor! More a problem with some Chinese imports, than quality US-made commercial motors!

[Rick Christopherson]

Only 8A/230V? These 2hp motors on machines in my shop....

Rockwell RC33 planer 12.0 FLA.
Rockwell HD shaper (Doerr motor) 12.0 FLA.
pFox moulder 12.0 FLA
Delta 1-Stage DC 10.2 FLA
8 amps is a little bit low for a 2 hp motor unless it has a high efficiency of 0.9 and operating with a high powerfactor of 0.9. You are correct. For a typical motor, 8-amps would equate to about 1.5 hp. 10 amps is a better approximation for an average motor. Your 12 amp motors are a little bit high, but I suspect that they are rated for an input voltage anywhere between 208 and 240, where the listed current is the worst-case situation with a 208 volt supply.

[Leo Graywacz]

8 amps is a little bit low for a 2 hp motor unless it has a high efficiency of 0.9 and operating with a high powerfactor of 0.9. You are correct. For a typical motor, 8-amps would equate to about 1.5 hp. 10 amps is a better approximation for an average motor. Your 12 amp motors are a little bit high, but I suspect that they are rated for an input voltage anywhere between 208 and 240, where the listed current is the worst-case situation with a 208 volt supply.

FLA is full locked amperage. I did calculate it on 100% efficiency 746w/HP

[glenn bradley]

A *dedicated* 15A circuit with 12ga wiring is sound. But that circuit is Full! Nothing else can be on that circuit but one 2hp motor at a any given time!

As usual Chip brings up a good point. I assumed you would run a dedicated circuit but, you know when we assume . . .

My G0513X is on a 20amp outlet with a 20 amp breaker, not exactly what was called for but at the time it went in I would plug my 3HP jointer in there now and again. Remember, your breaker is there to protect the circuit, not your tool.

[Rick Christopherson]

FLA is full locked amperage. I did calculate it on 100% efficiency 746w/HPNot that it is important, but FLA is Full Load Amperage, as Chip had listed. LRA is the Locked Rotor Amperage.

746 watts per horsepower is correct and does not change, ever. It is a direct conversion factor similar to 25.4 mm per inch. The powerfactor is one term in the equation and it (in a sense) converts volt-amperes (total power) to watts (real power). The other term is efficiency and it is related to any losses in the process of converting electrical power into mechanical power, and includes aspects such as friction, windage (air resistance) and electrical losses such as hysterysis in the iron core.

Many people mistakenly refer to the powerfactor as "efficiency", but this is incorrect. The powerfactor relates to how far out of phase the current and voltage are, which is a function of any capacitive or inductive device. The powerfactor changes depending on the load of the motor. An ideling motor will have a power factor as low as 0.1 to 0.2.

For most small motors, the powerfactor (pf) can be approximated at 0.8 at full rated load unless the motor nameplate states otherwise. The efficiency can be anywhere between 0.8 and 0.95, with 0.85 being pretty common for small motors.

hp = I x V x pf x eff / 746

[Chris Padilla]

Not that it is important, but FLA is Full Load Amperage, as Chip had listed. LRA is the Locked Rotor Amperage.

746 watts per horsepower is correct and does not change, ever. It is a direct conversion factor similar to 25.4 mm per inch. The powerfactor is one term in the equation and it (in a sense) converts volt-amperes (total power) to watts (real power). The other term is efficiency and it is related to any losses in the process of converting electrical power into mechanical power, and includes aspects such as friction, windage (air resistance) and electrical losses such as hysterysis in the iron core.

Many people mistakenly refer to the powerfactor as "efficiency", but this is incorrect. The powerfactor relates to how far out of phase the current and voltage are, which is a function of any capacitive or inductive device. The powerfactor changes depending on the load of the motor. An ideling motor will have a power factor as low as 0.1 to 0.2.

For most small motors, the powerfactor (pf) can be approximated at 0.8 at full rated load unless the motor nameplate states otherwise. The efficiency can be anywhere between 0.8 and 0.95, with 0.85 being pretty common for small motors.

hp = I x V x pf x eff / 746

Ah, I always forget about pf and eff so that helps me understand the rule of thumb of about 5 A per hp:

I/hp = 746/V*pf*eff

Now set 746/V*pf*eff = 5 and assuming a V of 240 V, the pf*eff terms would need to equal 0.62, which is about 0.8 * 0.8.

Cool. :)

[Chip Lindley]

Yes, FLA is max amps before a motor fries! I just choose to err on the Larger/Heavier side, unless there is a huge $$ difference. Why undersize a circuit to bare minimum?

Breaker and wire for a 20A circuit is only a couple of bucks more than that for a 15A circuit. Looking ahead, even a 3Hp, 16 FLA machine could be run on a 20A circuit! Install a 30A/10ga. circuit and you are good for 5Hp, 21 FLA down the road....